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<p>Let a triangle be bounded by the lines L<sub>1</sub> : 2x + 5y = 10; L<sub>2</sub> : $$-$$4x + 3y = 12 and the line L<sub>3</sub>, which passes through the point P(2, 3), intersects L<sub>2</sub> at A and L<sub>1</sub> at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :</p>
Options:
[{"identifier": "A", "content": "$${{110} \\over {13}}$$"}, {"identifier": "B", "content": "$${{132} \\over {13}}$$"}, {"identifier": "C", "content": "$${{142} \\over {13}}$$"}, {"identifier": "D", "content": "$${{151} \\over {13}}$$"}] | ["B"]
Explanation:
$L_{1}: 2 x+5 y=10$<br><br>
$L_{2}: -4 x+ 3 y=12$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l99tn43d/ebd74024-c280-415b-bc94-0859668a0f7c/89bb0690-4c7a-11ed-b94d-45a8040c2a81/file-1l99tn43e.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l99tn43d/ebd74024-c280-415b-bc94-0859668a0f7c/89bb0690-4c7a-11ed-b94d-45a8040c2a81/file-1l99tn43e.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 52 English Explanation"><br>
Solving $L_{1}$ and $L_{2}$ we get
<br><br>
$$
C \equiv\left(\frac{-15}{13}, \frac{32}{13}\right)
$$
<br><br>
Now, Let $A\left(x_{1}, \frac{1}{3}\left(12+4 x_{1}\right)\right)$ and
<br><br>
$$
\begin{aligned}
&B\left(x_{2}, \frac{1}{5}\left(10-2 x_{2}\right)\right) \\\\
&\therefore \quad \frac{3 x_{1}+x_{2}}{4}=2 \\\\
&\text { and } \frac{\left(12+4 x_{1}\right)+\frac{10-2 x_{2}}{5}}{4}=3
\end{aligned}
$$
<br><br>
So, $3 x_{1}+x_{2}=8$ and $10 x_{1}-x_{2}=-5$
<br><br>
So, $\left(x_{1}, x_{2}\right)=\left(\frac{3}{13}, \frac{95}{13}\right)$
<br><br>
$$
\begin{aligned}
&A=\left(\frac{3}{13}, \frac{56}{13}\right) \text { and } B=\left(\frac{95}{13}, \frac{-12}{13}\right) \\\\
&=\left|\frac{1}{2}\left(\frac{3}{13}\left(\frac{-44}{13}\right) \frac{-56}{13}\left(\frac{110}{13}\right)+1\left(\frac{2860}{169}\right)\right)\right| \\\\
&=\frac{132}{13} \text { sq. units }
\end{aligned}
$$ |
<p>Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of $$\Delta$$PQR is :</p>
Options:
[{"identifier": "A", "content": "$${{25} \\over {4\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${{25\\sqrt 3 } \\over 2}$$"}, {"identifier": "C", "content": "$${{25} \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${{25} \\over {2\\sqrt 3 }}$$"}] | ["D"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l59j3d6d/8a4b3248-26d5-486b-9475-fbc8a7d508b1/6bc27690-fd20-11ec-8035-4341fba75e09/file-1l59j3d6h.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l59j3d6d/8a4b3248-26d5-486b-9475-fbc8a7d508b1/6bc27690-fd20-11ec-8035-4341fba75e09/file-1l59j3d6h.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th June Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 49 English Explanation"></p>
<p>Let, side of triangle = a.</p>
<p>$$h = {{|3 + 7 - 5|} \over {\sqrt {{1^2} + {1^2}} }}$$</p>
<p>$$ = {5 \over {\sqrt 2 }}$$</p>
<p>From figure, $$h = a\sin 60^\circ $$</p>
<p>$$ \Rightarrow a = {{2h} \over {\sqrt 3 }}$$</p>
<p>$$ = {2 \over {\sqrt 3 }} \times {5 \over {\sqrt 2 }}$$</p>
<p>$$ = {{10} \over {\sqrt 6 }}$$</p>
<p>$$\therefore$$ Area $$ = {3 \over 4}{\left( {{{10} \over {\sqrt 6 }}} \right)^2}$$</p>
<p>$$ = {{25} \over {2\sqrt 3 }}$$</p> |
<p>Let the area of the triangle with vertices A(1, $$\alpha$$), B($$\alpha$$, 0) and C(0, $$\alpha$$) be 4 sq. units. If the points ($$\alpha$$, $$-$$$$\alpha$$), ($$-$$$$\alpha$$, $$\alpha$$) and ($$\alpha$$<sup>2</sup>, $$\beta$$) are collinear, then $$\beta$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "64"}, {"identifier": "B", "content": "$$-$$8"}, {"identifier": "C", "content": "$$-$$64"}, {"identifier": "D", "content": "512"}] | ["C"]
Explanation:
<p>$$\because$$ A(1, $$\alpha$$), B($$\alpha$$, 0) and C(0, $$\alpha$$) are the vertices of $$\Delta$$ABC and area of $$\Delta$$ABC = 4</p>
<p>$$\therefore$$ $$\left| {{1 \over 2}\left| {\matrix{
1 & \alpha & 1 \cr
\alpha & 0 & 1 \cr
0 & \alpha & 1 \cr
} } \right|} \right| = 4$$</p>
<p>$$ \Rightarrow \left| {1(1 - \alpha ) - \alpha (\alpha ) + {\alpha ^2}} \right| = 8$$</p>
<p>$$ \Rightarrow \alpha = \, \pm \,8$$</p>
<p>Now, $$(\alpha ,\, - \alpha ),\,( - \alpha ,\alpha )$$ and $$({\alpha ^2},\beta )$$ are collinear</p>
<p>$$\therefore$$ $$\left| {\matrix{
8 & { - 8} & 1 \cr
{ - 8} & 8 & 1 \cr
{64} & \beta & 1 \cr
} } \right| = 0 = \left| {\matrix{
{ - 8} & 8 & 1 \cr
8 & { - 8} & 1 \cr
{64} & \beta & 1 \cr
} } \right|$$</p>
<p>$$ \Rightarrow 8(8 - \beta ) + 8( - 8 - 64) + 1( - 8\beta - 8 \times 64) = 0$$</p>
<p>$$ \Rightarrow 8 - \beta - 72 - \beta - 64 = 0$$</p>
<p>$$ \Rightarrow \beta = - 64$$</p> |
<p>Let $$A\left( {{3 \over {\sqrt a }},\sqrt a } \right),\,a > 0$$, be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If $$D(3\cos \theta ,a\sin \theta )$$ is a point in the fourth quadrant such that the maximum area of $$\Delta$$ACD is 12 square units, then a is equal to ____________.</p>
Options:
[] | 8
Explanation:
Clearly $B$ is $\left(-\frac{3}{\sqrt{a}},+\sqrt{a}\right)$ and $C$ is $\left(-\frac{3}{\sqrt{a}},-\sqrt{a}\right)$
<br/><br/>
$$
\begin{aligned}
&\text { Area of } \triangle A C D=\frac{1}{2}\left|\begin{array}{ccc}
\frac{3}{\sqrt{a}} & \sqrt{a} & 1 \\\\
-\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\\\
3 \cos \theta & a \sin \theta & 1
\end{array}\right| \\\\
&\Rightarrow \quad \Delta=\left|\begin{array}{ccc}
0 & 0 & 1 \\\\
-\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\\\
3 \cos \theta & a \sin \theta & 1
\end{array}\right| \\\\
&\Rightarrow \quad \Delta=|3 \sqrt{a} \sin \theta+3 \sqrt{a} \cos \theta|=3 \sqrt{a}|\sin \theta+\cos \theta| \\\\
&\Rightarrow \quad \Delta_{\max }=3 \sqrt{a} \cdot \sqrt{2}=12 \Rightarrow a=(2 \sqrt{2})^{2}=8
\end{aligned}
$$ |
<p>Let $$\alpha$$<sub>1</sub>, $$\alpha$$<sub>2</sub> ($$\alpha$$<sub>1</sub> < $$\alpha$$<sub>2</sub>) be the values of $$\alpha$$ fo the points ($$\alpha$$, $$-$$3), (2, 0) and (1, $$\alpha$$) to be collinear. Then the equation of the line, passing through ($$\alpha$$<sub>1</sub>, $$\alpha$$<sub>2</sub>) and making an angle of $${\pi \over 3}$$ with the positive direction of the x-axis, is :</p>
Options:
[{"identifier": "A", "content": "$$x - \\sqrt 3 y - 3\\sqrt 3 + 1 = 0$$"}, {"identifier": "B", "content": "$$\\sqrt 3 x - y + \\sqrt 3 + 3 = 0$$"}, {"identifier": "C", "content": "$$x - \\sqrt 3 y + 3\\sqrt 3 + 1 = 0$$"}, {"identifier": "D", "content": "$$\\sqrt 3 x - y + \\sqrt 3 - 3 = 0$$"}] | ["B"]
Explanation:
<p>Points A($$\alpha$$, $$-$$3), B(2, 0) and C(1, $$\alpha$$) are collinear.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5yktc03/7dba7f27-1a1c-4810-bc97-87e79ff18cdf/bf409530-0ae6-11ed-a51c-73986e88f75f/file-1l5yktc04.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5yktc03/7dba7f27-1a1c-4810-bc97-87e79ff18cdf/bf409530-0ae6-11ed-a51c-73986e88f75f/file-1l5yktc04.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 46 English Explanation"></p>
<p>$$\therefore$$ Slope of AB = Slope of BC</p>
<p>$$ \Rightarrow {{0 + 3} \over {2 - \alpha }} = {{\alpha - 0} \over {1 - 2}}$$</p>
<p>$$ \Rightarrow - 3 = \alpha (2 - \alpha )$$</p>
<p>$$ \Rightarrow - 3 = 2\alpha - {\alpha ^2}$$</p>
<p>$$ \Rightarrow {\alpha ^2} - 2\alpha - 3 = 0$$</p>
<p>$$ \Rightarrow {\alpha ^2} - 3\alpha + \alpha - 3 = 0$$</p>
<p>$$ \Rightarrow \alpha (\alpha - 3) + 1(\alpha - 3) = 0$$</p>
<p>$$ \Rightarrow (\alpha + 1)(\alpha - 3) = 0$$</p>
<p>$$ \Rightarrow \alpha = - 1,\,3$$</p>
<p>Given, $${\alpha _1} < {\alpha _2}$$</p>
<p>$$\therefore$$ $${\alpha _1} = -1$$ and $${\alpha _2} = 3$$</p>
<p>$$\therefore$$ $$\left( {{\alpha _1},\,{\alpha _2}} \right) = ( - 1,\,3)$$</p>
<p>Now, equation of the line passing through ($$-$$1, 3) and making angle $${\pi \over 3}$$ with positive x-axis is</p>
<p>$$(y - {y_1}) = m(x - {x_1})$$</p>
<p>$$ \Rightarrow y - 3 = \left( {\tan {\pi \over 3}} \right)(x + 1)$$</p>
<p>$$ \Rightarrow y - 3 = \sqrt 3 (x + 1)$$</p>
<p>$$ \Rightarrow \sqrt 3 x - y + \sqrt 3 + 3 = 0$$</p> |
<p>Let $$A(1,1), B(-4,3), C(-2,-5)$$ be vertices of a triangle $$A B C, P$$ be a point on side $$B C$$, and $$\Delta_{1}$$ and $$\Delta_{2}$$ be the areas of triangles $$A P B$$ and $$A B C$$, respectively. If $$\Delta_{1}: \Delta_{2}=4: 7$$, then the area enclosed by the lines $$A P, A C$$ and the $$x$$-axis is :</p>
Options:
[{"identifier": "A", "content": "$$\\frac{1}{4}$$"}, {"identifier": "B", "content": "$$\\frac{3}{4}$$"}, {"identifier": "C", "content": "$$\\frac{1}{2}$$"}, {"identifier": "D", "content": "1"}] | ["C"]
Explanation:
<p>$${{{\Delta _1}} \over {{\Delta _2}}} = {{{1 \over 2} \times BP \times AH} \over {{1 \over 2} \times BC \times AH}} = {4 \over 7}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7psetoj/75213353-b6a6-48ee-b558-8da28250e29e/d5b7d130-2da9-11ed-8542-f96181a425b5/file-1l7psetok.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7psetoj/75213353-b6a6-48ee-b558-8da28250e29e/d5b7d130-2da9-11ed-8542-f96181a425b5/file-1l7psetok.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th July Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 41 English Explanation"></p>
<p>$$P\left( {{{ - 20} \over 7},{{ - 11} \over 7}} \right)$$</p>
<p>Line $$AC:y - 1 = 2(x - 1)$$</p>
<p>Intersection with x-axis $$ = \left( {{1 \over 2},0} \right)$$</p>
<p>Line $$AP:y - 1 = {2 \over 3}(x - 1)$$</p>
<p>Intersection with x-axis $$\left( {{{ - 1} \over 2},0} \right)$$</p>
<p>Vertices are $$(1,1),\left( {{1 \over 2},0} \right)$$ and $$\left( {{{ - 1} \over 2},0} \right)$$</p>
<p>Area $$ = {1 \over 2}$$ sq. unit</p> |
<p>Let $$B$$ and $$C$$ be the two points on the line $$y+x=0$$ such that $$B$$ and $$C$$ are symmetric with respect to the origin. Suppose $$A$$ is a point on $$y-2 x=2$$ such that $$\triangle A B C$$ is an equilateral triangle. Then, the area of the $$\triangle A B C$$ is :</p>
Options:
[{"identifier": "A", "content": "$$\\frac{10}{\\sqrt{3}}$$"}, {"identifier": "B", "content": "$$2 \\sqrt{3}$$"}, {"identifier": "C", "content": "$$3 \\sqrt{3}$$"}, {"identifier": "D", "content": "$$\\frac{8}{\\sqrt{3}}$$"}] | ["D"]
Explanation:
Origin $(O)$ is mid-point of $B C(x+y=0)$.
<br/><br/>
$A$ lies on perpendicular bisector of $B C$, which is $x-y=0$
<br/><br/>
A is point of intersection of $x-y=0$ and $y-2 x=2$
<br/><br/>
$\therefore A \equiv(-2,-2)$
<br/><br/>
Let $h=A O=\frac{-2-2}{\sqrt{1^{2}+1^{2}}}=2 \sqrt{2}$
<br/><br/>
$$
\text { Area }=\frac{h^{2}}{\sqrt{3}}=\frac{8}{\sqrt{3}}
$$ |
<p>A variable line $$\mathrm{L}$$ passes through the point $$(3,5)$$ and intersects the positive coordinate axes at the points $$\mathrm{A}$$ and $$\mathrm{B}$$. The minimum area of the triangle $$\mathrm{OAB}$$, where $$\mathrm{O}$$ is the origin, is :</p>
Options:
[{"identifier": "A", "content": "35"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "30"}, {"identifier": "D", "content": "40"}] | ["C"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw3avn6c/d5ed79f8-059d-4be3-a5b7-c41490773adc/6f4ef630-103d-11ef-94ca-5b39b659c816/file-1lw3avn6d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw3avn6c/d5ed79f8-059d-4be3-a5b7-c41490773adc/6f4ef630-103d-11ef-94ca-5b39b659c816/file-1lw3avn6d.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 11 English Explanation"></p>
<p>$$\begin{aligned}
& \frac{x}{a}+\frac{y}{b}=1 \\
& \frac{3}{a}+\frac{5}{b}=1 \\
& 3 b+5 a=a b \\
& 5 a-a b=-3 b \\
& a(5-b)=-3 b \\
& a=\frac{3 b}{b-5}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Area of triangle }=\left|\frac{1}{2} \times a \times b\right| \\
& =\frac{1}{2} \times \frac{3 b}{b-5} \times b \\
& \Rightarrow f(b)=\frac{3 b^2}{2 b-10} \\
& \Rightarrow f^{\prime}(b)=0,(2 b-10) 6 b-2\left(3 b^2\right)=0 \\
& 12 b^2-60 b-6 b^2=0 \\
& 6 b^2-60 b=0 \\
& b^2-10 b=0 \\
& b(b-10)=0 \\
& b=0 \text { or } b=10
\end{aligned}$$</p>
<p>So for minimum area, $$b=10$$</p>
<p>then $$\frac{1}{2} \times a \times b=30$$</p> |
If a vertex of a triangle is $$(1, 1)$$ and the mid points of two sides through this vertex are $$(-1, 2)$$ and $$(3, 2)$$ then the centroid of the triangle is :
Options:
[{"identifier": "A", "content": "$$\\left( { - 1,{7 \\over 3}} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {{{ - 1} \\over 3},{7 \\over 3}} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { 1,{7 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{{ 1} \\over 3},{7 \\over 3}} \\right)$$"}] | ["C"]
Explanation:
Vertex of triangle is $$\left( {1,\,1} \right)$$ and midpoint of sides through -
<br><br>this vertex is $$\left( { - 1,\,2} \right)$$ and $$\left( {3,2} \right)$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263971/exam_images/d94fpagqlokdwur1hypq.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Straight Lines and Pair of Straight Lines Question 138 English Explanation">
<br><br>$$ \Rightarrow $$ vertex $$B$$ and $$C$$ come out to be $$\left( { - 3,3} \right)$$ and $$\left( {5,3} \right)$$
<br><br>$$\therefore$$ centroid is $${{1 - 3 + 5} \over 3},{{1 + 3 + 5} \over 3} \Rightarrow \left( {1,{7 \over 3}} \right)$$ |
The $$x$$-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as $$(0, 1) (1, 1)$$ and $$(1, 0)$$ is :
Options:
[{"identifier": "A", "content": "$$2 + \\sqrt 2 $$ "}, {"identifier": "B", "content": "$$2 - \\sqrt 2 $$"}, {"identifier": "C", "content": "$$1 + \\sqrt 2 $$"}, {"identifier": "D", "content": "$$1 - \\sqrt 2 $$"}] | ["B"]
Explanation:
From the figure, we have
<br><br>$$a = 2,b = 2\sqrt 2 ,c = 2$$
<br><br>$${x_1} = 0,\,{x^2} = 0,\,{x_3} = 2$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264731/exam_images/mfh7nksrzqack2i9ykdk.webp" loading="lazy" alt="JEE Main 2013 (Offline) Mathematics - Straight Lines and Pair of Straight Lines Question 125 English Explanation">
<br><br>Now, $$x$$-co-ordinate of incenter is given as
<br><br>$${{a{x_1} + b{x_2} + c{x_3}} \over {a + b + c}}$$
<br><br>$$ \Rightarrow x$$-coordinate of incentre
<br><br>$$ = {{2 \times 0 + 2\sqrt 2 .0 + 2.2} \over {2 + 2 + 2\sqrt 2 }}$$
<br><br>$$=$$ $${2 \over {2 + \sqrt 2 }} = 2 - \sqrt 2 $$ |
Let k be an integer such that the triangle with vertices (k, – 3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point :
Options:
[{"identifier": "A", "content": "$$\\left( {1,{3 \\over 4}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {1, - {3 \\over 4}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {2,{1 \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {2, - {1 \\over 2}} \\right)$$"}] | ["C"]
Explanation:
Given, vertices of triangle are (k, – 3k), (5, k) and (–k, 2).
<br><br>$${1 \over 2}\left| {\matrix{
k & { - 3k} & 1 \cr
5 & k & 1 \cr
{ - k} & 2 & 1 \cr
} } \right| = \pm 28$$
<br><br>$$ \Rightarrow $$ k(k - 2) + 3k(5 + k) + 1(10 + k<sup>2</sup>) = $$ \pm $$ 56
<br><br>$$ \Rightarrow $$ 5k<sup>2</sup> + 13k + 10 = $$ \pm $$ 56
<br><br>$$ \Rightarrow $$ 5k<sup>2</sup> + 13k - 66 = 0
<br><br>$$ \Rightarrow $$ k = $${{ - 13 \pm \sqrt { - 1151} } \over {10}}$$
<br><br>So no real solution exist.
<br><br>or 5k<sup>2</sup> + 13k - 46 = 0
<br><br>$$ \therefore $$ k = $${{ - 23} \over 5}$$ or k = 2
<br><br>since k is an integer $$ \therefore $$ k = 2
<br><br>Thus, the coordinate of vertices of triangle are
<br><br>A(2, -6), B(5, 2) and C(-2, 2).
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265540/exam_images/uwujwuudzsbrhbcutpgm.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Offline) Mathematics - Straight Lines and Pair of Straight Lines Question 118 English Explanation">
<br><br>Now, equation of altitude from vertex A is
<br><br>y - (-6) = $${{ - 1} \over {\left( {{{2 - 2} \over { - 2 - 5}}} \right)}}\left( {x - 2} \right)$$
<br><br>$$ \Rightarrow $$ x = 2 .......(1)
<br><br>Equation of altitude from vertex B is
<br><br>y - 2 = $${{ - 1} \over {\left( {{{2 + 6} \over { - 2 - 2}}} \right)}}\left( {x - 5} \right)$$
<br><br>$$ \Rightarrow $$ 2y - 4 = x - 5
<br><br>$$ \Rightarrow $$ x - 2y = 1 .......(2)
<br><br>Point H($$\alpha $$, $$\beta $$) lies on both (1) and (2),
<br><br>$$ \therefore $$ $$\alpha $$ = 2 .........(3)
<br><br>$$\alpha $$ - 2$$\beta $$ = 1 ......(4)
<br><br>Solving (3) and (4), we get
<br><br>$$\alpha $$ = 2 , $$\beta $$ = $${1 \over 2}$$
<br><br>$$ \therefore $$ Orthocentre is $$\left( {2,{1 \over 2}} \right)$$. |
Let the equations of two sides of a triangle be 3x $$-$$ 2y + 6 = 0 and 4x + 5y $$-$$ 20 = 0. If the orthocentre of this triangle is at (1, 1), then the equation of its third side is :
Options:
[{"identifier": "A", "content": "122y $$-$$ 26x $$-$$ 1675 = 0"}, {"identifier": "B", "content": "122y + 26x + 1675 = 0"}, {"identifier": "C", "content": "26x + 61y + 1675 = 0"}, {"identifier": "D", "content": "26x $$-$$ 122y $$-$$ 1675 = 0"}] | ["D"]
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264905/exam_images/ubp63holkjbubzmm73xt.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Mathematics - Straight Lines and Pair of Straight Lines Question 104 English Explanation">
<br><br>4x + 5y $$-$$ 20 = 0 . . .(1)
<br><br>3x $$-$$ 2y + 6 = 0 . . . (2)
<br><br>orthocentre is (1, 1)
<br><br>line perpendicular to 4x + 5y $$-$$ 20 = 0
<br><br>and passes through (1, 1) is
<br><br>(y $$-$$ 1) = $${5 \over 4}$$(x $$-$$ 1)
<br><br>$$ \Rightarrow $$ 5x $$-$$ 4y = 1 . . .(3)
<br><br>and line $$ \bot $$ to 3x $$-$$ 2y + 6 = 0
<br><br>and passes through (1, 1)
<br><br>y $$-$$ 1 = $$-$$ $${2 \over 3}$$ (x $$-$$ 1)
<br><br>$$ \Rightarrow $$ 2x + 3y = 5 . . .(4)
<br><br>Solving (1) and (4) we get C$$\left( {{{35} \over 2}, - 10} \right)$$
<br><br>Solving (2) and (3) we get A $$\left( { - 13,{{ - 33} \over 2}} \right)$$
<br><br>Side BC is y + 10 = $${{{{ - 33} \over 2} + 10} \over { - 13 - {{35} \over 2}}}\left( {x - {{35} \over 2}} \right)$$
<br><br>$$ \Rightarrow $$ y + 10 = $${{13} \over {61}}\left( {x - {{35} \over 2}} \right)$$
<br><br>$$ \Rightarrow $$ 26x $$-$$ 122y $$-$$ 1675 = 0 |
A point P moves on the line 2x – 3y + 4 = 0. If Q(1, 4) and R (3, – 2) are fixed points, then the locus of the centroid of $$\Delta $$PQR is a line :
Options:
[{"identifier": "A", "content": "parallel to y-axis"}, {"identifier": "B", "content": "with slope $${2 \\over 3}$$"}, {"identifier": "C", "content": "parallel to x-axis"}, {"identifier": "D", "content": "with slope $${3 \\over 2}$$"}] | ["B"]
Explanation:
Let the centroid of $$\Delta $$PQR is (h, k) & P is ($$\alpha $$, $$\beta $$), then
<br><br>$${{\alpha + 1 + 3} \over 3} = h\,$$ and $${{\beta + 4 - 2} \over 3} = k$$
<br><br>$$\alpha = \left( {3h - 4} \right)$$ $$\beta = \left( {3k - 4} \right)$$
<br><br>Point P($$\alpha $$, $$\beta $$) lies on the line 2x $$-$$ 3y + 4 = 0
<br><br>$$ \therefore $$ 2(3h $$-$$ 4) $$-$$ 3 (3k $$-$$ 2) + 4 = 0
<br><br>$$ \Rightarrow $$ locus is 6x $$-$$ 9y + 2 = 0 |
If the line 3x + 4y – 24 = 0 intersects the x-axis at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is :
Options:
[{"identifier": "A", "content": "(3, 4)"}, {"identifier": "B", "content": "(2, 2)"}, {"identifier": "C", "content": "(4, 4)"}, {"identifier": "D", "content": "(4, 3)"}] | ["B"]
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265477/exam_images/migtvndabwdpqcwl9h0o.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 101 English Explanation">
<br><br>$$\left| {{{3r + 4r - 24} \over 5}} \right| = r$$
<br><br>$$7r - 24 = \pm 5r$$
<br><br>$$2r = 24$$ or $$12r + 24$$
<br><br>$$r = 14,\,\,\,r = 2$$
<br><br> then incentre is $$(2,2)$$ |
Two vertices of a triangle are (0, 2) and (4, 3). If its orthocenter is at the origin, then its third vertex lies in which quadrant :
Options:
[{"identifier": "A", "content": "third"}, {"identifier": "B", "content": "fourth"}, {"identifier": "C", "content": "second"}, {"identifier": "D", "content": "first"}] | ["C"]
Explanation:
m<sub>BD</sub> $$ \times $$ m<sub>AD</sub> = $$-$$ 1
<br><br>$$ \Rightarrow $$ $$\left( {{{3 - 2} \over {4 - 0}}} \right) \times \left( {{{b - 0} \over {a - 0}}} \right) = - 1$$
<br><br>$$ \Rightarrow $$ b + 4a = 0 . . . . (i)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267693/exam_images/loayka4cghxne7uqkofa.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Mathematics - Straight Lines and Pair of Straight Lines Question 99 English Explanation">
<br>m<sub>AB</sub> $$ \times $$ m<sub>CF</sub> = $$-$$ 1
<br><br>$$ \Rightarrow $$ $$\left( {{{\left( {b - 2} \right)} \over {a - 0}}} \right) \times \left( {{3 \over 4}} \right) = - 1$$
<br><br>$$ \Rightarrow $$ 3b $$-$$ 6 = $$-$$ 4a
<br><br>$$ \Rightarrow $$ 4a + 3b = 6 . . . . .(ii)
<br><br>From (i) and (ii)
<br><br>a = $${{ - 3} \over 4}$$, b = 3
<br><br>$$ \therefore $$ II<sup>nd</sup> quadrant. |
Let A(1, 0), B(6, 2) and C $$\left( {{3 \over 2},6} \right)$$ be the vertices of a triangle ABC. If P is a Point inside the triangle ABC such that the triangles APC, APB and BPC have equal areas, then the length of the line segment PQ, where Q is the point $$\left( { - {7 \over 6}, - {1 \over 3}} \right)$$, is ________.
Options:
[] | 5
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265627/exam_images/uw3nb7jjk16aarugxdjz.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 86 English Explanation">
<br> P is centroid of the triangle ABC.
<br><br>P = $$\left( {{{1 + 6 + {3 \over 2}} \over 3},{{0 + 2 + 6} \over 3}} \right)$$
<br><br>= $$\left( {{{17} \over 6},{8 \over 3}} \right)$$
<br><br>Given Q $$\left( { - {7 \over 6}, - {1 \over 3}} \right)$$.
<br><br>$$ \therefore $$ PQ = $$\sqrt {{{\left( {{{24} \over 6}} \right)}^2} + {{\left( {{9 \over 3}} \right)}^2}} $$ = 5 |
Let C be the centroid of the triangle with
vertices (3, –1), (1, 3) and (2, 4). Let P be the
point of intersection of the lines x + 3y – 1 = 0
and 3x – y + 1 = 0. Then the line passing through
the points C and P also passes through the
point :
Options:
[{"identifier": "A", "content": "(\u20139, \u20137)"}, {"identifier": "B", "content": "(9, 7)"}, {"identifier": "C", "content": "(7, 6)"}, {"identifier": "D", "content": "(\u20139, \u20136)"}] | ["D"]
Explanation:
Centroid C $$\left( {{{3 + 1 + 2} \over 3},{{ - 1 + 3 + 4} \over 3}} \right)$$ = (2, 2)
<br><br>Point of intersection of lines x + 3y – 1 = 0 <br><br>and 3x – y + 1 = 0 is P $$\left( { - {1 \over 5},{2 \over 5}} \right)$$
<br><br>So, equation of line CP is 8x – 11y + 6 = 0
<br><br>Point (–9, –6) satisfy this equation. |
If a $$\Delta $$ABC has vertices A(–1, 7), B(–7, 1) and
C(5, –5), then its orthocentre has coordinates :
Options:
[{"identifier": "A", "content": "(\u20133, 3)"}, {"identifier": "B", "content": "(3, \u20133)"}, {"identifier": "C", "content": "$$\\left( {{3 \\over 5}, - {3 \\over 5}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - {3 \\over 5},{3 \\over 5}} \\right)$$"}] | ["A"]
Explanation:
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267134/exam_images/ubvfpgl2agbziakghok1.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265216/exam_images/cip9scai14rag6uy6gdh.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267104/exam_images/idr7pfamynl59pilcm5t.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Evening Slot Mathematics - Straight Lines and Pair of Straight Lines Question 81 English Explanation"></picture>
<br><br>equation of CD
<br> <br>y + 5 = -1 (x - 5)
<br><br>$$ \Rightarrow $$ x + y = 0 .....(1)
<br><br> equation of AE
<br><br>y - 7 = 2 (x + 1)
<br><br>$$ \Rightarrow $$ 2x - y = -9 ......(2)
<br><br> from (1) & (2)
<br><br> x = -3, y = 3
<br><br> Othocentre = (-3, 3) |
In a triangle PQR, the co-ordinates of the points P and Q are ($$-$$2, 4) and (4, $$-$$2) respectively. If the equation of the perpendicular bisector of PR is 2x $$-$$ y + 2 = 0, then the centre of the circumcircle of the $$\Delta$$PQR is :
Options:
[{"identifier": "A", "content": "($$-$$1, 0)"}, {"identifier": "B", "content": "(1, 4)"}, {"identifier": "C", "content": "(0, 2)"}, {"identifier": "D", "content": "($$-$$2, $$-$$2)"}] | ["D"]
Explanation:
Mid point of $$PQ \equiv \left( {{{ - 2 + 4} \over 2},{{4 - 2} \over 2}} \right) \equiv (1,1)$$<br><br>Slope of $$PQ = {{4 + 2} \over { - 2 - 4}} = - 1$$<br><br>Slope of perpendicular bisector of PQ = 1<br><br>Equation of perpendicular bisector of PQ <br><br>$$y - 1 = 1(x - 1)$$<br><br>$$ \Rightarrow y = x$$<br><br>Solving with perpendicular bisector of PR,
<br><br>2x $$-$$ y + 2 = 0<br><br>Circumcentre is ($$-$$2, $$-$$2) |
Let tan$$\alpha$$, tan$$\beta$$ and tan$$\gamma$$; $$\alpha$$, $$\beta$$, $$\gamma$$ $$\ne$$ $${{(2n - 1)\pi } \over 2}$$, n$$\in$$N be the slopes of three line segments OA, OB and OC, respectively, where O is origin. If circumcentre of $$\Delta$$ABC coincides with origin and its orthocentre lies on y-axis, then the value of $${\left( {{{\cos 3\alpha + \cos 3\beta + \cos 3\gamma } \over {\cos \alpha \cos \beta \cos \gamma }}} \right)^2}$$ is equal to ____________.
Options:
[] | 144
Explanation:
Since orthocentre and circumcentre both lies on y-axis.<br><br>$$ \Rightarrow $$ Centroid also lies on y-axis.<br><br>$$ \Rightarrow $$ $$\sum {\cos \alpha = 0} $$<br><br>cos$$\alpha$$ + cos$$\beta$$ + cos$$\gamma$$ = 0<br><br>$$ \Rightarrow $$ cos<sup>3</sup> $$\alpha$$ + cos<sup>3</sup> $$\beta$$ + cos<sup>3</sup> $$\gamma$$ = 3cos$$\alpha$$cos$$\beta$$cos$$\gamma$$<br><br>$$ \therefore $$ $${{\cos 3\alpha + \cos 3\beta + \cos 3\gamma } \over {\cos \alpha \cos \beta \cos \gamma }}$$<br><br>$$ = {{4({{\cos }^3}\alpha + {{\cos }^3}\beta + {{\cos }^3}\gamma ) - 3(\cos \alpha + \cos \beta + \cos \gamma )} \over {\cos \alpha \cos \beta \cos \gamma }} = 12$$<br><br>then, $${\left( {{{\cos 3\alpha + \cos 3\beta + \cos 3\gamma } \over {\cos \alpha \cos \beta \cos \gamma }}} \right)^2} = 144$$ |
Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line x + y = 3. If R and r be the radius of circumcircle and incircle respectively of $$\Delta$$ABC, then (R + r) is equal to :
Options:
[{"identifier": "A", "content": "$$7\\sqrt 2 $$"}, {"identifier": "B", "content": "$${9 \\over {\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$$2\\sqrt 2 $$"}, {"identifier": "D", "content": "$$3\\sqrt 2 $$"}] | ["B"]
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264797/exam_images/ajzrnczcpsrafxwpsdbk.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 65 English Explanation">
<br><br>$$r = \left| {{{0 + 0 - 3} \over {\sqrt 2 }}} \right| = {3 \over {\sqrt 2 }}$$<br><br>$$\sin 30^\circ = {r \over R} = {1 \over 2}$$<br><br>R = 2r<br><br>So, $$r + R = 3r = 3 \times \left( {{3 \over {\sqrt 2 }}} \right) = {9 \over {\sqrt 2 }}$$ |
Consider a triangle having vertices A($$-$$2, 3), B(1, 9) and C(3, 8). If a line L passing through the circum-centre of triangle ABC, bisects line BC, and intersects y-axis at point $$\left( {0,{\alpha \over 2}} \right)$$, then the value of real number $$\alpha$$ is ________________.
Options:
[] | 9
Explanation:
<br>$${\left( {\sqrt {50} } \right)^2} = {\left( {\sqrt {45} } \right)^2} + {\left( {\sqrt 5 } \right)^2}$$<br><br>$$\angle B = 90^\circ $$<br><br>Circum-center $$ = \left( {{1 \over 2},{{11} \over 2}} \right)$$<br><br>Mid point of BC $$ = \left( {2,{{17} \over 2}} \right)$$<br><br>Line : $$\left( {y - {{11} \over 2}} \right) = 2\left( {x - {1 \over 2}} \right) \Rightarrow y = 2x + {9 \over 2}$$<br><br>Passing through $$\left( {0,{\alpha \over 2}} \right)$$<br><br>$${\alpha \over 2} = {9 \over 2} \Rightarrow \alpha = 9$$<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwjjv06e/aee1807a-0488-47e6-945f-12a768cd6dab/5074a460-5075-11ec-8f7b-45f8e5a35226/file-1kwjjv06f.jpeg" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 20th July Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 64 English Explanation"> |
<p>The distance of the origin from the centroid of the triangle whose two sides have the equations $$x - 2y + 1 = 0$$ and $$2x - y - 1 = 0$$ and whose orthocenter is $$\left( {{7 \over 3},{7 \over 3}} \right)$$ is :</p>
Options:
[{"identifier": "A", "content": "$$\\sqrt 2 $$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "2$$\\sqrt 2 $$"}, {"identifier": "D", "content": "4"}] | ["C"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5gmnar6/e2444175-9dd2-4655-a827-d624ec162bae/ce00a120-0107-11ed-a5c5-a7461ac77783/file-1l5gmnar7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5gmnar6/e2444175-9dd2-4655-a827-d624ec162bae/ce00a120-0107-11ed-a5c5-a7461ac77783/file-1l5gmnar7.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 53 English Explanation"></p>
<p>For point A,</p>
<p>2x $$-$$ y $$-$$ 1 = 0 ...... (1)</p>
<p>x $$-$$ 2y + 1 = 0 ...... (2)</p>
<p>Solving (1) and (2), we get</p>
<p>x = 1, y = 1.</p>
<p>$$\therefore$$ Point A = (1, 1)</p>
<p>Altitude from B to line AC is perpendicular to line AC.</p>
<p>$$\therefore$$ Equator of altitude BH is</p>
<p>2x + y + $$\lambda$$ = 0 ...... (3)</p>
<p>It passes through point $$H\left( {{7 \over 3},{7 \over 3}} \right)$$ so it satisfy the equation (3).</p>
<p>$${{14} \over 3} + {7 \over 3} + \lambda = 0$$</p>
<p>$$\Rightarrow$$ $$\alpha$$ = $$-$$7</p>
<p>$$\therefore$$ Altitude BH = 2x + y $$-$$ 7 = 0 ...... (4)</p>
<p>Solving equation (1) and (4), we get</p>
<p>x = 2, y = 3.</p>
<p>$$\therefore$$ Point B = (2, 3)</p>
<p>Altitude from C to line AB is perpendicular to line AB.</p>
<p>$$\therefore$$ Equation of altitude CH is</p>
<p>x + 2y + $$\lambda$$ = 0 ...... (5)</p>
<p>It passes through point $$H\left( {{7 \over 3},{7 \over 3}} \right)$$ so it satisfy equation (5).</p>
<p>$${7 \over 3} + {{14} \over 3} + \lambda = 0$$</p>
<p>$$\Rightarrow$$ $$\lambda$$ = $$-$$7</p>
<p>$$\therefore$$ Altitude CH = x + 2y $$-$$ 7 = 0 ...... (6)</p>
<p>Solving equation (2) and (6), we get</p>
<p>x = 3, y = 2</p>
<p>$$\therefore$$ Point C = (3, 2)</p>
<p>Centroid G (x, y) of triangle A (1, 1), B (2, 3) and C (3, 2) is</p>
<p>$$x = {{1 + 2 + 3} \over 3} = 2$$, $$y = {{1 + 2 + 3} \over 3} = 2$$</p>
<p>Now, distance of point G (2, 2) from center O (0, 0) is</p>
<p>$$OG = \sqrt {{2^2} + {2^2}} = 2\sqrt 2 $$</p> |
<p>In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If ($$\alpha$$, $$\beta$$) is the centroid of $$\Delta$$ABC, then 15($$\alpha$$ + $$\beta$$) is equal to :</p>
Options:
[{"identifier": "A", "content": "39"}, {"identifier": "B", "content": "41"}, {"identifier": "C", "content": "51"}, {"identifier": "D", "content": "63"}] | ["C"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5q9xuo0/da6e885c-cdde-4480-b9d1-c22df8a701e5/f34a6800-0655-11ed-903e-c9687588b3f3/file-1l5q9xuo1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5q9xuo0/da6e885c-cdde-4480-b9d1-c22df8a701e5/f34a6800-0655-11ed-903e-c9687588b3f3/file-1l5q9xuo1.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 50 English Explanation"></p>
<p>$$\left. \matrix{
2x + y = 4 \hfill \cr
2x + 6y = 14 \hfill \cr} \right\}y = 2,\,x = 3$$</p>
<p>B(1, 2)</p>
<p>Let C(k, 4 $$-$$ 2k)</p>
<p>Now $$A{B^2} = A{C^2}$$</p>
<p>$${5^2} + {( - 1)^2} = {(6 - k)^2} + {( - 3 + 2k)^2}$$</p>
<p>$$ \Rightarrow 5{k^2} - 24k + 19 = 0$$</p>
<p>$$(5k - 19)(k - 1) = 0 \Rightarrow k = {{19} \over 5}$$</p>
<p>$$C\left( {{{19} \over 5}, - {{18} \over 5}} \right)$$</p>
<p>Centroid ($$\alpha$$, $$\beta$$)</p>
<p>$$\alpha = {{6 + 1 + {{19} \over 5}} \over 3} = {{18} \over 5}$$</p>
<p>$$\beta = {{1 + 2 - {{18} \over 5}} \over 3} = - {1 \over 5}$$</p>
<p>Now $$15(\alpha + \beta )$$</p>
<p>$$15\left( {{{17} \over 5}} \right) = 51$$</p> |
<p>The equations of the sides $$\mathrm{AB}, \mathrm{BC}$$ and $$\mathrm{CA}$$ of a triangle $$\mathrm{ABC}$$ are $$2 x+y=0, x+\mathrm{p} y=15 \mathrm{a}$$ and $$x-y=3$$ respectively. If its orthocentre is $$(2, a),-\frac{1}{2}<\mathrm{a}<2$$, then $$\mathrm{p}$$ is equal to ______________.</p>
Options:
[] | 3
Explanation:
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbzt6e/130dc0b4-c7c5-40cb-813c-5bded9b079b4/10d39160-2c50-11ed-9dc0-a1792fcc650d/file-1l7nbzt6f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nbzt6e/130dc0b4-c7c5-40cb-813c-5bded9b079b4/10d39160-2c50-11ed-9dc0-a1792fcc650d/file-1l7nbzt6f.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th July Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 42 English Explanation"> </p>
<p>Slope of $$AH = {{a + 2} \over 1}$$</p>
<p>Slope of $$BC = - {1 \over p}$$</p>
<p>$$\therefore$$ $$p = a + 2$$ ...... (i)</p>
<p>Coordinate of $$C = \left( {{{18p - 30} \over {p + 1}},\,{{15p - 33} \over {p + 1}}} \right)$$</p>
<p>Slope of $$HC = {{{{15P - 33} \over {p + 1}} - a} \over {{{18p - 30} \over {p + 1}} - 2}} = {{15p - 33 - (p - 2)(p + 1)} \over {18p - 30 - 2p - 2}}$$</p>
<p>$$ = {{16p - {p^2} - 31} \over {16p - 32}}$$</p>
<p>$$\because$$ $${{16p - {p^2} - 31} \over {16p - 32}} \times - 2 = - 1$$</p>
<p>$$\therefore$$ $${p^2} - 8p + 15 = 0$$</p>
<p>$$\therefore$$ $$p = 3$$ or $$5$$</p>
<p>But if $$p = 5$$ then $$a = 3$$ not acceptable</p>
<p>$$\therefore$$ $$p = 3$$</p> |
<p>The equations of the sides $$\mathrm{AB}, \mathrm{BC}$$ and CA of a triangle ABC are $$2 x+y=0, x+\mathrm{p} y=39$$ and $$x-y=3$$ respectively and $$\mathrm{P}(2,3)$$ is its circumcentre. Then which of the following is NOT true?</p>
Options:
[{"identifier": "A", "content": "$$(\\mathrm{AC})^{2}=9 \\mathrm{p}$$"}, {"identifier": "B", "content": "$$(\\mathrm{AC})^{2}+\\mathrm{p}^{2}=136$$"}, {"identifier": "C", "content": "$$32<\\operatorname{area}\\,(\\Delta \\mathrm{ABC})<36$$"}, {"identifier": "D", "content": "$$34<\\operatorname{area}\\,(\\triangle \\mathrm{ABC})<38$$"}] | ["D"]
Explanation:
<p>Intersection of $$2x + y = 0$$ and $$x - y = 3\,:\,A(1, - 2)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7qa6xtq/139436a6-31d3-4607-a05c-66e2b0bc9d4d/5ec55de0-2def-11ed-a744-1fb8f3709cfa/file-1l7qa6xtr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7qa6xtq/139436a6-31d3-4607-a05c-66e2b0bc9d4d/5ec55de0-2def-11ed-a744-1fb8f3709cfa/file-1l7qa6xtr.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th July Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 40 English Explanation"></p>
<p>Equation of perpendicular bisector of AB is</p>
<p>$$x - 2y = - 4$$</p>
<p>Equation of perpendicular bisector of AC is</p>
<p>$$x + y = 5$$</p>
<p>Point B is the image of A in line $$x - 2y + 4 = 0$$ which is obtained as $$B\left( {{{ - 13} \over 5},{{26} \over 5}} \right)$$</p>
<p>Similarly vertex $$C:(7,4)$$</p>
<p>Equation of line $$BC:x + 8y = 39$$</p>
<p>So, $$p = 8$$</p>
<p>$$AC = \sqrt {{{(7 - 1)}^2} + {{(4 + 2)}^2}} = 6\sqrt 2 $$</p>
<p>Area of triangle $$ABC = 32.4$$</p> |
<p>Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1,1). If the line AP intersects the line BC at the point Q$$\left(k_{1}, k_{2}\right)$$, then $$k_{1}+k_{2}$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$\\frac{4}{7}$$"}, {"identifier": "C", "content": "$$\\frac{2}{7}$$"}, {"identifier": "D", "content": "4"}] | ["B"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7ssjo3c/b55f47a7-9a89-4d60-875b-d115e046f61f/b6651380-2f50-11ed-85dd-19dc023e9ad1/file-1l7ssjo3d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7ssjo3c/b55f47a7-9a89-4d60-875b-d115e046f61f/b6651380-2f50-11ed-85dd-19dc023e9ad1/file-1l7ssjo3d.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 39 English Explanation"></p>
<p>Let D be mid-point of AC, then</p>
<p>$${{b + 3} \over 2} = 1 \Rightarrow b = - 1$$</p>
<p>Let E be mid-point of BC,</p>
<p>$${{5 - b} \over {b - a}}\,.\,{{{{(3 + b)} \over 2}} \over {{{a + b} \over 2} - 1}} = - 1$$</p>
<p>On putting $$b = - 1$$, we get $$a = 5$$ or $$-3$$</p>
<p>But $$a = 5$$ is rejected as $$ab > 0$$</p>
<p>$$A( - 3,3),\,B( - 1,5),\,C( - 3, - 1),\,P(1,1)$$</p>
<p>Line $$BC \Rightarrow y = 3x + 8$$</p>
<p>Line $$AP \Rightarrow y = {{3 - x} \over 2}$$</p>
<p>Point of intersection $$\left( {{{ - 13} \over 7},{{17} \over 7}} \right)$$</p> |
<p>Let $$\mathrm{A}(\alpha,-2), \mathrm{B}(\alpha, 6)$$ and $$\mathrm{C}\left(\frac{\alpha}{4},-2\right)$$ be vertices of a $$\triangle \mathrm{ABC}$$. If $$\left(5, \frac{\alpha}{4}\right)$$ is the circumcentre of $$\triangle \mathrm{ABC}$$, then which of the following is NOT correct about $$\triangle \mathrm{ABC}$$?</p>
Options:
[{"identifier": "A", "content": "area is 24"}, {"identifier": "B", "content": "perimeter is 25"}, {"identifier": "C", "content": "circumradius is 5"}, {"identifier": "D", "content": "inradius is 2"}] | ["B"]
Explanation:
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7xrag3x/1171f510-f750-4a4d-8bf7-c934d96c9030/b8102bd0-320b-11ed-b25b-a342e8d3e498/file-1l7xrag3y.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7xrag3x/1171f510-f750-4a4d-8bf7-c934d96c9030/b8102bd0-320b-11ed-b25b-a342e8d3e498/file-1l7xrag3y.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 37 English Explanation">
<br><br>Circumcentre of $\triangle A B C$
<br><br>$$
\begin{aligned}
&=\left(\frac{\alpha+\frac{\alpha}{4}}{2}, \frac{6-2}{2}\right) \\\\
&=\left(\frac{5 \alpha}{8}, 2\right) \\\\
&=\left(5, \frac{\alpha}{4}\right) \\\\
&\Rightarrow \alpha=8
\end{aligned}
$$
<br><br>$\operatorname{area}(\triangle A B C)=\frac{1}{2} \cdot \frac{3 \alpha}{4} \times 8=24$ sq. units
<br><br>$$
\begin{aligned}
\text { Perimeter } &=8+\frac{3 \alpha}{4}+\sqrt{8^{2}+\left(\frac{3 \alpha}{4}\right)^{2}} \\\\
&=8+6+10=24
\end{aligned}
$$
<br><br>Circumradius $=\frac{10}{2}=5$
<br><br> inradius $(r)=\frac{\Delta}{s}=\frac{24}{12}=2$ |
<p>If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is $$(\alpha,\beta)$$, then the quadratic equation whose roots are $$\alpha+4\beta$$ and $$4\alpha+\beta$$, is :</p>
Options:
[{"identifier": "A", "content": "$$x^2-20x+99=0$$"}, {"identifier": "B", "content": "$$x^2-22x+120=0$$"}, {"identifier": "C", "content": "$$x^2-19x+90=0$$"}, {"identifier": "D", "content": "$$x^2-18x+80=0$$"}] | ["A"]
Explanation:
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lf41l718/e00587c4-d3cd-4c76-ac3c-9d17149d6abd/35cb41c0-c016-11ed-89c7-f9c9b186ec3c/file-1lf41l719.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lf41l718/e00587c4-d3cd-4c76-ac3c-9d17149d6abd/35cb41c0-c016-11ed-89c7-f9c9b186ec3c/file-1lf41l719.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 35 English Explanation">
<br><br>$$
\mathrm{m}=-\frac{1}{2}
$$
<br><br>Here $\mathrm{m_BH} \times \mathrm{m_AC}=-1$
<br><br>$$
\begin{array}{ll}
& \left(\frac{\beta-3}{\alpha-2}\right)\left(\frac{1}{-2}\right)=-1 \\\\
& \beta-3=2 \alpha-4 \\\\
& \beta=2 \alpha-1 \\\\
& \mathrm{~m}_{\mathrm{AH}} \times \mathrm{m}_{\mathrm{BC}}=-1 \\\\
\Rightarrow & \left(\frac{\beta-2}{\alpha-1}\right)(-2)=-1 \\\\
\Rightarrow & 2 \beta-4=\alpha-1 \\\\
\Rightarrow & 2(2 \alpha-1)=\alpha+3 \\\\
\Rightarrow & 3 \alpha=5 \\\\
& \alpha=\frac{5}{3}, \beta=\frac{7}{3} \Rightarrow \mathrm{H}\left(\frac{5}{3}, \frac{7}{3}\right) \\\\
& \alpha+4 \beta=\frac{5}{3}+\frac{28}{3}=\frac{33}{3}=11 \\\\
& \beta+4 \alpha=\frac{7}{3}+\frac{20}{3}=\frac{27}{3}=9 \\\\
& \mathrm{x}^2-20 \mathrm{x}+99=0
\end{array}
$$ |
<p>The equations of the sides AB, BC and CA of a triangle ABC are : $$2x+y=0,x+py=21a,(a\pm0)$$ and $$x-y=3$$ respectively. Let P(2, a) be the centroid of $$\Delta$$ABC. Then (BC)$$^2$$ is equal to ___________.</p>
Options:
[] | 122
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5hxc6e/38f5f2ee-c758-49c2-a1e2-99e79ae87394/aad05860-ad16-11ed-8a8c-4d67f5492755/file-1le5hxc6f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5hxc6e/38f5f2ee-c758-49c2-a1e2-99e79ae87394/aad05860-ad16-11ed-8a8c-4d67f5492755/file-1le5hxc6f.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 32 English Explanation"></p>
$$
\because \frac{21 a}{1-2 p}+1+\frac{3 p+21 a}{p+1}=6
$$
<br><br>
$$
\begin{aligned}
& \therefore 4 p^{2}-21 a p+8 p+42 a-5=0\quad...(1)
\end{aligned}
$$
<br><br>
And $\frac{-42 a}{1-2 p}-2+\frac{21 a-3}{p+1}=3 a$
<br><br>
$$
\therefore 4 p^{2}-81 a p+6 a p^{2}-24 a+8 p-5=0 \quad...(2)
$$
<br><br>
From equation (1) - equation (2) we get;
<br><br>
$$
60 a p+66 a-6 a p^{2}=0
$$
<br><br>
$$
\begin{aligned}
\because a \neq 0 \Rightarrow p^{2}-10 p-11=0 \\\\
p=-1 \text { or } 11 \Rightarrow p=11 .
\end{aligned}
$$
<br><br>
When $p=11$ then $a=3$
<br><br>
Coordinate of $B=(-3,6)$
<br><br>
And coordinate of $C=(8,5)$
<br><br>
$\therefore B C^{2}=122$ |
If $(\alpha, \beta)$ is the orthocenter of the triangle $\mathrm{ABC}$ with vertices $A(3,-7), B(-1,2)$ and $C(4,5)$, then $9 \alpha-6 \beta+60$ is equal to :
Options:
[{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "40"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "35"}] | ["C"]
Explanation:
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgxuayzx/c68e4c1e-f2ed-4189-96d3-77fa5b78bc40/7473b5d0-e445-11ed-97cc-4f9b4bf32610/file-1lgxuayzy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgxuayzx/c68e4c1e-f2ed-4189-96d3-77fa5b78bc40/7473b5d0-e445-11ed-97cc-4f9b4bf32610/file-1lgxuayzy.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 31 English Explanation">
<br><br>$$
\begin{aligned}
& \text { Altitude of BC: } y+7=\frac{-5}{3}(x-3) \\\\
& 3 y+21=-5 x+15 \\\\
& 5 x+3 y+6=0 \\\\
& \text { Altitude of AC: } y-2=\frac{-1}{12}(x+1) \\\\
& 12 y-24=-x-1 \\\\
& x+12 y=23 \\\\
& \alpha=\frac{-47}{19}, \quad \beta=\frac{121}{57} \\\\
& 9 \alpha-6 \beta+60=25
\end{aligned}
$$ |
<p>Let $$(\alpha, \beta)$$ be the centroid of the triangle formed by the lines $$15 x-y=82,6 x-5 y=-4$$ and $$9 x+4 y=17$$. Then $$\alpha+2 \beta$$ and $$2 \alpha-\beta$$ are the roots of the equation :</p>
Options:
[{"identifier": "A", "content": "$$x^{2}-7 x+12=0$$"}, {"identifier": "B", "content": "$$x^{2}-13 x+42=0$$"}, {"identifier": "C", "content": "$$x^{2}-14 x+48=0$$"}, {"identifier": "D", "content": "$$x^{2}-10 x+25=0$$"}] | ["B"]
Explanation:
<ol>
<li>Solve the equations $15x - y = 82$ and $6x - 5y = -4$</li>
</ol>
<p>Multiply the first equation by 5 and the second by 1 and then subtract the second from the first:</p>
<p>$75x - 5y = 410$</p>
<p>$6x - 5y = -4$</p>
<p>Subtracting these gives $69x = 414$ which leads to $x = 6$</p>
<p>Substitute $x = 6$ into the first equation to get $y = 8$</p>
<p>So, the first vertex is $(6,8)$.</p>
<ol>
<li>Solve the equations $15x - y = 82$ and $9x + 4y = 17$</li>
</ol>
<p>Multiply the first equation by 4 and the second by 1 and then add:</p>
<p>$60x - 4y = 328$</p>
<p>$9x + 4y = 17$</p>
<p>Adding these gives $69x = 345$ which leads to $x = 5$</p>
<p>Substitute $x = 5$ into the second equation to get $y = -7$</p>
<p>So, the second vertex is $(5,-7)$.</p>
<ol>
<li>Solve the equations $6x - 5y = -4$ and $9x + 4y = 17$</li>
</ol>
<p>Multiply the first equation by 4 and the second by 5 and then add:</p>
<p>$24x - 20y = -16$</p>
<p>$45x + 20y = 85$</p>
<p>Adding these gives $69x = 69$ which leads to $x = 1$</p>
<p>Substitute $x = 1$ into the first equation to get $y = 2$</p>
<p>So, the third vertex is $(1,2)$.</p>
<p>So, the vertices of the triangle are $(6,8)$, $(5,-7)$, and $(1,2)$.</p>
<p>Now that we have the vertices of the triangle, we can find the centroid. The centroid $(\alpha, \beta)$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by :</p>
<p>$$
\alpha=\frac{x_1+x_2+x_3}{3}$$,
<br/><br/>$$
\beta=\frac{y_1+y_2+y_3}{3}
$$</p>
<p>Substituting the coordinates of the vertices into these equations, we get:</p>
<p>$$
\alpha=\frac{6+5+1}{3}=4 $$,
<br/><br/>$$
\beta=\frac{8-7+2}{3}=1
$$</p>
<p>Then, we find $\alpha+2\beta$ and $2\alpha-\beta$:</p>
<p>$$
\alpha+2\beta=4+2(1)=6 \
2\alpha-\beta=2(4)-1=7
$$</p>
<p>So, $\alpha+2\beta=6$ and $2\alpha-\beta=7$ are the roots of the quadratic equation. </p>
<p>We can write the quadratic equation as </p>
<p>$$x^2-(\alpha+2\beta+2\alpha-\beta)x+(\alpha+2\beta)(2\alpha-\beta)=0$$</p>
<p>Substituting $\alpha=4$, $\beta=1$ gives us:</p>
<p>$$x^2-13x+42=0$$</p> |
<p>Let $$C(\alpha, \beta)$$ be the circumcenter of the triangle formed by the lines</p>
<p>$$4 x+3 y=69$$</p>
<p>$$4 y-3 x=17$$, and</p>
<p>$$x+7 y=61$$.</p>
<p>Then $$(\alpha-\beta)^{2}+\alpha+\beta$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "17"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "18"}] | ["B"]
Explanation:
We have,
<br><br>$$
\begin{aligned}
& 4 x+3 y=69 .......(i) \\\\
& 4 y-3 x=17 .......(ii) \\\\
& x+7 y=61 .......(iii)
\end{aligned}
$$
<br><br>On solving (i) and (iii), we get,
<br><br>$$
\begin{aligned}
& x=12, \text { and } y=7 \\\\
& \text { So, } A \equiv(12,7)
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ljg2jcjv/40b71be8-c597-4834-92a0-31af71c5e4d0/30e031b0-15e4-11ee-a5f8-af44f69476e0/file-6y3zli1ljg2jcjw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ljg2jcjv/40b71be8-c597-4834-92a0-31af71c5e4d0/30e031b0-15e4-11ee-a5f8-af44f69476e0/file-6y3zli1ljg2jcjw.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 8th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 26 English Explanation">
<br>On solving (ii) and (iii), we get,
<br><br>$$
\begin{aligned}
& x=5 \text { and } y=8 \\\\
& \text { So, B } \equiv(5,8)
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& \text { Hence, circumcentre } \equiv\left(\frac{12+5}{2}, \frac{7+8}{2}\right) \\\\
& \equiv\left(\frac{17}{2}, \frac{15}{2}\right)
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& \therefore \alpha=\frac{17}{2}, \beta=\frac{15}{2} \\\\
& \therefore(\alpha-\beta)^2+(\alpha+\beta)=\left(\frac{17}{2}-\frac{15}{2}\right)^2+\left(\frac{17}{2}+\frac{15}{2}\right) \\\\
& =(1)^2+(16)=17
\end{aligned}
$$ |
<p>If the orthocentre of the triangle formed by the lines $$2 x+3 y-1=0, x+2 y-1=0$$ and $$a x+b y-1=0$$, is the centroid of another triangle, whose circumcentre and orthocentre respectively are $$(3,4)$$ and $$(-6,-8)$$, then the value of $$|a-b|$$ is _________.</p>
Options:
[] | 16
Explanation:
<p>Let circumcentre, orthocentre and centroid of a triangle $$P Q R$$ are $$C_1, H_1$$ and $$G_1$$ respectively</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw8vj8jo/f80a9888-77a1-4003-a6a8-8ff4c790fc36/21421350-134e-11ef-8bbe-1b4949638519/file-1lw8vj8jp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw8vj8jo/f80a9888-77a1-4003-a6a8-8ff4c790fc36/21421350-134e-11ef-8bbe-1b4949638519/file-1lw8vj8jp.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 6 English Explanation 1"></p>
<p>$$\Rightarrow G_1 \equiv(0,0)$$ orthocentre of $$\triangle A B C$$ is $$(0,0)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw8vkatq/d9a6291d-037a-470f-95f4-e32e235ad67c/3ed3cee0-134e-11ef-8bbe-1b4949638519/file-1lw8vkatr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw8vkatq/d9a6291d-037a-470f-95f4-e32e235ad67c/3ed3cee0-134e-11ef-8bbe-1b4949638519/file-1lw8vkatr.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 6 English Explanation 2"></p>
<p>$$\begin{aligned}
& m_{A H_2}=+\frac{b}{a} \Rightarrow a+b=0 \\
& \text { eq }{ }^{\text {n }} \text { of lines } H_2 C \text { is } y=\frac{3}{2} x \\
& \Rightarrow \text { point } C \equiv\left(\frac{1}{4}, \frac{3}{8}\right) \text { lies on } a x+b y-1=0 \\
& \Rightarrow \frac{a}{4}+\frac{3}{8} b-1=0 \Rightarrow \frac{a}{4}-\frac{3}{8} a-1=0 \\
& \Rightarrow a=-8, b=8 \\
& |a-b|=16
\end{aligned}$$</p> |
A triangle with vertices $$\left( {4,0} \right),\left( { - 1, - 1} \right),\left( {3,5} \right)$$ is :
Options:
[{"identifier": "A", "content": "isosceles and right angled"}, {"identifier": "B", "content": "isosceles but not right angled"}, {"identifier": "C", "content": "right angled but not isosceles "}, {"identifier": "D", "content": "neither right angled nor isosceles "}] | ["A"]
Explanation:
$$AB = \sqrt {{{\left( {4 + 1} \right)}^2} + {{\left( {0 + 1} \right)}^2}} = \sqrt {26} ;$$
<br><br>$$BC = \sqrt {{{\left( {3 + 1} \right)}^2} + {{\left( {5 + 1} \right)}^2}} = \sqrt {52} $$
<br><br>$$CA = \sqrt {{{\left( {4 - 3} \right)}^2} + {{\left( {0 - 5} \right)}^2}} = \sqrt {26} ;$$
<br><br>In isosceles triangle side $$AB=CA$$
<br><br>For right angled triangle, $$B{C^2} = A{B^2} + A{C^2}$$
<br><br>So, here $$BC = \sqrt {52} $$ or $$B{C^2} = 52$$
<br><br>or $$\,\,\,\,\,\,\,\,\,{\left( {\sqrt {26} } \right)^2} + {\left( {\sqrt {26} } \right)^2} = 52$$
<br><br>So, the given triangle is right angled and also isosceles. |
A man starts walking from the point P($$-$$3, 4), touches the x-axis at R, and then turns to reach at the point Q(0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $$50\left( {{{(PR)}^2} + {{(RQ)}^2}} \right)$$ is equal to ____________.
Options:
[] | 1250
Explanation:
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwjjmzcw/b86a6d68-7b5e-434f-a89c-eef8431d96b0/715950f0-5074-11ec-8a49-997c456a58f9/file-1kwjjmzcx.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kwjjmzcw/b86a6d68-7b5e-434f-a89c-eef8431d96b0/715950f0-5074-11ec-8a49-997c456a58f9/file-1kwjjmzcx.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 1st September Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 56 English Explanation">
<br>For minimum $(P R+R Q)$
<br><br>$R$ lies on $P Q^{\prime}$ (where $Q^{\prime}$ is image of $Q$ in $X$-axis)
<br><br>$\Rightarrow$ Equation on $P Q^{\prime}$ is
<br><br>$$
2 x+y+2=0 \Rightarrow R(-1,0)
$$
<br><br>$$ \therefore $$ 50(PR<sup>2</sup> + RQ<sup>2</sup>)<br><br>= 50(20 + 5)<br><br>= 50(25)<br><br>= 1250 |
<p>The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle $${\pi \over 4}$$ at the origin, is equal to :</p>
Options:
[{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "$${48 \\over 5}$$"}, {"identifier": "C", "content": "$${52 \\over 5}$$"}, {"identifier": "D", "content": "3"}] | ["C"]
Explanation:
Let $A(\alpha, 2) \quad$ Given $B(2,3)$
<br/><br/>
$$
\begin{aligned}
& m_{O A}=\frac{2}{\alpha} \quad\&\quad m_{O B}=\frac{3}{2} \\\\
& \tan \frac{\pi}{4}=\left|\frac{\frac{2}{\alpha}-\frac{3}{2}}{1+\frac{2}{\alpha} \cdot \frac{3}{2}}\right| \Rightarrow \frac{4-3 \alpha}{2 \alpha+6}=\pm 1 \\\\
& 4-3 \alpha=2 \alpha+6 \quad \& 4-3 \alpha=-2 \alpha-6 \\\\
& \alpha=\frac{-2}{5} \& \alpha=10 \\\\
& A\left(-\frac{2}{5}, 2\right) \& A^{\prime}(10,2) \text { and } B(2,3) \\\\
& A A^{\prime}=10+\frac{2}{5}=\frac{52}{5}
\end{aligned}
$$ |
The shortest distance between the line $$y - x = 1$$ and the curve $$x = {y^2}$$ is :
Options:
[{"identifier": "A", "content": "$${{2\\sqrt 3 } \\over 8}$$ "}, {"identifier": "B", "content": "$${{3\\sqrt 2 } \\over 5}$$"}, {"identifier": "C", "content": "$${{\\sqrt 3 } \\over 4}$$ "}, {"identifier": "D", "content": "$${{3\\sqrt 2 } \\over 8}$$"}] | ["D"]
Explanation:
Let $$\left( {{a^2},a} \right)$$ be the point of shortest distance on $$x = {y^2}$$
<br><br>Then distance between $$\left( {{a^2},a} \right)$$ and line $$x - y + 1 = 0$$
<br><br>is given by
<br><br>$$\,\,\,\,\,\,\,\,D = {{{a^2} - a + 1} \over {\sqrt 2 }} = {1 \over {\sqrt 2 }}\left[ {{{\left( {a - {1 \over 2}} \right)}^2} + {3 \over 4}} \right]$$
<br><br>It is min when $$a = {1 \over 2}$$ and $$D{}_{\min } = {3 \over {4\sqrt 2 }} = {{3\sqrt 2 } \over 8}$$ |
The line $$L$$ given by $${x \over 5} + {y \over b} = 1$$ passes through the point $$\left( {13,32} \right)$$. The line K is parrallel to $$L$$ and has the equation $${x \over c} + {y \over 3} = 1.$$ Then the distance between $$L$$ and $$K$$ is :
Options:
[{"identifier": "A", "content": "$$\\sqrt {17} $$ "}, {"identifier": "B", "content": "$${{17} \\over {\\sqrt {15} }}$$ "}, {"identifier": "C", "content": "$${{23} \\over {\\sqrt {17} }}$$ "}, {"identifier": "D", "content": "$${{23} \\over {\\sqrt {15} }}$$ "}] | ["C"]
Explanation:
Slope of line $$L = - {b \over 5}$$
<br><br>Slope of line $$K = - {3 \over c}$$
<br><br>Line $$L$$ is parallel to line $$k.$$
<br><br>$$ \Rightarrow {b \over 5} = {3 \over c} \Rightarrow bc = 15$$
<br><br>$$(13,32)$$ is a point on $$L.$$
<br><br>$$\therefore$$ $${{13} \over 5} + {{32} \over b} = 1 \Rightarrow {{32} \over b} = - {8 \over 5}$$
<br><br>$$ \Rightarrow b = - 20 \Rightarrow c = - {3 \over 4}$$
<br><br>Equation of $$K:$$ $$y - 4x = 3$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,$$ $$ \Rightarrow 4x - y + 3 = 0$$
<br><br>Distance between $$L$$ and $$K$$
<br><br>$$ = {{\left| {52 - 32 + 3} \right|} \over {\sqrt {17} }} = {{23} \over {\sqrt {17} }}$$ |
The foot of the perpendicular drawn from the origin, on the line, 3x + y = $$\lambda $$ ($$\lambda $$ $$ \ne $$ 0) is P. If the line meets x-axis at A and y-axis at B, then the ratio BP : PA is :
Options:
[{"identifier": "A", "content": "1 : 3"}, {"identifier": "B", "content": "3 : 1"}, {"identifier": "C", "content": "1 : 9"}, {"identifier": "D", "content": "9 : 1"}] | ["D"]
Explanation:
Equation of the line, which is perpendicular to the line,
<br><br>3x + y = $$\lambda $$($$\lambda $$ $$ \ne $$0) and passing through origin ,
<br><br>is given by $${{x - 0} \over 3} = {{y - 0} \over 1} = r$$
<br><br>For foot of perpendicular
<br><br>r = $${{ - \left( {\left( {3 \times 0} \right) + \left( {1 \times 0} \right) - \lambda } \right)} \over {{3^2} + {1^2}}}$$ = $${\lambda \over {10}}$$
<br><br>So, foot of perpendicular P = $$\left( {{{3\lambda } \over {10}},{\lambda \over {10}}} \right)$$
<br><br> Given the line meets X-axis where y = 0, so 3x + 0 = $$\lambda $$
<br><br>$$ \Rightarrow $$ x = $${\lambda \over 3}$$
<br><br>Hence, coordinates of A = $$\left( {{\lambda \over 3},0} \right)$$ and meets
<br><br>Y-axis at B = (0, $$\lambda $$)
<br><br>So, BP = $$\sqrt {{{\left( {{{3\lambda } \over {10}}} \right)}^2} + {{\left( {{\lambda \over {10}} - \lambda } \right)}^2}} $$
<br><br>$$ \Rightarrow $$ BP = $$\sqrt {{{9{\lambda ^2}} \over {100}} + {{81{\lambda ^2}} \over {100}}} $$
<br><br>= BP = $$\sqrt {{{90{\lambda ^2}} \over {100}}} $$
<br><br>Now, PA = $$\sqrt {{{\left( {{\lambda \over 3} - {{3\lambda } \over {10}}} \right)}^2} + {{\left( {0 - {\lambda \over {10}}} \right)}^2}} $$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ PA = $$\sqrt {{{{\lambda ^2}} \over {900}} + {{{\lambda ^2}} \over {100}}} \Rightarrow PA$$ = $$\sqrt {{{10{\lambda ^2}} \over {900}}} $$
<br><br>Therefore BP : PA = 9 : 1
<br><br> |
Lines are drawn parallel to the line 4x – 3y + 2 = 0, at a distance
$${3 \over 5}$$
from the origin. Then which one of the
following points lies on any of these lines ?
Options:
[{"identifier": "A", "content": "$$\\left( {{1 \\over 4}, - {1 \\over 3}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - {1 \\over 4},{2 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over 4}, - {2 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 4},{1 \\over 3}} \\right)$$"}] | ["B"]
Explanation:
Line parallel to 4x – 3y + 2 = 0<br><br>
is given as 4x – 3y + $$\lambda $$ = 0<br><br>
distance from origin is <br><br>
$$\left| {{\lambda \over 5}} \right| = {3 \over 5}$$<br><br>
$$ \Rightarrow \lambda = \pm 3$$<br><br>
$$ \therefore $$ required lines are 4x – 3y + 3 = 0 & 4x – 3y – 3 = 0<br><br>
By Putting $$\left( { - {1 \over 4},{2 \over 3}} \right)$$ on both lines it satisfy. |
If the line, 2x - y + 3 = 0 is at a distance<br/> $${1 \over {\sqrt 5 }}$$
and $${2 \over {\sqrt 5 }}$$ from the lines 4x - 2y + $$\alpha $$ = 0 <br/>and 6x - 3y + $$\beta $$ = 0, respectively, then the sum of all possible values of $$\alpha $$ and $$\beta $$ is :
Options:
[] | 30
Explanation:
Apply distance between parallel line formula<br><br>$$4x - 2y + \alpha = 0$$<br><br>$$4x - 2y + 6 = 0$$<br><br>$$\left| {{{\alpha - 6} \over {25}}} \right| = {1 \over {55}}$$<br><br>$$|\alpha - 6|\, = 2 \Rightarrow \alpha = 8,4$$<br><br>sum = 12<br><br>Again <br><br>$$6x - 3y + \beta = 0$$<br><br>$$6x - 3y + 9 = 0$$<br><br>$$\left| {{{\beta - 9} \over {3\sqrt 5 }}} \right| = {2 \over {\sqrt 5 }}$$<br><br>$$|\beta - 9|\, = 6 \Rightarrow \beta = 15,3$$<br><br>sum = 18<br><br>$$ \therefore $$ Sum of all values of $$\alpha$$ and $$\beta$$ is = 30 |
If p and q are the lengths of the perpendiculars from the origin on the lines,<br/><br/>x cosec $$\alpha$$ $$-$$ y sec $$\alpha$$ = k cot 2$$\alpha$$ and<br/><br/>x sin$$\alpha$$ + y cos$$\alpha$$ = k sin2$$\alpha$$<br/><br/>respectively, then k<sup>2</sup> is equal to :
Options:
[{"identifier": "A", "content": "4p<sup>2</sup> + q<sup>2</sup>"}, {"identifier": "B", "content": "2p<sup>2</sup> + q<sup>2</sup>"}, {"identifier": "C", "content": "p<sup>2</sup> + 2q<sup>2</sup>"}, {"identifier": "D", "content": "p<sup>2</sup> + 4q<sup>2</sup>"}] | ["A"]
Explanation:
First line is $${x \over {\sin \alpha }} - {y \over {\cos \alpha }} = {{k\cos 2\alpha } \over {\sin 2\alpha }}$$<br><br>$$ \Rightarrow x\cos \alpha - y\sin \alpha = {k \over 2}\cos 2\alpha $$ <br><br>$$ \Rightarrow p = \left| {{k \over 2}\cos \alpha } \right| \Rightarrow 2p = \left| {k\cos 2\alpha } \right|$$ .... (i)<br><br>second line is $$x\sin \alpha + y\cos \alpha = k\sin 2\alpha $$<br><br>$$ \Rightarrow q = \left| {k\sin 2\alpha } \right|$$ .... (ii)<br><br>Hence, $$4{p^2} + {q^2} = {k^2}$$ (From (i) & (ii)) |
<p>Let the point $$P(\alpha, \beta)$$ be at a unit distance from each of the two lines $$L_{1}: 3 x-4 y+12=0$$, and $$L_{2}: 8 x+6 y+11=0$$. If $$P$$ lies below $$L_{1}$$ and above $${ }{L_{2}}$$, then $$100(\alpha+\beta)$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "$$-$$14"}, {"identifier": "B", "content": "42"}, {"identifier": "C", "content": "$$-$$22"}, {"identifier": "D", "content": "14"}] | ["D"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7bukhhd/39a595d1-0f44-4de9-a640-89709cea02f6/4c50df10-25ff-11ed-9c74-c5a04899a045/file-1l7bukhhe.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7bukhhd/39a595d1-0f44-4de9-a640-89709cea02f6/4c50df10-25ff-11ed-9c74-c5a04899a045/file-1l7bukhhe.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 44 English Explanation"></p>
<p>$${L_1}:3x - 4y + 12 = 0$$</p>
<p>$${L_2}:8x + 6y + 11 = 0$$</p>
<p>Equation of angle bisector of L<sub>1</sub> and L<sub>2</sub> of angle containing origin</p>
<p>$$2(3x - 4y + 12) = 8x + 6y + 11$$</p>
<p>$$2x + 14y - 13 = 0$$ ...... (i)</p>
<p>$${{3\alpha - 4\beta + 12} \over 5} = 1$$</p>
<p>$$ \Rightarrow 3\alpha - 4\beta + 7 = 0$$ ...... (ii)</p>
<p>Solution of $$2x + 14y - 13 = 0$$ and $$3x - 4y + 7 = 0$$ gives the required point $$P(\alpha ,\beta ),\,\alpha = {{ - 23} \over {25}},\beta = {{53} \over {50}}$$</p>
<p>$$100(\alpha + \beta ) = 14$$</p> |
<p>Let the equations of two adjacent sides of a parallelogram $$\mathrm{ABCD}$$ be $$2 x-3 y=-23$$ and $$5 x+4 y=23$$. If the equation of its one diagonal $$\mathrm{AC}$$ is $$3 x+7 y=23$$ and the distance of A from the other diagonal is $$\mathrm{d}$$, then $$50 \mathrm{~d}^{2}$$ is equal to ____________.</p>
Options:
[] | 529
Explanation:
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkfs1i7/86b3f2b6-ecef-40f3-8533-8213186f85af/f96ad2f0-677b-11ee-aeb9-a1910acfd49b/file-6y3zli1lnkfs1i8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnkfs1i7/86b3f2b6-ecef-40f3-8533-8213186f85af/f96ad2f0-677b-11ee-aeb9-a1910acfd49b/file-6y3zli1lnkfs1i8.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 27 English Explanation">
<br><br>We have, $A B C D$ is a parallelogram
<br><br>Let equation of $A B$ be $2 x-3 y=-23 \ldots$ (i)
<br><br>and equation of $B C$ be $5 x+4 y=23\ldots$ (ii)
<br><br>Equation of $A C$ is $3 x+7 y=23 \ldots$ (iii)
<br><br>Solving Eqs. (i) and (ii), we get
<br><br>$x=-1$, and $y=7$
<br><br>$\therefore$ Co-ordinate of $B$ is $(-1,7)$
<br><br>On solving Eqs. (ii) and (iii), we get
<br><br>$$
x=3, y=2
$$
<br><br>$\therefore$ Co-ordinate of $C$ is $(3,2)$
<br><br>On solving Eqs. (i) and (iii), we get $x=-4$ and $y=5$
<br><br>$\therefore$ Co-ordinate of $A$ is $(-4,5)$.
<br><br>Let $E$ be the intersection point of diagonal co-ordinate of
<br><br>$E$ is $\left(\frac{-4+3}{2}, \frac{5+2}{2}\right)$ or $\left(-\frac{1}{2}, \frac{7}{2}\right)$
<br><br>$\because E$ is mid-point of $A C$
<br><br>$$
\begin{aligned}
& \text { Equation of } B D \text { is } y-7=\left(\frac{7-\frac{7}{2}}{-1+\frac{1}{2}}\right)(x+1) \\\\
& \Rightarrow 7 x+y=0
\end{aligned}
$$
<br><br>Distance of $A$ from diagonal $B D=\frac{|7 \times(-4)+5|}{\sqrt{7^2+1^2}}$
<br><br>$$
\therefore d=\frac{23}{\sqrt{50}}
$$
<br><br>Hence, $50 d^2=(23)^2=529$ |
<p>If the sum of squares of all real values of $$\alpha$$, for which the lines $$2 x-y+3=0,6 x+3 y+1=0$$ and $$\alpha x+2 y-2=0$$ do not form a triangle is $$p$$, then the greatest integer less than or equal to $$p$$ is _________.</p>
Options:
[] | 32
Explanation:
<p>$$\begin{aligned}
& 2 x-y+3=0 \\
& 6 x+3 y+1=0 \\
& \alpha x+2 y-2=0
\end{aligned}$$</p>
<p>Will not form a $$\Delta$$ if $$\alpha x+2 y-2=0$$ is concurrent with $$2 x-y+3=0$$ and $$6 x+3 y+1=0$$ or parallel to either of them so</p>
<p>Case-1: Concurrent lines</p>
<p>$$\left|\begin{array}{ccc}
2 & -1 & 3 \\
6 & 3 & 1 \\
\alpha & 2 & -2
\end{array}\right|=0 \Rightarrow \alpha=\frac{4}{5}$$</p>
<p>Case-2 : Parallel lines</p>
<p>$$\begin{aligned}
& -\frac{\alpha}{2}=\frac{-6}{3} \text { or }-\frac{\alpha}{2}=2 \\
& \Rightarrow \alpha=4 \text { or } \alpha=-4 \\
& P=16+16+\frac{16}{25} \\
& {[P]=\left[32+\frac{16}{25}\right]=32}
\end{aligned}$$</p> |
<p>Let $$A(a, b), B(3,4)$$ and $$C(-6,-8)$$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $$P(2 a+3,7 b+5)$$ from the line $$2 x+3 y-4=0$$ measured parallel to the line $$x-2 y-1=0$$ is</p>
Options:
[{"identifier": "A", "content": "$$\\frac{17 \\sqrt{5}}{6}$$\n"}, {"identifier": "B", "content": "$$\\frac{15 \\sqrt{5}}{7}$$\n"}, {"identifier": "C", "content": "$$\\frac{17 \\sqrt{5}}{7}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\sqrt{5}}{17}$$"}] | ["C"]
Explanation:
<p>$$\mathrm{A}(\mathrm{a}, \mathrm{b}), \quad \mathrm{B}(3,4), \quad \mathrm{C}(-6,-8)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwjvlw/dec041c9-e406-4f98-a94d-783370c1f55d/ddd44840-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwjvlx.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsjwjvlw/dec041c9-e406-4f98-a94d-783370c1f55d/ddd44840-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwjvlx.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 31st January Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 19 English Explanation"></p>
<p>$$\Rightarrow \mathrm{a}=0, \mathrm{~b}=0 \quad \Rightarrow \mathrm{P}(3,5)$$</p>
<p>Distance from $$\mathrm{P}$$ measured along $$\mathrm{x}-2 \mathrm{y}-1=0$$</p>
<p>$$\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta$$</p>
<p>$$\begin{aligned}
& \text { Where } \tan \theta=\frac{1}{2} \\
& \mathrm{r}(2 \cos \theta+3 \sin \theta)=-17 \\
& \Rightarrow \mathrm{r}=\left|\frac{-17 \sqrt{5}}{7}\right|=\frac{17 \sqrt{5}}{7}
\end{aligned}$$</p> |
<p>Let $$\mathrm{A}$$ be the point of intersection of the lines $$3 x+2 y=14,5 x-y=6$$ and $$\mathrm{B}$$ be the point of intersection of the lines $$4 x+3 y=8,6 x+y=5$$. The distance of the point $$P(5,-2)$$ from the line $$\mathrm{AB}$$ is</p>
Options:
[{"identifier": "A", "content": "$$\\frac{13}{2}$$"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "$$\\frac{5}{2}$$"}, {"identifier": "D", "content": "6"}] | ["D"]
Explanation:
<p>Solving lines $$\mathrm{L}_1(3 \mathrm{x}+2 \mathrm{y}=14)$$ and $$\mathrm{L}_2(5 \mathrm{x}-\mathrm{y}=6)$$ to get $$\mathrm{A}(2,4)$$ and solving lines $$\mathrm{L}_3(4 \mathrm{x}+3 \mathrm{y}=8)$$ and $$\mathrm{L}_4(6 \mathrm{x}+\mathrm{y}=5)$$ to get $$\mathrm{B}\left(\frac{1}{2}, 2\right)$$.</p>
<p>Finding Eqn. of $$\mathrm{AB}: 4 \mathrm{x}-3 \mathrm{y}+4=0$$</p>
<p>Calculate distance PM</p>
<p>$$\Rightarrow\left|\frac{4(5)-3(-2)+4}{5}\right|=6$$</p> |
<p>The distance of the point $$(2,3)$$ from the line $$2 x-3 y+28=0$$, measured parallel to the line $$\sqrt{3} x-y+1=0$$, is equal to</p>
Options:
[{"identifier": "A", "content": "$$3+4 \\sqrt{2}$$\n"}, {"identifier": "B", "content": "$$6 \\sqrt{3}$$\n"}, {"identifier": "C", "content": "$$4+6 \\sqrt{3}$$\n"}, {"identifier": "D", "content": "$$4 \\sqrt{2}$$"}] | ["C"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8s48w/7106c58a-521f-4564-b530-e2a9d2152dfd/2d124800-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8s48x.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr8s48w/7106c58a-521f-4564-b530-e2a9d2152dfd/2d124800-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8s48x.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 14 English Explanation"></p>
<p>Writing $$P$$ in terms of parametric co-ordinates $$2+r$$</p>
<p>$$\begin{aligned}
& \cos \theta, 3+\mathrm{r} \sin \theta \text { as } \tan \theta=\sqrt{3} \\
& \mathrm{P}\left(2+\frac{\mathrm{r}}{2}, 3+\frac{\sqrt{3} \mathrm{r}}{2}\right)
\end{aligned}$$</p>
<p>$$\mathrm{P}$$ must satisfy $$2 \mathrm{x}-3 \mathrm{y}+28=0$$</p>
<p>So, $$2\left(2+\frac{r}{2}\right)-3\left(3+\frac{\sqrt{3} \mathrm{r}}{2}\right)+28=0$$</p>
<p>We find $$r=4+6 \sqrt{3}$$</p> |
<p>If $$x^2-y^2+2 h x y+2 g x+2 f y+c=0$$ is the locus of a point, which moves such that it is always equidistant from the lines $$x+2 y+7=0$$ and $$2 x-y+8=0$$, then the value of $$g+c+h-f$$ equals</p>
Options:
[{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "29"}, {"identifier": "D", "content": "6"}] | ["B"]
Explanation:
<p>Cocus of point $$\mathrm{P}(\mathrm{x}, \mathrm{y})$$ whose distance from Gives
$$X+2 y+7=0$$ & $$2 x-y+8=0$$ are equal is $$\frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}$$</p>
<p>$$(x+2 y+7)^2-(2 x-y+8)^2=0$$</p>
<p>Combined equation of lines</p>
<p>$$\begin{aligned}
& (x-3 y+1)(3 x+y+15)=0 \\
& 3 x^2-3 y^2-8 x y+18 x-44 y+15=0 \\
& x^2-y^2-\frac{8}{3} x y+6 x-\frac{44}{3} y+5=0 \\
& x^2-y^2+2 h x y+2 g x 2+2 f y+c=0 \\
& h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5 \\
& g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14
\end{aligned}$$</p> |
<p>The vertices of a triangle are $$\mathrm{A}(-1,3), \mathrm{B}(-2,2)$$ and $$\mathrm{C}(3,-1)$$. A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :</p>
Options:
[{"identifier": "A", "content": "$$-x+y-(2-\\sqrt{2})=0$$\n"}, {"identifier": "B", "content": "$$x+y-(2-\\sqrt{2})=0$$\n"}, {"identifier": "C", "content": "$$x+y+(2-\\sqrt{2})=0$$\n"}, {"identifier": "D", "content": "$$x-y-(2+\\sqrt{2})=0$$"}] | ["B"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwk4845t/9c9451b9-d46c-4c1c-ab08-287bac97c8b0/bd468810-197c-11ef-8e15-2107d8c35500/file-1lwk4845u.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwk4845t/9c9451b9-d46c-4c1c-ab08-287bac97c8b0/bd468810-197c-11ef-8e15-2107d8c35500/file-1lwk4845u.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 9 English Explanation"></p>
<p>Equation of $$A C: x+y=2$$</p>
<p>Equation of $$A B: x-y+4=0$$</p>
<p>Equation of $$B C: 3 x+5 y=4$$</p>
<p>The line nearest to origin is parallel to $$A C$$ and inward. Let its equation is $$x+y=C$$.</p>
<p>$$\therefore\left|\frac{C-2}{\sqrt{2}}\right|=1$$</p>
<p>$$\therefore \quad C=2-\sqrt{2}$$</p>
<p>$$\therefore$$ required equation line is :</p>
<p>$$x+y-(2-\sqrt{2})=0$$</p> |
Consider the set of all lines px + qy + r = 0 such that 3p + 2q + 4r = 0. Which one of the following statements
is true?
Options:
[{"identifier": "A", "content": "The lines are not concurrent"}, {"identifier": "B", "content": "The lines are concurrent at the point $$\\left( {{3 \\over 4},{1 \\over 2}} \\right)$$"}, {"identifier": "C", "content": "The lines are all parallel "}, {"identifier": "D", "content": "Each line passes through the origin"}] | ["B"]
Explanation:
Equation of lines;
<br><br>px + qy + r = 0 . . . . . (1)
<br><br>Also given
<br><br>3p + 2q + 4r = 0 . . . . . . (2)
<br><br>divide equation (2) by 4, we get
<br><br>$${3 \over 4}P + {2 \over 4}q + r = 0$$ . . . . (3)
<br><br>By comparing (1) and (3) we get,
<br><br>x = $${3 \over 4}$$ and y = $${2 \over 4}$$ = $${1 \over 2}$$
<br><br>For any value of p,q and r, the equation of set of lines will pan through $$\left( {{3 \over 4},{1 \over 2}} \right)$$ |
Locus of mid point of the portion between the axes of
<br/><br/>$$x$$ $$cos$$ $$\alpha + y\,\sin \alpha = p$$ where $$p$$ is constant is :
Options:
[{"identifier": "A", "content": "$${x^2} + {y^2} = {4 \\over {{p^2}}}$$ "}, {"identifier": "B", "content": "$${x^2} + {y^2} = 4{p^2}$$ "}, {"identifier": "C", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} = {2 \\over {{p^2}}}$$ "}, {"identifier": "D", "content": "$${1 \\over {{x^2}}} + {1 \\over {{y^2}}} = {4 \\over {{p^2}}}$$ "}] | ["D"]
Explanation:
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264686/exam_images/ug1vcgctwnpzuwlv3nqn.webp" loading="lazy" alt="AIEEE 2002 Mathematics - Straight Lines and Pair of Straight Lines Question 116 English Explanation">
<br><br>Equation of $$AB$$ is
<br><br>$$x\cos \alpha + y\sin \alpha = p;$$
<br><br>$$ \Rightarrow {{x\cos \alpha } \over p} + {{y\sin \alpha } \over p} = 1;$$
<br><br>$$ \Rightarrow {x \over {p/\cos \alpha }} + {y \over {p/\sin \alpha }} = 1$$
<br><br>So co-ordinates of $$A$$ and $$B$$ are
<br><br>$$\left( {{p \over {\cos \alpha }},0} \right)$$ and $$\left( {0,{p \over {\sin \alpha }}} \right);$$
<br><br>So coordinates of midpoint of $$AB$$ are
<br><br>$$\left( {{p \over {2\cos \,\alpha }},{p \over {2\sin \alpha }}} \right) = \left( {{x_1},{y_1}} \right)\left( {let} \right);$$
<br><br>$${x_1} = {p \over {2\,\cos \,\alpha }}\,\,\& \,\,{y_1} = {p \over {2\sin \alpha }};$$
<br><br>$$ \Rightarrow \cos \alpha = p/2{x_1}$$ and $$\sin \alpha = p/2{y_1};$$
<br><br>$${\cos ^2}\alpha + {\sin ^2}\alpha = 1 \Rightarrow {{{p^2}} \over 4}\left( {{1 \over {{x_1}^2}} + {1 \over {{y_1}^2}}} \right) = 1$$
<br><br>Locus of $$\left( {{x_1},{y_1}} \right)$$ is $${1 \over {{x^2}}} + {1 \over {{y^2}}} = {4 \over {{p^2}}}.$$ |
Locus of centroid of the triangle whose vertices are $$\left( {a\cos t,a\sin t} \right),\left( {b\sin t, - b\cos t} \right)$$ and $$\left( {1,0} \right),$$ where $$t$$ is a parameter, is :
Options:
[{"identifier": "A", "content": "$${\\left( {3x + 1} \\right)^2} + {\\left( {3y} \\right)^2} = {a^2} - {b^2}$$ "}, {"identifier": "B", "content": "$${\\left( {3x - 1} \\right)^2} + {\\left( {3y} \\right)^2} = {a^2} - {b^2}$$"}, {"identifier": "C", "content": "$${\\left( {3x - 1} \\right)^2} + {\\left( {3y} \\right)^2} = {a^2} + {b^2}$$"}, {"identifier": "D", "content": "$${\\left( {3x + 1} \\right)^2} + {\\left( {3y} \\right)^2} = {a^2} + {b^2}$$"}] | ["C"]
Explanation:
$$x = {{a\cos t + b\sin t + 1} \over 3}$$
<br><br>$$ \Rightarrow a\cos t + b\sin t = 3x - 1$$
<br><br>$$y = {{a\sin t - b\cos t} \over 3}$$
<br><br>$$ \Rightarrow a\sin t - b\cos t = 3y$$ |
If the equation of the locus of a point equidistant from the point $$\left( {{a_{1,}}{b_1}} \right)$$ and $$\left( {{a_{2,}}{b_2}} \right)$$ is
<br/>$$\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0$$ , then the value of $$'c'$$ is :
Options:
[{"identifier": "A", "content": "$$\\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2} $$ "}, {"identifier": "B", "content": "$${1 \\over 2}\\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \\right)$$ "}, {"identifier": "C", "content": "$${{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}$$ "}, {"identifier": "D", "content": "$${1 \\over 2}\\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \\right)$$."}] | ["B"]
Explanation:
Since, the points $\left(a_1, b_1\right)$ and $\left(a_2, b_2\right)$ satisfy the equation. So, that
<br/><br/>$$
\begin{aligned}
& a_1\left(a_1-a_2\right)+b_1\left(b_1-b_2\right)+c=0 ~~........(1) \\\\
& \text { and } a_2\left(a_1-a_2\right)+b_2\left(b_1-b_2\right)+c=0 ~~.........(2) \\\\
& \text { On adding Eqs. (i) and (ii), we get } \\\\
& \left(a_1+a_2\right)\left(a_1-a_2\right)+\left(b_1+b_2\right)\left(b_1-b_2\right)+2 c=0 \\\\
& \Rightarrow 2 c=-\left(a_1^2-a_2^2+b_1^2-b_2^2\right) \\\\
& \Rightarrow c=\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right)
\end{aligned}
$$ |
Let $$A\left( {2, - 3} \right)$$ and $$B\left( {-2, 1} \right)$$ be vertices of a triangle $$ABC$$. If the centroid of this triangle moves on the line $$2x + 3y = 1$$, then the locus of the vertex $$C$$ is the line :
Options:
[{"identifier": "A", "content": "$$3x - 2y = 3$$"}, {"identifier": "B", "content": "$$2x - 3y = 7$$"}, {"identifier": "C", "content": "$$3x + 2y = 5$$"}, {"identifier": "D", "content": "$$2x + 3y = 9$$"}] | ["D"]
Explanation:
Let the vertex $$C$$ be $$(h,k),$$ then the
<br><br>centroid of $$\Delta ABC$$ is $$\left( {{{2 + (- 2) + h} \over 3},{{ - 3 + 1 + k} \over 3}} \right)$$
<br><br>or $$\left( {{h \over 3},{{ - 2 + k} \over 3}} \right).$$ It lies on $$2x+3y=1$$
<br><br>$$ \Rightarrow {{2h} \over 3} - 2 + k = 1$$
<br><br>$$ \Rightarrow 2h + 3k = 9$$
<br><br>$$ \therefore $$ Locus of $$C$$ is $$2x+3y=9$$ |
If a variable line drawn through the intersection of the lines $${x \over 3} + {y \over 4} = 1$$ and $${x \over 4} + {y \over 3} = 1,$$ meets the coordinate axes at A and B, (A $$ \ne $$ B), then the locus of the midpoint of AB is :
Options:
[{"identifier": "A", "content": "6xy = 7(x + y)"}, {"identifier": "B", "content": "4(x + y)<sup>2</sup> \u2212 28(x + y) + 49 = 0"}, {"identifier": "C", "content": "7xy = 6(x + y)\n"}, {"identifier": "D", "content": "14(x + y)<sup>2</sup> \u2212 97(x + y) + 168 = 0"}] | ["C"]
Explanation:
L<sub>1</sub> : 4x + 3y $$-$$ 12 = 0
<br><br>L<sub>2</sub> : 3x + 4y $$-$$ 12 = 0
<br><br>Equation of line passing through the intersection of these two lines L<sub>1</sub> and L<sub>2</sub> is
<br><br>L<sub>1</sub> + $$\lambda $$L<sub>2</sub> = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$(4x + 3y $$-$$ 12) + $$\lambda $$(3x + 4y $$-$$ 12) = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x(4 + 3$$\lambda $$) + y(3 + 4$$\lambda $$) $$-$$ 12(1 + $$\lambda $$) = 0
<br><br>this line meets x coordinate at point A and y coordinate at point B.
<br><br>$$\therefore\,\,\,$$ Point A = $$\left( {{{12\left( {1 + \lambda } \right)} \over {4 + 3\lambda }},0} \right)$$
<br><br>and Point B = $$\left( {0,\,\,{{12\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}} \right)$$
<br><br>Let coordinate of midpoint of line AB is (h, k).
<br><br>$$\therefore\,\,\,$$ h = $${{6\left( {1 + \lambda } \right)} \over {4 + 3\lambda }}$$ . . . . . (1)
<br><br>and k = $${{6\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}$$ . . . . (2)
<br><br>Eliminate $$\lambda $$ from (1) and (2), then we get
<br><br>6(h + k) = 7 hk
<br><br>$$\therefore\,\,\,$$ Locus of midpoint of line AB is ,
<br><br>6(x + y) = 7xy |
A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is
the origin and the rectangle OPRQ is completed, then the locus of R is :
Options:
[{"identifier": "A", "content": "3x + 2y = 6xy"}, {"identifier": "B", "content": "3x + 2y = 6"}, {"identifier": "C", "content": "2x + 3y = xy"}, {"identifier": "D", "content": "3x + 2y = xy"}] | ["D"]
Explanation:
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265351/exam_images/bpha9tfo2ojwynaftia1.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Straight Lines and Pair of Straight Lines Question 119 English Explanation">
<br><br>Let coordinate of point R = (h, k).
<br><br>Equation of line PQ,
<br><br>(y $$-$$ 3) = m (x $$-$$ 2).
<br><br>Put y = 0 to get coordinate of point p,
<br><br>0 $$-$$ 3 = (x $$-$$ 2)
<br><br>$$ \Rightarrow $$ x = 2 $$-$$ $${3 \over m}$$
<br><br>$$\therefore\,\,\,$$ p = (2 $$-$$ $${3 \over m}$$, 0)
<br><br>As p = (h, 0) then
<br><br>h = 2 $$-$$ $${3 \over m}$$
<br><br>$$ \Rightarrow $$ $${3 \over m}$$ = 2 $$-$$ h
<br><br>$$ \Rightarrow $$ m = $${3 \over {2 - h}}$$ . . . . . . (1)
<br><br>Put x = 0 to get coordinate of point Q,
<br><br>y $$-$$ 3 $$=$$ m (0 $$-$$ 2)
<br><br> $$ \Rightarrow $$ y = 3 $$-$$ 2m
<br><br>$$\therefore\,\,\,$$ point Q = (0, 3 $$-$$ 2m)
<br><br>And From the graph you can see Q = (0, k).
<br><br>$$\therefore\,\,\,$$ k = 3 $$-$$ 2m
<br><br>$$ \Rightarrow $$ m = $${{3 - k} \over 2}$$ . . . . (2)
<br><br>By comparing (1) and (2) get
<br><br>$${3 \over {2 - h}} = {{3 - k} \over 2}$$
<br><br>$$ \Rightarrow $$ (2 $$-$$ h)(3 $$-$$ k) = 6
<br><br>$$ \Rightarrow $$ 6 $$-$$ 3h $$-$$ 2K + hk = 6
<br><br>$$ \Rightarrow $$ 3 h + 2K = hk
<br><br>$$\therefore\,\,\,$$ locus of point R is 3x + 2y= xy |
Let O(0, 0) and A(0, 1) be two fixed points. Then
the locus of a point P such that the perimeter of
$$\Delta $$AOP is 4, is :
Options:
[{"identifier": "A", "content": "9x<sup>2</sup> + 8y<sup>2</sup> \u2013 8y = 16\n"}, {"identifier": "B", "content": "8x<sup>2</sup> \u2013 9y<sup>2</sup> + 9y = 18"}, {"identifier": "C", "content": "8x<sup>2</sup> + 9y<sup>2</sup> \u2013 9y = 18\n"}, {"identifier": "D", "content": "9x<sup>2</sup> \u2013 8y<sup>2</sup> + 8y = 16"}] | ["A"]
Explanation:
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267479/exam_images/wjrxugubrcihnruos7c4.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265282/exam_images/uonftnxl941nxalen4tg.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 94 English Explanation"></picture>
<br>Let point C(h, k)
<br><br>Given, AB + BC + AC = 4
<br><br>From graph, AB = 1
<br><br>$$ \therefore $$ BC + AC = 3
<br><br>$$ \Rightarrow $$ $$\sqrt {{h^2} + {k^2}} + \sqrt {{h^2} + {{\left( {k - 1} \right)}^2}} = 3$$
<br><br>$$ \Rightarrow $$ $${{h^2} + {{\left( {k - 1} \right)}^2}}$$ = 9 + $${{h^2} + {k^2}}$$ - $$6\sqrt {{h^2} + {k^2}} $$
<br><br>$$ \Rightarrow $$ $$6\sqrt {{h^2} + {k^2}} = 2k + 8$$
<br><br>$$ \Rightarrow $$ $$9\left( {{h^2} + {k^2}} \right) = {k^2} + 8k + 16$$
<br><br>$$ \Rightarrow $$ $$9{h^2} + 8{k^2} - 8k - 16 = 0$$
<br><br>$$ \therefore $$ Locus of point C will be,
<br><br>9x<sup>2</sup> + 8y<sup>2</sup> – 8y = 16 |
The locus of the mid-points of the perpendiculars drawn from points on the line, x = 2y to the line
x = y is :
Options:
[{"identifier": "A", "content": "3x - 2y = 0"}, {"identifier": "B", "content": "7x - 5y = 0\n"}, {"identifier": "C", "content": "2x - 3y = 0"}, {"identifier": "D", "content": "5x - 7y = 0"}] | ["D"]
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263683/exam_images/w8qnmvdmubr7jibhtirj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Evening Slot Mathematics - Straight Lines and Pair of Straight Lines Question 85 English Explanation">
<br><br>Slope of line y = x is 1
<br><br>Line AB is perpendicular to line y = x so
<br><br>Slope of AB = -1
<br><br>Also slope of AB = $${{\alpha - \beta } \over {2\alpha - \beta }}$$
<br><br>$$ \therefore $$ $${{\alpha - \beta } \over {2\alpha - \beta }}$$ = -1
<br><br>$$ \Rightarrow $$ 3$$\alpha $$ = 2$$\beta $$
<br><br>h = $${{2\alpha + \beta } \over 2}$$
<br><br>$$ \Rightarrow $$ 2h = $${{4\alpha + 2\beta } \over 2}$$ = $${{4\alpha + 3\alpha } \over 2}$$ = $${{7\alpha } \over 2}$$
<br><br>Also k = $${{\alpha + \beta } \over 2}$$
<br><br>$$ \Rightarrow $$ 2k = $${{5\alpha } \over 2}$$
<br><br>So $${h \over k} = {7 \over 5}$$
<br><br>$$ \Rightarrow $$ 5h = 7k
<br><br>$$ \Rightarrow $$ 5x = 7y |
A square ABCD has all its vertices on the curve x<sup>2</sup>y<sup>2</sup> = 1. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is _________.
Options:
[] | 80
Explanation:
x<sup>2</sup>y<sup>2</sup> = 1
<br><br>$$ \Rightarrow $$ y<sup>2</sup> = $${1 \over {{x^2}}}$$
<br><br>$$ \Rightarrow $$ y = $$ \pm {1 \over x}$$
<br><br>Graph of this equation,
<br><br>$$OA \bot OB$$<br><br>$$ \Rightarrow \left( {{1 \over {{p^2}}}} \right)\left( { - {1 \over {{q^2}}}} \right) = - 1$$<br><br>$$ \Rightarrow {p^2}{q^2} = 1$$<br><br>$$P\left( {{{p + q} \over 2},{{{1 \over p} - {1 \over q}} \over 2}} \right)$$ midpoint of AB lies<br><br>On $${x^2}{y^2} = 1$$<br><br>$$ \Rightarrow {(p + q)^2}{\left( {{1 \over p} - {1 \over q}} \right)^2} = 16$$<br><br>$$ \Rightarrow {(p + q)^2}{(p - q)^2} = 16$$<br><br>$$ \Rightarrow {({p^2} - {q^2})^2} = 16$$<br><br>$$ \Rightarrow {P^2} - {1 \over {{P^2}}} = \pm 4$$<br><br>$$ \Rightarrow {p^4} \pm 4{p^2} - 1 = 0$$<br><br>$$ \Rightarrow {p^2} = {{ \pm 4 \pm \sqrt {20} } \over 2} = \pm 2 \pm \sqrt 5 $$<br><br>$$ \Rightarrow {p^2} = 2 + \sqrt 5 $$ or $$ - 2 + \sqrt 5 $$<br><br>$$O{B^2} = {p^2} + {1 \over {{p^2}}} = 2 + \sqrt 5 + {1 \over {2 + \sqrt 5 }}$$ or $$ - 2 + \sqrt 5 + {1 \over { - 2 + \sqrt 5 }} = 2\sqrt 5 $$<br><br>Area $$ = 4\left( {{1 \over 2}} \right)(OA)(OB) = 2{(OB)^2} = 4\sqrt 5 $$<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwjjyhu6/768a07f6-81f2-4c0b-ad88-e9817b38c6d7/b186a5f0-5075-11ec-8f7b-45f8e5a35226/file-1kwjjyhu7.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 50vh" alt="JEE Main 2021 (Online) 18th March Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 66 English Explanation"> |
Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is :
Options:
[{"identifier": "A", "content": "3x<sup>2</sup> $$-$$ 2y $$-$$ 6 = 0"}, {"identifier": "B", "content": "3x<sup>2</sup> + 2y $$-$$ 6 = 0"}, {"identifier": "C", "content": "2x<sup>2</sup> + 3y $$-$$ 9 = 0"}, {"identifier": "D", "content": "2x<sup>2</sup> $$-$$ 3y + 9 = 0"}] | ["C"]
Explanation:
A(0, 6) and B(2t, 0)<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265928/exam_images/t38wivroym0l1mk7udq7.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 59 English Explanation"><br><br>Perpendicular bisector of AB is <br><br>$$(y - 3) = {t \over 3}(x - t)$$<br><br>So, $$C = \left( {0,3 - {{{t^2}} \over 3}} \right)$$<br><br>Let P be (h, k)<br><br>$$h = {t \over 2};k = \left( {3 - {{{t^2}} \over 6}} \right)$$<br><br>$$ \Rightarrow k = 3 - {{4{h^2}} \over 6} \Rightarrow 2{x^2} + 3y - 9 = 0$$ |
<p>A point $$P$$ moves so that the sum of squares of its distances from the points $$(1,2)$$ and $$(-2,1)$$ is 14. Let $$f(x, y)=0$$ be the locus of $$\mathrm{P}$$, which intersects the $$x$$-axis at the points $$\mathrm{A}$$, $$\mathrm{B}$$ and the $$y$$-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to :</p>
Options:
[{"identifier": "A", "content": "$${9 \\over 2}$$"}, {"identifier": "B", "content": "$${{3\\sqrt {17} } \\over 2}$$"}, {"identifier": "C", "content": "$${{3\\sqrt {17} } \\over 4}$$"}, {"identifier": "D", "content": "9"}] | ["B"]
Explanation:
<p>Let point $$P:(h,\,k)$$</p>
<p>$${(h - 1)^2} + {(k - 2)^2} + {(h + 2)^2} + {(k - 1)^2} = 14$$</p>
<p>$$2{h^2} + 2{k^2} + 2h - 6k - 4 = 0$$</p>
<p>Locus of $$P:{x^2} + {y^2} + x - 3y - 2 = 0$$</p>
<p>Intersection with x-axis,</p>
<p>$${x^2} + x - 2 = 0$$</p>
<p>$$ \Rightarrow x = - 2,\,1$$</p>
<p>Intersection with y-axis,</p>
<p>$${y^2} - 3y - 2 = 0$$</p>
<p>$$ \Rightarrow y = {{3\, \pm \,\sqrt {17} } \over 2}$$</p>
<p>Area of the quadrilateral ACBD is</p>
<p>$$ = {1 \over 2}(|{x_1}| + |{x_2}|)(|{y_1}| + |{y_2}|)$$</p>
<p>$$ = {1 \over 2} \times 3 \times \sqrt {17} = {{3\sqrt {17} } \over 2}$$</p> |
<p>If the point $$\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$$ lies on the curve traced by the mid-points of the line segments of the lines $$x \cos \theta+y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)$$ between the co-ordinates axes, then $$\alpha$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "$$-$$7"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "$$-$$7$$\\sqrt3$$"}, {"identifier": "D", "content": "7$$\\sqrt3$$"}] | ["B"]
Explanation:
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lhu7fnjs/31373c2a-3f37-47d1-bed5-e74cffaeb86b/e21d2070-f611-11ed-8eab-4b3832376002/file-1lhu7fnjt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lhu7fnjs/31373c2a-3f37-47d1-bed5-e74cffaeb86b/e21d2070-f611-11ed-8eab-4b3832376002/file-1lhu7fnjt.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 12th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 29 English Explanation">
<br><br>$$
\begin{gathered}
x \cos \theta+y \sin \theta=7 \\\\
x-\text { intercept }=\frac{7}{\cos \theta} \\\\
y-\text { intercept }=\frac{7}{\sin \theta} \\\\
\mathrm{A}:\left(\frac{7}{\cos \theta}, 0\right) \mathrm{B}:\left(0, \frac{7}{\sin \theta}\right)
\end{gathered}
$$
<br><br>Locus of mid point M : (h, k)
<br><br>$$
\begin{aligned}
& \mathrm{h}=\frac{7}{2 \cos \theta}, \mathrm{k}=\frac{7}{2 \sin \theta} \\\\
& \frac{7}{2 \sin \theta}=\frac{7 \sqrt{3}}{3} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{3} \\\\
& \alpha=\frac{7}{2 \cos \theta}=7
\end{aligned}
$$ |
<p>Let $$\mathrm{A}(-1,1)$$ and $$\mathrm{B}(2,3)$$ be two points and $$\mathrm{P}$$ be a variable point above the line $$\mathrm{AB}$$ such that the area of $$\triangle \mathrm{PAB}$$ is 10. If the locus of $$\mathrm{P}$$ is $$\mathrm{a} x+\mathrm{by}=15$$, then $$5 \mathrm{a}+2 \mathrm{~b}$$ is :</p>
Options:
[{"identifier": "A", "content": "$$-\\frac{12}{5}$$"}, {"identifier": "B", "content": "$$-\\frac{6}{5}$$"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "4"}] | ["A"]
Explanation:
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lvnxhk5n/61340bd7-15c2-4d49-8787-6b1d30c56e5c/4f3a81b0-07c9-11ef-afdb-fb69ed1077de/file-6y3zli1lvnxhk5o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lvnxhk5n/61340bd7-15c2-4d49-8787-6b1d30c56e5c/4f3a81b0-07c9-11ef-afdb-fb69ed1077de/file-6y3zli1lvnxhk5o.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 3 English Explanation">
<br>$\begin{aligned} & \frac{1}{2}\left|\begin{array}{ccc}h & k & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1\end{array}\right|=10 \\\\ & -2 x+3 y=25 \\\\ & -\frac{6}{5} x+\frac{9}{5} y=15\end{aligned}$
<br><br>$\begin{aligned} & a=-\frac{6}{5}, b=\frac{9}{5} \\\\ & 5 a=-6,2 b=\frac{18}{5}\end{aligned}$
<br><br>$$ \therefore $$ $$5 \mathrm{a}+2 \mathrm{~b}$$ = $$-\frac{12}{5}$$ |
<p>If the locus of the point, whose distances from the point $$(2,1)$$ and $$(1,3)$$ are in the ratio $$5: 4$$, is $$a x^2+b y^2+c x y+d x+e y+170=0$$, then the value of $$a^2+2 b+3 c+4 d+e$$ is equal to :</p>
Options:
[{"identifier": "A", "content": "37"}, {"identifier": "B", "content": "$$-27$$"}, {"identifier": "C", "content": "437"}, {"identifier": "D", "content": "5"}] | ["A"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwagdtgz/adc2b178-39c7-4610-ae95-7cf96da59f71/72edac30-142c-11ef-983f-65b4bd1ca415/file-1lwagdth0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwagdtgz/adc2b178-39c7-4610-ae95-7cf96da59f71/72edac30-142c-11ef-983f-65b4bd1ca415/file-1lwagdth0.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 2 English Explanation"></p>
<p>$$\begin{aligned}
& \therefore \frac{(h-2)^2+(k-1)^2}{(h-1)^2+(k-3)^2}=\frac{25}{16} \\
& \frac{h^2+k^2-4 h-2 k+5}{h^2+k^2-2 h-6 k+10}=\frac{25}{16} \\
\end{aligned}$$</p>
<p>Replacing $$h \rightarrow x$$ and $$k \rightarrow y$$.</p>
<p>$$\begin{aligned}
& 16 x^2+16 y^2-64 x-32 y+80 \\
& =25 x^2+25 y^2-50 x-150 y+250 \\
& 9 x^2+9 y^2+14 x-118 y+170=0 \\
& \Rightarrow a=9, b=9, c=0, d=14, e=-118 \\
& a^2+2 b+3 c+4 d+e \\
& 81+18+56-118 \\
& \Rightarrow 37
\end{aligned}$$</p> |
If the pair of lines
<br/><br/>$$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$
<br/><br/>intersect on the $$y$$-axis then :
Options:
[{"identifier": "A", "content": "$$2fgh = b{g^2} + c{h^2}$$ "}, {"identifier": "B", "content": "$$b{g^2} \\ne c{h^2}$$ "}, {"identifier": "C", "content": "$$abc = 2fgh$$ "}, {"identifier": "D", "content": "none of these "}] | ["A"]
Explanation:
Put $$x=0$$ in the given equation
<br><br>$$ \Rightarrow b{y^2} + 2fy + c = 0.$$
<br><br>For unique point of intersection $${f^2} - bc = 0$$
<br><br>$$ \Rightarrow a{f^2} - abc = 0.$$
<br><br>Since $$abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0$$
<br><br>$$ \Rightarrow 2fgh - b{g^2} - c{h^2} = 0$$ |
The pair of lines represented by
$$$3a{x^2} + 5xy + \left( {{a^2} - 2} \right){y^2} = 0$$$
<br/><br/>are perpendicular to each other for :
Options:
[{"identifier": "A", "content": "two values of $$a$$"}, {"identifier": "B", "content": "$$\\forall \\,a$$ "}, {"identifier": "C", "content": "for one value of $$a$$ "}, {"identifier": "D", "content": "for no values of $$a$$ "}] | ["A"]
Explanation:
$$3a + {a^2} - 2 = 0 \Rightarrow {a^2} + 3a - 2 = 0;$$
<br><br>$$ \Rightarrow a = {{ - 3 \pm \sqrt {9 + 8} } \over 2} = {{ - 3 \pm \sqrt {17} } \over 2}$$ |
If the pair of straight lines $${x^2} - 2pxy - {y^2} = 0$$ and $${x^2} - 2qxy - {y^2} = 0$$ be such that each pair bisects the angle between the other pair, then :
Options:
[{"identifier": "A", "content": "$$pq = -1$$ "}, {"identifier": "B", "content": "$$p = q$$ "}, {"identifier": "C", "content": "$$p = -q$$ "}, {"identifier": "D", "content": "$$pq = 1$$."}] | ["A"]
Explanation:
Equation of bisectors of second pair of straight lines is,
<br><br>$$q{x^2} + 2xy - q{y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>It must be identical to the first pair
<br><br>$${x^2} - 2\,pxy - {y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(... 2 \right)$$
<br><br>from $$(1)$$ and $$(2)$$ $${q \over 1} = {2 \over { - 2p}} = {{ - q} \over { - 1}}$$
<br><br>$$ \Rightarrow pq = - 1.$$ |
If the sum of the slopes of the lines given by $${x^2} - 2cxy - 7{y^2} = 0$$ is four times their product $$c$$ has the value :
Options:
[{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "$$-1$$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$1$$"}] | ["C"]
Explanation:
Let the lines be $$y = {m_1}x$$ and $$y = {m_2}x$$ then
<br><br>$${m_1} + {m_2} = - {{2c} \over 7}$$ and $${m_1}{m_2} = - {1 \over 7}$$
<br><br>Given $${m_1} + {m_2} = 4m{}_1{m_2}$$
<br><br>$$ \Rightarrow {{2c} \over 7} = - {4 \over 7} \Rightarrow c = 2$$ |
If one of the lines given by $$6{x^2} - xy + 4c{y^2} = 0$$ is $$3x + 4y = 0,$$ then $$c$$ equals :
Options:
[{"identifier": "A", "content": "$$-3$$ "}, {"identifier": "B", "content": "$$-1$$"}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "content": "$$1$$ "}] | ["A"]
Explanation:
$$3x+4y=0$$ is one of the lines of the pair
<br><br>$$6{x^2} - xy + 4c{y^2} = 0,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$
<br><br>Put $$y = - {3 \over 4}x,$$
<br><br>we get $$6{x^2} + {3 \over 4}{x^2} + 4c{\left( { - {3 \over 4}x} \right)^2} = 0$$
<br><br>$$ \Rightarrow 6 + {3 \over 4} + {{9c} \over 4} = 0 \Rightarrow c = - 3$$ |
If one of the lines of $$m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0$$ is a bisector of angle between the lines $$xy = 0,$$ then $$m$$ is :
Options:
[{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$2$$ "}, {"identifier": "C", "content": "$$-1/2$$ "}, {"identifier": "D", "content": "$$-2$$"}] | ["A"]
Explanation:
Equation of bisectors of lines, $$xy=0$$ are $$y = \pm x$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264578/exam_images/cudjhzff4ynokmmza2f9.webp" loading="lazy" alt="AIEEE 2007 Mathematics - Straight Lines and Pair of Straight Lines Question 133 English Explanation">
<br><br>$$\therefore$$ Put $$y = \pm \,x$$ in the given equation
<br><br>$$m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0$$
<br><br>$$\therefore$$ $$m{x^2} + \left( {1 - {m^2}} \right){x^2} - m{x^2} = 0$$
<br><br>$$ \Rightarrow 1 - {m^2} = 0 \Rightarrow m = \pm 1$$ |
Let the equation of the pair of lines, y = px and y = qx, can be written as (y $$-$$ px) (y $$-$$ qx) = 0. Then the equation of the pair of the angle bisectors of the lines x<sup>2</sup> $$-$$ 4xy $$-$$ 5y<sup>2</sup> = 0 is :
Options:
[{"identifier": "A", "content": "x<sup>2</sup> $$-$$ 3xy + y<sup>2</sup> = 0"}, {"identifier": "B", "content": "x<sup>2</sup> + 4xy $$-$$ y<sup>2</sup> = 0"}, {"identifier": "C", "content": "x<sup>2</sup> + 3xy $$-$$ y<sup>2</sup> = 0"}, {"identifier": "D", "content": "x<sup>2</sup> $$-$$ 3xy $$-$$ y<sup>2</sup> = 0"}] | ["C"]
Explanation:
Equation of angle bisector of homogeneous <br>equation of pair of straight line ax<sup>2</sup>
+ 2hxy + by<sup>2</sup>
is
<br><br>$${{{x^2} - {y^2}} \over {a - b}} = {{xy} \over h}$$
<br><br>for x<sup>2</sup> – 4xy – 5y<sup>2</sup>
= 0
<br><br> a = 1, h = – 2, b = – 5
<br><br>So, equation of angle bisector is
<br><br>$${{{x^2} - {y^2}} \over {1 - ( - 5)}} = {{xy} \over { - 2}}$$<br><br>$${{{x^2} - {y^2}} \over 6} = {{xy} \over { - 2}}$$<br><br>$$ \Rightarrow {x^2} - {y^2} = - 3xy$$<br><br>So, combined equation of angle bisector is $$ {x^2} + 3xy - {y^2} = 0$$ |
The lines $\mathrm{L}_1, \mathrm{~L}_2, \ldots, \mathrm{L}_{20}$ are distinct. For $\mathrm{n}=1,2,3, \ldots, 10$ all the lines $\mathrm{L}_{2 \mathrm{n}-1}$ are parallel to each other and all the lines $L_{2 n}$ pass through a given point $P$. The maximum number of points of intersection of pairs of lines from the set $\left\{\mathrm{L}_1, \mathrm{~L}_2, \ldots, \mathrm{L}_{20}\right\}$ is equal to ___________.
Options:
[] | 101
Explanation:
<p>To find the maximum number of points of intersection of pairs of lines from the given set, we need to consider how the lines are arranged based on the given conditions.</p><p>Firstly, there are 10 lines (${L}_1, {L}_3, ..., {L}_{19}$) that are parallel to each other. Since parallel lines do not intersect with each other, these 10 lines will not contribute to the number of intersection points among themselves.</p><p>Secondly, there are 10 lines (${L}_2, {L}_4, ..., {L}_{20}$) that all pass through a given point $P$. Although these lines intersect at $P$, they only contribute one unique point of intersection to the total count.</p><p>To calculate the maximum number of intersection points, we need to consider the total number of ways to pick pairs of lines from the 20 lines available without restrictions, then subtract the combinations that do not result in intersections, which includes the combinations of parallel lines among themselves and the concurrent lines through point $P$.</p><p>This calculation is represented as:</p><p>$$Total = ^{20}C_2 - ^{10}C_2 - ^{10}C_2 + 1$$</p><p>Here, $^{20}C_2$ calculates the total number of ways to pick any two lines out of 20, which includes intersecting and non-intersecting lines. $^{10}C_2$ is subtracted twice: once for the set of parallel lines (${L}_1, {L}_3, ..., {L}_{19}$) that don't intersect among themselves and once more for the set of concurrent lines (${L}_2, {L}_4, ..., {L}_{20}$) intersecting only at point $P$. Since all the concurrent lines intersect at the same point, we add 1 back to include this intersection point.</p><p>Carrying out this calculation gives us the total number of distinct intersection points as $101$.</p> |
If $$\left( {a,{a^2}} \right)$$ falls inside the angle made by the lines $$y = {x \over 2},$$ $$x > 0$$ and $$y = 3x,$$ $$x > 0,$$ then a belong to :
Options:
[{"identifier": "A", "content": "$$\\left( {0,{1 \\over 2}} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {3,\\infty } \\right)$$ "}, {"identifier": "C", "content": "$$\\left( {{1 \\over 2},3} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( {-3,-{1 \\over 2}} \\right)$$"}] | ["C"]
Explanation:
Clearly for point $$P,$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266344/exam_images/ctbame8e1bgmyuficejz.webp" loading="lazy" alt="AIEEE 2006 Mathematics - Straight Lines and Pair of Straight Lines Question 136 English Explanation">
<br><br>$${a^2} - 3a < 0$$ <b>and</b> $${a^2} - {a \over 2} > 0 \Rightarrow {1 \over 2} < a < 3$$ |
The set of all possible values of
$$\theta $$ in the interval
<br/>(0, $$\pi $$) for which the points (1, 2) and (sin
$$\theta $$, cos $$\theta $$) lie <br/>on the same side of the line x + y =
1 is :
Options:
[{"identifier": "A", "content": "$$\\left( {0,{\\pi \\over 4}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {0,{{3\\pi } \\over 4}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{\\pi \\over 4},{{3\\pi } \\over 4}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {0,{\\pi \\over 2}} \\right)$$"}] | ["D"]
Explanation:
Let f(x, y) = x + y - 1<br><br>
$$ \because f\left( {1,2} \right).f\left( {\sin \theta ,\cos \theta } \right) > 0$$<br><br>
$$ \Rightarrow 2\left[ {\sin \theta + \cos \theta - 1} \right] > 0$$<br><br>
$$ \Rightarrow \sin \theta + \cos \theta > 1$$<br><br>
$$ \Rightarrow \sin \left( {\theta + {\pi \over 4}} \right) > {1 \over {\sqrt 2 }}$$<br><br>
$$ \Rightarrow \theta + {\pi \over 4} \in \left( {{\pi \over 4},{{3\pi } \over 4}} \right)$$<br><br>
$$ \Rightarrow \theta \in \left( {0,{\pi \over 2}} \right)$$ |
Let L denote the line in the xy-plane with x and
y intercepts as 3 and 1 respectively. Then the
image of the point (–1, –4) in this line is :
Options:
[{"identifier": "A", "content": "$$\\left( {{{11} \\over 5},{{28} \\over 5}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{{29} \\over 5},{{11} \\over 5}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{{29} \\over 5},{8 \\over 5}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{8 \\over 5},{{29} \\over 5}} \\right)$$"}] | ["A"]
Explanation:
Line is $${x \over 3} + {y \over 1} = 1$$
<br><br>$$ \Rightarrow $$ x + 3y – 3 = 0
<br><br>Let Image of point (–1, –4) is ($$\alpha $$, $$\beta $$)
<br><br>Hence, $${{\alpha + 1} \over 1} = {{\beta + 4} \over 3} = - 2\left( {{{ - 1 - 12 - 3} \over {10}}} \right)$$
<br><br>$$ \Rightarrow $$ $${{\alpha + 1} \over 1} = {{\beta + 4} \over 3} = {{16} \over 5}$$
<br><br>$$ \Rightarrow $$ $$\alpha $$ = $${{11} \over 5}$$, $$\beta $$ = $${{28} \over 5}$$ |
The image of the point (3, 5) in the line x $$-$$ y + 1 = 0, lies on :
Options:
[{"identifier": "A", "content": "(x $$-$$ 4)<sup>2</sup> + (y $$-$$ 4)<sup>2</sup> = 8"}, {"identifier": "B", "content": "(x $$-$$ 4)<sup>2</sup> + (y $$+$$ 2)<sup>2</sup> = 16"}, {"identifier": "C", "content": "(x $$-$$ 2)<sup>2</sup> + (y $$-$$ 2)<sup>2</sup> = 12"}, {"identifier": "D", "content": "(x $$-$$ 2)<sup>2</sup> + (y $$-$$ 4)<sup>2</sup> = 4"}] | ["D"]
Explanation:
So, let the image is (x, y)<br><br>So, we have<br><br>$${{x - 3} \over 1} = {{y - 5} \over { - 1}} = - {{2(3 - 5 + 1)} \over {1 + 1}}$$<br><br>$$ \Rightarrow $$ x = 4, y = 4<br><br>$$ \Rightarrow $$ Point (4, 4)<br><br>Which will satisfy the curve <br><br>(x $$-$$ 2)<sup>2</sup> + (y $$-$$ 4)<sup>2</sup> = 4<br><br>as (4 $$-$$ 2)<sup>2</sup> + (4 $$-$$ 4)<sup>2</sup> = 4 + 0 = 4 |
<p>Let $$\mathrm{R}$$ be the interior region between the lines $$3 x-y+1=0$$ and $$x+2 y-5=0$$ containing the origin. The set of all values of $$a$$, for which the points $$\left(a^2, a+1\right)$$ lie in $$R$$, is :</p>
Options:
[{"identifier": "A", "content": " $$(-3,0) \\cup\\left(\\frac{2}{3}, 1\\right)$$\n"}, {"identifier": "B", "content": "$$(-3,0) \\cup\\left(\\frac{1}{3}, 1\\right)$$\n"}, {"identifier": "C", "content": "$$(-3,-1) \\cup\\left(\\frac{1}{3}, 1\\right)$$\n"}, {"identifier": "D", "content": "$$(-3,-1) \\cup\\left(-\\frac{1}{3}, 1\\right)$$"}] | ["B"]
Explanation:
<p>$$\begin{aligned}
& \mathrm{P}\left(\mathrm{a}^2, \mathrm{a}+1\right) \\
& \mathrm{L}_1=3 \mathrm{x}-\mathrm{y}+1=0
\end{aligned}$$</p>
<p>Origin and $$\mathrm{P}$$ lies same side w.r.t. $$\mathrm{L}_1$$</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{L}_1(0) \cdot \mathrm{L}_1(\mathrm{P})>0 \\
& \therefore 3\left(\mathrm{a}^2\right)-(\mathrm{a}+1)+1>0
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1u0w1z/5e71e275-9890-4f83-b0e3-dce0674de6b5/173b5570-d40a-11ee-b9d5-0585032231f0/file-1lt1u0w20.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1u0w1z/5e71e275-9890-4f83-b0e3-dce0674de6b5/173b5570-d40a-11ee-b9d5-0585032231f0/file-1lt1u0w20.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 21 English Explanation"></p>
<p>$$\begin{aligned}
& \Rightarrow 3 \mathrm{a}^2-\mathrm{a}>0 \\
& \mathrm{a} \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \quad \text{..... (1)}
\end{aligned}$$</p>
<p>Let $$L_2: x+2 y-5=0$$</p>
<p>Origin and $$\mathrm{P}$$ lies same side w.r.t. $$\mathrm{L}_2$$</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{L}_2(0) \cdot \mathrm{L}_2(\mathrm{P})>0 \\
& \Rightarrow \mathrm{a}^2+2(\mathrm{a}+1)-5<0 \\
& \Rightarrow \mathrm{a}^2+2 \mathrm{a}-3<0 \\
& \Rightarrow(\mathrm{a}+3)(\mathrm{a}-1)<0
\end{aligned}$$</p>
<p>$$\therefore \mathrm{a} \in(-3,1)\quad \text{..... (2)}$$</p>
<p>Intersection of (1) and (2)</p>
<p>$$\mathrm{a} \in(-3,0) \cup\left(\frac{1}{3}, 1\right)$$</p> |
The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept -4. Then a possible value of k is :
Options:
[{"identifier": "A", "content": "1 "}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "-2"}, {"identifier": "D", "content": "-4"}] | ["D"]
Explanation:
Slope of $$PQ = {{3 - 4} \over {k - 1}} = {{ - 1} \over {k - 1}}$$
<br><br>$$\therefore$$ Slope of perpendicular bisector of
<br><br>$$PQ = \left( {k - 1} \right)$$
<br><br>Also mid point of
<br><br>$$PQ\left( {{{k + 1} \over 2},{7 \over 2}} \right).$$
<br><br>Equation of perpendicular bisector is
<br><br>$$y - {7 \over 2} = \left( {k - 1} \right)\left( {x - {{k + 1} \over 2}} \right)$$
<br><br>$$ \Rightarrow 2y - 7 = 2\left( {k - 1} \right)x - \left( {{k^2} - 1} \right)$$
<br><br>$$ \Rightarrow 2\left( {k - 1} \right)x - 2y + \left( {8 - {k^2}} \right) = 0$$
<br><br>$$\therefore$$ $$y$$-intercept $$ = {{8 - {k^2}} \over { - 2}} = - 4$$
<br><br>$$ \Rightarrow $$ $$8 - {k^2} = - 8$$ or $${k^2} = 16 \Rightarrow k = \pm 4$$ |
If the line $$2x + y = k$$ passes through the point which divides the line segment joining the points $$(1, 1)$$ and $$(2, 4)$$ in the ratio $$3 : 2$$, then $$k$$ equals :
Options:
[{"identifier": "A", "content": "$${{29 \\over 5}}$$"}, {"identifier": "B", "content": "$$5$$"}, {"identifier": "C", "content": "$$6$$ "}, {"identifier": "D", "content": "$${{11 \\over 5}}$$"}] | ["C"]
Explanation:
<p>The point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2 is</p>
<p>$$ = \left( {{{3 \times 2 + 2 \times 1} \over {3 + 2}},{{3 \times 4 + 2 \times 1} \over {3 + 2}}} \right)$$</p>
<p>$$ = \left( {{{6 + 2} \over 5},{{12 + 2} \over 5}} \right) = \left( {{8 \over 5},{{14} \over 5}} \right)$$</p>
<p>Since the line 2x + y = k passes through this point,</p>
<p>$$\therefore$$ $$2 \times {8 \over 5} + {{14} \over 5} = k$$ or $${{30} \over 5} = k$$ or, k = 6</p> |
Let $$PS$$ be the median of the triangle with vertices $$P(2, 2)$$, $$Q(6, -1)$$ and $$R(7, 3)$$. The equation of the line passing through $$(1, -1)$$ band parallel to PS is :
Options:
[{"identifier": "A", "content": "$$4x + 7y + 3 = 0$$ "}, {"identifier": "B", "content": "$$2x - 9y - 11 = 0$$"}, {"identifier": "C", "content": "$$4x - 7y - 11 = 0$$"}, {"identifier": "D", "content": "$$2x + 9y + 7 = 0$$"}] | ["D"]
Explanation:
Let $$P,Q,R,$$ be the vertices of $$\Delta PQR$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263525/exam_images/hrqnkqyxxwu0ays0itme.webp" loading="lazy" alt="JEE Main 2014 (Offline) Mathematics - Straight Lines and Pair of Straight Lines Question 124 English Explanation">
<br><br>Since $$PS$$ is the median, $$S$$ is mid-point of $$QR$$
<br><br>So, $$S = \left( {{{7 + 6} \over 2},{{3 - 1} \over 2}} \right) = \left( {{{13} \over 2},1} \right)$$
<br><br>Now, slope of $$PS$$ $$ = {{2 - 1} \over {2 - {{13} \over 2}}} = - {2 \over 9}$$
<br><br>Since, required line is parallel to $$PS$$ therefore slope of required line $$=$$ slope of
$$PS$$ <br><br>Now, equation of line passing through $$(1, -1)$$ and having slope $$ - {2 \over 9}$$ is
<br><br>$$y - \left( { - 1} \right) = - {2 \over 9}\left( {x - 1} \right)$$
<br><br>$$9y + 9 = - 2x + 2$$
<br><br>$$ \Rightarrow 2x + 9y + 7 = 0$$ |
A straight line through origin O meets the lines 3y = 10 − 4x and 8x + 6y + 5 = 0 at points A and B respectively. Then O divides the segment AB in the ratio :
Options:
[{"identifier": "A", "content": "2 : 3 "}, {"identifier": "B", "content": "1 : 2"}, {"identifier": "C", "content": "4 : 1"}, {"identifier": "D", "content": "3 : 4"}] | ["C"]
Explanation:
The lines 4x + 3y $$-$$ 10 = 0 and
<br><br>8x + 6y + 5 = 0 , are parallel as
<br><br> $${4 \over 8}$$ = $${3 \over 6}$$
<br><br>Now length of perpendicular from
<br><br>(0, 0, 0) to 4x + 3y $$-$$ 10 = 0 is,
<br><br>P<sub>1</sub> = $$\left| {{{4\left( 0 \right) + 3\left( 0 \right) - 10} \over {\sqrt {{4^2} + {3^2}} }}} \right|$$ = $${{10} \over 5}$$ = 2
<br><br>Length of perpendicular from
<br><br>0 (0, 0) to 8x + 6y + 5 = 0 is
<br><br>P<sub>2</sub> = $$\left| {{{8\left( 0 \right) + 6\left( 0 \right) + 5} \over {\sqrt {{6^2} + {8^2}} }}} \right|$$ = $${5 \over {10}}$$ = $${1 \over 2}$$
<br><br>$$\therefore\,\,\,$$ P<sub>1</sub> : P<sub>2</sub> = 2 : $${1 \over 2}$$ = 4 : 1 |
Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A
and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m)
above the line AC is :
Options:
[{"identifier": "A", "content": "10/3"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "20/3"}, {"identifier": "D", "content": "6"}] | ["D"]
Explanation:
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264412/exam_images/wsbio4fzzawvxpgqqiq9.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265203/exam_images/ovvgpzy8at9xuqen3rrd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 4th September Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 80 English Explanation"></picture>
<br>Equation of AD : $$y = {{10} \over a}x$$<br><br>Equation of BC : $${x \over a} + {y \over {15}} = 1$$<br><br>$$ \Rightarrow {{ay} \over {10a}} + {y \over {15}} = 1$$<br><br>$$ \Rightarrow {{3y + 2y} \over {30}} = 1$$<br><br>$$ \Rightarrow y = 6$$<br><br>$$ \therefore $$ y coordinate of point P = 6 = height of point P above the line AC. |
<p>A ray of light passing through the point P(2, 3) reflects on the x-axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ratio 2 : 1. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be ($$\alpha$$, $$\beta$$). Then, the value of 7$$\alpha$$ + 3$$\beta$$ is equal to ____________.</p>
Options:
[] | 31
Explanation:
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5oc3byi/c98d69fd-398a-42cd-a87f-3ee77ee795a4/cca73aa0-0544-11ed-987f-3938cfc0f7f1/file-1l5oc3byj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5oc3byi/c98d69fd-398a-42cd-a87f-3ee77ee795a4/cca73aa0-0544-11ed-987f-3938cfc0f7f1/file-1l5oc3byj.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 51 English Explanation"> </p>
<p>$${4 \over {5 - \alpha }} = {3 \over {\alpha - 2}} \Rightarrow 4\alpha - 8 = 15 - 3\alpha $$</p>
<p>$$\alpha = {{23} \over 7}$$</p>
<p>$$A = \left( {{{23} \over 7},0} \right)\,Q = (5,4)$$</p>
<p>$$R = \left( {{{10 + {{23} \over 7}} \over 3},{8 \over 3}} \right)$$</p>
<p>$$ = \left( {{{31} \over 7},{8 \over 3}} \right)$$</p>
<p>Bisector of angle PAQ is $$X = {{23} \over 7}$$</p>
<p>$$ \Rightarrow M = \left( {{{23} \over 7},{8 \over 3}} \right)$$</p>
<p>So, $$7\alpha + 3\beta = 31$$</p> |
<p>The equations of two sides $$\mathrm{AB}$$ and $$\mathrm{AC}$$ of a triangle $$\mathrm{ABC}$$ are $$4 x+y=14$$ and $$3 x-2 y=5$$, respectively. The point $$\left(2,-\frac{4}{3}\right)$$ divides the third side $$\mathrm{BC}$$ internally in the ratio $$2: 1$$, the equation of the side $$\mathrm{BC}$$ is</p>
Options:
[{"identifier": "A", "content": "$$x+6 y+6=0$$\n"}, {"identifier": "B", "content": "$$x-3 y-6=0$$\n"}, {"identifier": "C", "content": "$$x+3 y+2=0$$\n"}, {"identifier": "D", "content": "$$x-6 y-10=0$$"}] | ["C"]
Explanation:
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw8pr3ta/992548a6-ea30-4544-b8d1-c7f041deb6d3/854cb2e0-1337-11ef-9f8d-838c388c326d/file-1lw8pr3tb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw8pr3ta/992548a6-ea30-4544-b8d1-c7f041deb6d3/854cb2e0-1337-11ef-9f8d-838c388c326d/file-1lw8pr3tb.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 5 English Explanation"></p>
<p>$$\begin{aligned}
& 2=\frac{2 x_2+x_1}{3}, \frac{-4}{3}=\frac{2 y_2+y_1}{3} \\
& 2 x_2+x_1=6,2 y_2+y_1=-4
\end{aligned}$$</p>
<p>$$\begin{aligned}
& x_1=6-2 x_2 \quad \text{.... (1)}\\
& y_1=-4-2 y_2 \quad \text{.... (2)}\\
& 4 x_1+y_1=14 \quad \text{.... (3)}\\
& 3 x_2-2 y_2=5 \quad \text{.... (4)}
\end{aligned}$$</p>
<p>From here, $$x_2=1, y_2=-1, x_1=4, y_1=-2$$</p>
<p>$$\begin{aligned}
& B(4,-2) C(1,-1) \\
& y+2=\frac{-1+2}{1-4}(x-4) \\
& -3 y-6=x-4 \\
& x+3 y+2=0
\end{aligned}$$</p> |
If $${x_1},{x_2},{x_3}$$ and $${y_1},{y_2},{y_3}$$ are both in G.P. with the same common ratio, then the points $$\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$$ and $$\left( {{x_3},{y_3}} \right)$$ :
Options:
[{"identifier": "A", "content": "are vertices of a triangle"}, {"identifier": "B", "content": "lie on a straight line "}, {"identifier": "C", "content": "lie on an ellipse "}, {"identifier": "D", "content": "lie on a circle "}] | ["B"]
Explanation:
Taking co-ordinates as
<br><br>$$\left( {{x \over r},{y \over r}} \right);\left( {x,y} \right)\,\,\& \,\,\left( {xr,yr} \right)$$
<br><br>Then slope of line joining
<br><br>$$\left( {{x \over r},{y \over r}} \right),\left( {x,y} \right) = {{y\left( {1 - {1 \over r}} \right)} \over {x\left( {1 - {1 \over r}} \right)}} = {y \over x}$$
<br><br>and slope of line joining $$(x,y)$$ and $$(xr, yr)$$
<br><br>$$ = {{y\left( {r - 1} \right)} \over {x\left( {r - 1} \right)}} = {y \over x}$$
<br><br>$$\therefore$$ $${m_1} = {m_2}$$
<br><br>$$ \Rightarrow $$ Points lie on the straight line. |
A square of side a lies above the $$x$$-axis and has one vertex at the origin. The side passing through the origin makes an angle $$\alpha \left( {0 < \alpha < {\pi \over 4}} \right)$$ with the positive direction of x-axis. The equation of its diagonal not passing through the origin is :
Options:
[{"identifier": "A", "content": "$$y\\left( {\\cos \\alpha + \\sin \\alpha } \\right) + x\\left( {\\cos \\alpha - \\sin \\alpha } \\right) = a$$ "}, {"identifier": "B", "content": "$$y\\left( {\\cos \\alpha - \\sin \\alpha } \\right) - x\\left( {\\sin \\alpha - \\cos \\alpha } \\right) = a$$ "}, {"identifier": "C", "content": "$$y\\left( {\\cos \\alpha + \\sin \\alpha } \\right) + x\\left( {\\sin \\alpha - \\cos \\alpha } \\right) = a$$ "}, {"identifier": "D", "content": "$$y\\left( {\\cos \\alpha + \\sin \\alpha } \\right) + x\\left( {\\sin \\alpha + \\cos \\alpha } \\right) = a$$ "}] | ["A"]
Explanation:
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265012/exam_images/xq65o2ooxotlgbornwg4.webp" loading="lazy" alt="AIEEE 2003 Mathematics - Straight Lines and Pair of Straight Lines Question 148 English Explanation">
<br><br>Co-ordinate of $$A = \left( {a\,\cos \,\alpha ,\,\,a\,\sin \,\alpha } \right)$$
<br><br>Equation of $$OB,$$
<br><br>$$y = \tan \left( {{\pi \over 4} + \alpha } \right)x$$
<br><br>$$CA{ \bot ^r}$$ to $$OB$$
<br><br>$$\therefore$$ slope of $$CA=-$$ $$\cot \left( {{\pi \over 4} + \alpha } \right)$$
<br><br>Equation of $$CA$$
<br><br>$$y - a\sin \alpha = - cot\left( {{\pi \over 4} + \alpha } \right)\left( {x - a\,\cos \,\alpha } \right)$$
<br><br>$$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {\tan \left( {{\pi \over 4} + \alpha } \right)} \right) = \left( {a\,\cos \,\alpha - x} \right)$$
<br><br>$$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {{{\tan {\pi \over 4} + \tan \alpha } \over {1 - \tan {\pi \over 4}\tan \alpha }}} \right)\left( {a\,\cos \,\alpha - x} \right)$$
<br><br>$$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {1 + \tan \alpha } \right) = \left( {a\cos \alpha - x} \right)\left( {1 - \tan \alpha } \right)$$
<br><br>$$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {\cos \alpha + \sin \alpha } \right) = \left( {a\cos \alpha - x} \right)\left( {\cos \alpha - \sin \alpha } \right)$$
<br><br>$$ \Rightarrow y\left( {\cos + \sin \alpha } \right) - a\sin \alpha \cos \alpha - a{\sin ^2}\alpha $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = a{\cos ^2}\alpha - a\cos \alpha \sin \alpha - x\left( {\cos \alpha - \sin \alpha } \right)$$
<br><br>$$ \Rightarrow y\left( {\cos \alpha + sin\alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a$$
<br><br>$$y\left( {\sin \alpha + \cos \alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a.$$ |
The equation of the straight line passing through the point $$(4, 3)$$ and making intercepts on the co-ordinate axes whose sum is $$-1$$ is :
Options:
[{"identifier": "A", "content": "$${x \\over 2} - {y \\over 3} = 1$$ and $${x \\over -2} +{y \\over 1} = 1$$"}, {"identifier": "B", "content": "$${x \\over 2} - {y \\over 3} = -1$$ and $${x \\over -2} +{y \\over 1} = -1$$"}, {"identifier": "C", "content": "$${x \\over 2} + {y \\over 3} = 1$$ and $${x \\over 2} +{y \\over 1} = 1$$ "}, {"identifier": "D", "content": "$${x \\over 2} + {y \\over 3} = -1$$ and $${x \\over -2} +{y \\over 1} = -1$$"}] | ["A"]
Explanation:
Let the required line be $${x \over a} + {y \over b} = 1.......\left( 1 \right)$$
<br><br>then $$a+b=-1$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.........\left( 2 \right)$$
<br><br>$$(1)$$ passes through $$(4,3), $$ $$ \Rightarrow {4 \over a} + {3 \over b} = 1$$
<br><br>$$ \Rightarrow 4b + 3a = ab\,\,...............\left( 3 \right)$$
<br><br>Eliminating $$b$$ from $$(2)$$ and $$(3),$$ we get
<br><br>$${a^2} - 4 = 0 \Rightarrow a = \pm 2 \Rightarrow b = - 3$$, $$1$$
<br><br>$$\therefore$$ Equation of straight lines are
<br><br>$${x \over 2} + {y \over { - 3}} = 1$$
<br><br>or $${x \over { - 2}} + {y \over 1} = 1$$ |
The line parallel to the $$x$$ - axis and passing through the intersection of the lines $$ax + 2by + 3b = 0$$ and $$bx - 2ay - 3a = 0,$$ where $$(a, b)$$ $$ \ne $$ $$(0, 0)$$ is :
Options:
[{"identifier": "A", "content": "below the $$x$$ - axis at a distance of $${3 \\over 2}$$ from it "}, {"identifier": "B", "content": "below the $$x$$ - axis at a distance of $${2 \\over 3}$$ from it "}, {"identifier": "C", "content": "above the $$x$$ - axis at a distance of $${3 \\over 2}$$ from it "}, {"identifier": "D", "content": "above the $$x$$ - axis at a distance of $${2 \\over 3}$$ from it "}] | ["A"]
Explanation:
The line passing through the intersection of lines
<br><br>$$ax + 2by = 3b = 0$$
<br><br>and $$bx - 2ay - 3a = 0$$ is
<br><br>$$ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0$$
<br><br>$$ \Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0$$
<br><br>As this line is parallel to $$x$$-axis.
<br><br>$$\therefore$$ $$a + b\lambda = 0 \Rightarrow \lambda = - a/b$$
<br><br>$$ \Rightarrow ax + 2by + 3b - {a \over b}\left( {bx - 2ay - 3a} \right) = 0$$
<br><br>$$ \Rightarrow ax + 2by + 3b - ax + {{2{a^2}} \over b}y + {{3{a^2}} \over b} = 0$$
<br><br>$$y\left( {2b + {{2{a^2}} \over b}} \right) + 3b + {{3{a^2}} \over b} = 0$$
<br><br>$$y\left( {{{2{b^2} + 2{a^2}} \over b}} \right) = - \left( {{{3{b^2} + 3{a^2}} \over b}} \right)$$
<br><br>$$y = {{ - 3\left( {{a^2} + {b^2}} \right)} \over {2\left( {{b^2} + {a^2}} \right)}} = {{ - 3} \over 2}$$
<br><br>So it is $$3/2$$ units below $$x$$-axis. |
If non zero numbers $$a, b, c$$ are in $$H.P.,$$ then the straight line $${x \over a} + {y \over b} + {1 \over c} = 0$$ always passes through a fixed point. That point is :
Options:
[{"identifier": "A", "content": "$$(-1,2)$$ "}, {"identifier": "B", "content": "$$(-1, -2)$$ "}, {"identifier": "C", "content": "$$(1, -2)$$ "}, {"identifier": "D", "content": "$$\\left( {1, - {1 \\over 2}} \\right)$$ "}] | ["C"]
Explanation:
$$a,b,c$$ are in $$H.P. \Rightarrow {1 \over a}.{1 \over b},{1 \over c}$$ are in $$A.P.$$
<br><br>$$ \Rightarrow {2 \over b} = {1 \over a} + {1 \over c}$$
<br><br>$$ \Rightarrow {1 \over a} - {2 \over b} + {1 \over c} = 0$$
<br><br>$$\therefore$$ $${x \over a} + {y \over a} + {1 \over c} = 0$$ passes through $$\left( {1, - 2} \right)$$ |
A straight line through the point $$A (3, 4)$$ is such that its intercept between the axes is bisected at $$A$$. Its equation is :
Options:
[{"identifier": "A", "content": "$$x + y = 7$$ "}, {"identifier": "B", "content": "$$3x - 4y + 7 = 0$$ "}, {"identifier": "C", "content": "$$4x + 3y = 24$$ "}, {"identifier": "D", "content": "$$3x + 4y = 25$$ "}] | ["C"]
Explanation:
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266358/exam_images/znwolrv8chfin230lej7.webp" loading="lazy" alt="AIEEE 2006 Mathematics - Straight Lines and Pair of Straight Lines Question 137 English Explanation">
<br><br>As is the mid point of $$PQ,$$ therefore
<br><br>$${{a + 0} \over 2} = 3,{{0 + b} \over 2} = 4 \Rightarrow a = 6,b = 8$$
<br><br>$$\therefore$$ Equation of line is $${x \over 6} + {y \over 8} = 1$$
<br><br>or $$4x + 3y = 24$$ |
Let $$a, b, c$$ and $$d$$ be non-zero numbers. If the point of intersection of the lines $$4ax + 2ay + c = 0$$ and $$5bx + 2by + d = 0$$ lies in the fourth quadrant and is equidistant from the two axes then :
Options:
[{"identifier": "A", "content": "$$3bc - 2ad = 0$$ "}, {"identifier": "B", "content": "$$3bc + 2ad = 0$$"}, {"identifier": "C", "content": "$$2bc - 3ad = 0$$"}, {"identifier": "D", "content": "$$2bc + 3ad = 0$$"}] | ["A"]
Explanation:
<p>Since the point of intersection lies on fourth quadrant and equidistant from the two axes,</p>
<p>i.e., let the point be (k, $$-$$k) and this point satisfies the two equations of the given lines.</p>
<p>$$\therefore$$ 4ak $$-$$ 2ak + c = 0 ......... (1)</p>
<p>and 5bk $$-$$ 2bk + d = 0 ..... (2)</p>
<p>From (1) we get, $$k = {{ - c} \over {2a}}$$</p>
<p>Putting the value of k in (2) we get,</p>
<p>$$5b\left( { - {c \over {2a}}} \right) - 2b\left( { - {c \over {2a}}} \right) + d = 0$$</p>
<p>or, $$ - {{5bc} \over {2a}} + {{2bc} \over {2a}} + d = 0$$ or, $$ - {{3bc} \over {2a}} + d = 0$$</p>
<p>or, $$ - 3bc + 2ad = 0$$ or, $$3bc - 2ad = 0$$</p> |
Two sides of a rhombus are along the lines, $$x - y + 1 = 0$$ and $$7x - y - 5 = 0$$. If its diagonals intersect at $$(-1, -2)$$, then which one of the following is a vertex of this rhombus?
Options:
[{"identifier": "A", "content": "$$\\left( {{{ 1} \\over 3}, - {8 \\over 3}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( - {{{ 10} \\over 3}, - {7 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - 3, - 9} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( { - 3, - 8} \\right)$$"}] | ["A"]
Explanation:
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264559/exam_images/k1bzxqvdfbjwl4zlissa.webp" loading="lazy" alt="JEE Main 2016 (Offline) Mathematics - Straight Lines and Pair of Straight Lines Question 121 English Explanation">
<br><br>Let other two sides of rhombus are
<br><br>$$x - y + \lambda = 0$$
<br><br>and $$7x - y + \mu = 0$$
<br><br>then $$O$$ is equidistant from $$AB$$ and $$DC$$ and from $$AD$$ and $$BC$$
<br><br>$$\therefore$$ $$\left| { - 1 + 2 + 1} \right| = \left| { - 1 + 2 + \lambda } \right| \Rightarrow \lambda = - 3$$
<br><br>and $$\left| { - 7 + 2 - 5} \right| = \left| { - 7 + 2 + \mu } \right| \Rightarrow \mu = 15$$
<br><br>$$\therefore$$ Other two sides are $$x-y-3=0$$ and $$7x-y+15=0$$
<br><br>On solving the equations of sides pairwise, we get
<br><br>the vertices as $$\left( {{1 \over 3},{{ - 8} \over 3}} \right),\left( {1,2} \right),\left( {{{ - 7} \over 3},{{ - 4} \over 3}} \right),\left( { - 3, - 6} \right)$$ |
The point (2, 1) is translated parallel to the line L : x− y = 4 by $$2\sqrt 3 $$ units. If the newpoint Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is :
Options:
[{"identifier": "A", "content": "x + y = 2 $$-$$ $$\\sqrt 6 $$"}, {"identifier": "B", "content": "x + y = 3 $$-$$ 3$$\\sqrt 6 $$"}, {"identifier": "C", "content": "x + y = 3 $$-$$ 2$$\\sqrt 6 $$"}, {"identifier": "D", "content": "2x + 2y = 1 $$-$$ $$\\sqrt 6 $$"}] | ["C"]
Explanation:
x $$-$$ y = 4
<br><br>To find equation of R
<br><br>slope of L = 0 is 1
<br><br>$$ \Rightarrow $$ slope of QR = $$-$$ 1
<br><br>Let QR is y = mx + c
<br><br>y = $$-$$ x + c
<br><br>x + y $$-$$ c = 0
<br><br>distance of QR from (2, 1) is 2$$\sqrt 3 $$
<br><br>2$$\sqrt 3 $$ = $${{\left| {2 + 1 - c} \right|} \over {\sqrt 2 }}$$
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266026/exam_images/irq1wiy4pougehf2jvx7.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 9th April Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 108 English Explanation">
<br><br>2$$\sqrt 6 $$ = $$\left| {3 - c} \right|$$
<br><br>c $$-$$ 3 = $$ \pm 2\sqrt 6 $$ c = 3 $$ \pm $$ 2$$\sqrt 6 $$
<br><br>Line can be x + y = 3 $$ \pm $$ 2$$\sqrt 6 $$
<br><br>x + y = 3 $$-$$ 2$$\sqrt 6 $$ |
A square, of each side 2, lies above the x-axis and has one vertex at the origin. If
one of the sides passing through the origin makes an angle 30<sup>o</sup> with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square is :
Options:
[{"identifier": "A", "content": "$$2\\sqrt 3 - 1$$ "}, {"identifier": "B", "content": "$$2\\sqrt 3 - 2$$"}, {"identifier": "C", "content": "$$\\sqrt 3 - 2$$"}, {"identifier": "D", "content": "$$\\sqrt 3 - 1$$"}] | ["B"]
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265670/exam_images/k7tcjadeqglnae8jaifx.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Online) 9th April Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 110 English Explanation">
<br><br>Let, coordinate of point A = (x, y).
<br><br>$$\therefore\,\,\,$$ For point A,
<br><br>$${x \over {\cos {{30}^ \circ }}}$$ = $${y \over {\sin {{30}^ \circ }}}$$ = 2
<br><br>$$ \Rightarrow $$ x = $$\sqrt 3 $$
<br><br>and y = 1
<br><br>Similarly, For point B,
<br><br>$${x \over {\cos {{75}^ \circ }}}$$ = $${y \over {\sin {{75}^ \circ }}}$$ = 2$$\sqrt 2 $$
<br><br>$$\therefore\,\,\,$$ x = $$\sqrt 3 - 1$$
<br><br>y = $$\sqrt 3 + 1$$
<br><br>For point C,
<br><br>$${x \over {cos{{120}^ \circ }}}$$ = $${y \over {sin{{120}^ \circ }}}$$ = 2
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x = $$-$$1
<br><br>y = $$\sqrt 3 $$
<br><br>$$\therefore\,\,\,$$ Sum of the x - coordinate of the vertices
<br><br>= 0 + $$\sqrt 3 $$ + $$\sqrt 3 $$ $$-$$ 1 + ($$-$$ 1) = 2$$\sqrt 3 $$ $$-$$ 2 |
The sides of a rhombus ABCD are parallel to the lines, x $$-$$ y + 2 = 0 and 7x $$-$$ y + 3 = 0. If the diagonals of the rhombus intersect P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the coordinate of A is :
Options:
[{"identifier": "A", "content": "$${5 \\over 2}$$"}, {"identifier": "B", "content": "$${7 \\over 4}$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$${7 \\over 2}$$"}] | ["A"]
Explanation:
Let the coordinate A be (0, c)
<br><br>Equations of the given lines are
<br><br>x $$-$$ y + 2 = 0 and 7x $$-$$ y + 3 = 0
<br><br>We know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; y = x + 2 and y = 7x + 3
<br><br>$$\therefore\,\,\,$$ equation of angle bisectors is given as :
<br><br>$${{x - y + 2} \over {\sqrt 2 }} = \pm {{7x - y + 3} \over {5\sqrt 2 }}$$
<br><br>5x $$-$$ 5y + 10 = $$ \pm $$ (7x $$-$$ y + 3)
<br><br>$$\therefore\,\,\,$$ Parallel equations of the diagonals are 2x + 4y $$-$$ 7 = 0
<br><br>and 12x $$-$$ 6y + 13 = 0
<br><br>$$\therefore\,\,\,$$ slopes of diagonals are $${{ - 1} \over 2}$$ and 2.
<br><br>Now, slope of the diagonal from A(0, c) and passing through P(1, 2) is (2 $$-$$ c)
<br><br>$$\therefore\,\,\,$$ 2 $$-$$ c = 2 $$ \Rightarrow $$ c = 0 (not possible)
<br><br>$$ \therefore $$$$\,\,\,$$ 2 $$-$$ c = $${{ - 1} \over 2}$$ $$ \Rightarrow $$ c = $${5 \over 2}$$
<br><br>$$\therefore\,\,\,$$ Coordinate of A is $${5 \over 2}$$. |
A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and
the perpendicular from the origin to this line makes an angle of 60<sup>o</sup> with the line x + y = 0. Then an equation
of the line L is :
Options:
[{"identifier": "A", "content": "x + $$\\sqrt 3 $$y = 8"}, {"identifier": "B", "content": "$$\\sqrt 3 $$x + y = 8"}, {"identifier": "C", "content": "( $$\\sqrt 3 $$ + 1)x + ( $$\\sqrt 3 $$ \u2013 1)y = 8 $$\\sqrt 2 $$"}, {"identifier": "D", "content": "( $$\\sqrt 3 $$ - 1)x + ( $$\\sqrt 3 $$ + 1)y = 8 $$\\sqrt 2 $$"}] | ["D"]
Explanation:
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263835/exam_images/thzbx3newaxtuygg2q1m.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Straight Lines and Pair of Straight Lines Question 87 English Explanation"><br><br>
The equation of line is<br>
x cos $$\theta $$ + y sin $$\theta $$ = p<br><br>
$$ \Rightarrow $$ x cos (75<sup>o</sup>) + y sin (75<sup>o</sup>) = 4<br><br>
$$ \Rightarrow $$ $$x\left( {{{\sqrt 3 - 1} \over {2\sqrt 2 }}} \right) + y\left( {{{\sqrt 3 + 1} \over {2\sqrt 2 }}} \right) = 4$$<br><br>
$$ \Rightarrow $$ $$x\left( {\sqrt 3 - 1} \right) + y\left( {\sqrt 3 + 1} \right) = 8\sqrt 2 $$ |
The region represented by| x – y | $$ \le $$ 2 and | x + y| $$ \le $$ 2 is bounded by a :
Options:
[{"identifier": "A", "content": "rhombus of area 8$$\\sqrt 2 $$ sq. units"}, {"identifier": "B", "content": "square of side length 2$$\\sqrt 2 $$ units"}, {"identifier": "C", "content": "square of area 16 sq. units"}, {"identifier": "D", "content": "rhombus of side length 2 units "}] | ["B"]
Explanation:
$${C_1}{\rm{ }}:{\rm{ }}\left| {y{\rm{ }}-{\rm{ }}x} \right|{\rm{ }} \le {\rm{ }}2$$<br><br>
$${C_2}{\rm{ }}:{\rm{ }}\left| {y{\rm{ + }}x} \right|{\rm{ }} \le {\rm{ }}2$$<br><br>
Now region is square<br><br>
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shown figure is square with side length 2$$\sqrt 2$$ . |
Slope of a line passing through P(2, 3) and
intersecting the line, x + y = 7 at a distance of
4 units from P, is :
Options:
[{"identifier": "A", "content": "$${{\\sqrt 7 - 1} \\over {\\sqrt 7 + 1}}$$"}, {"identifier": "B", "content": "$${{\\sqrt 5 - 1} \\over {\\sqrt 5 + 1}}$$"}, {"identifier": "C", "content": "$${{1 - \\sqrt 5 } \\over {1 + \\sqrt 5 }}$$"}, {"identifier": "D", "content": "$${{1 - \\sqrt 7 } \\over {1 + \\sqrt 7 }}$$"}] | ["D"]
Explanation:
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266822/exam_images/vodxfa1zmymjb9pd25x3.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263362/exam_images/jeb4k12c2xkbyeykmyvx.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266409/exam_images/r46trc585juw1pxqfahv.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 91 English Explanation"></picture>
<br><br>We know parametric form of straight line is
<br><br>$${{x - {x_1}} \over {\cos \theta }} = {{y - {y_1}} \over {\sin \theta }} = r$$
<br><br>$$ \Rightarrow $$ $${{x - 2} \over {\cos \theta }} = {{y - 3} \over {\sin \theta }} = 4$$
<br><br>$$ \therefore $$ x = 4 + 2cos$$\theta $$
<br><br>y = 4 + 3sin$$\theta $$
<br><br>$$ \therefore $$ Point A = (4 + 2cos$$\theta $$, 4 + 3sin$$\theta $$)
<br><br>Point A lies on line x + y = 7,
<br><br>$$ \therefore $$ (4 + 2cos$$\theta $$) + (4 + 3sin$$\theta $$) = 7
<br><br>$$ \Rightarrow $$ sin$$\theta $$ + cos$$\theta $$ = $${1 \over 2}$$
<br><br>$$ \Rightarrow $$ $${\sin ^2}\theta + {\cos ^2}\theta $$ + 2sin$$\theta $$cos$$\theta $$ = $${1 \over 4}$$
<br><br>$$ \Rightarrow $$ 1 + sin 2$$\theta $$ = $${1 \over 4}$$
<br><br>$$ \Rightarrow $$ sin 2$$\theta $$ = $$ - {3 \over 4}$$
<br><br>$$ \Rightarrow $$ $${{2\tan \theta } \over {1 + {{\tan }^2}\theta }}$$ = $$ - {3 \over 4}$$
<br><br>$$ \Rightarrow $$ 3tan<sup>2</sup>$$\theta $$ + 8tan$$\theta $$ + 3 = 0
<br><br>$$ \Rightarrow $$ tan$$\theta $$ = $${{ - 8 \pm 2\sqrt 7 } \over 6}$$
<br><br>So slope = $${{ - 8 \pm 2\sqrt 7 } \over 6}$$
<br><br>By checking each options,
<br><br>Slope = $${{1 - \sqrt 7 } \over {1 + \sqrt 7 }}$$
<br><br>As $${{1 - \sqrt 7 } \over {1 + \sqrt 7 }}$$ = $${{{{\left( {1 - \sqrt 7 } \right)}^2}} \over {1 - 7}}$$ = $${{8 - 2\sqrt 7 } \over { - 6}}$$ = $${{ - 8 + 2\sqrt 7 } \over 6}$$ |
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