Straight Lines JEE Main previous year questions with solutions

5 solved JEE Main questions on Straight Lines, free to read — no sign-in needed. The full chapter has 196 questions; sign in to attempt the remaining 191 in the exam simulator.

1. Q1JEE Main 2026 (05 Apr, Shift 2)Angle between Lines
   From the point $(-1, -1)$, two rays are sent making angles of $45°$ with the line $x + y = 0$. These rays get reflected from the mirror $x + 2y = 1$. If the equations of the reflected rays are $ax + by = 9$ and $cx + dy = 7$, $a, b, c, d \in \mathbf{Z}$, then the value of $ad + bc$ is _______.
   Show answer & solution

   Answer: 7

   The given line is $x + y = 0$, which has a slope of $m_1 = -1$. Let the slope of an incident ray be $m$. Since it makes an angle of $45^{\circ}$ with the line $x + y = 0$, we have: $\left| \dfrac{m - (-1)}{1 + m(-1)} \right| = \tan 45^{\circ} = 1$ $\Rightarrow \dfrac{m + 1}{1 - m} = \pm 1$ Taking $+1$: $m + 1 = 1 - m \Rightarrow 2m = 0 \Rightarrow m = 0$ Taking $-1$: $m + 1 = -1 + m \Rightarrow 1 = -1$ (which is not possible, implying $m \to \infty$). The incident rays pass through $P(-1, -1)$. Thus, their equations are: Ray 1: $y = -1$ Ray 2: $x = -1$ Next, we find the points of intersection of these incident rays with the mirror $x + 2y = 1$. For Ray 1 ($y = -1$): $x + 2(-1) = 1 \Rightarrow x = 3$. The point of incidence is $A(3, -1)$. For Ray 2 ($x = -1$): $-1 + 2y = 1 \Rightarrow 2y = 2 \Rightarrow y = 1$. The point of incidence is $B(-1, 1)$. The reflected rays will appear to originate from the image of $P(-1, -1)$ in the mirror $x + 2y - 1 = 0$. Let the image be $P'(x', y')$. $\dfrac{x' - (-1)}{1} = \dfrac{y' - (-1)}{2} = -2 \dfrac{1(-1) + 2(-1) - 1}{1^2 + 2^2}$ $\dfrac{x' + 1}{1} = \dfrac{y' + 1}{2} = -2 \left(\dfrac{-4}{5}\right) = \dfrac{8}{5}$ $\Rightarrow x' = \dfrac{8}{5} - 1 = \dfrac{3}{5}$ and $y' = \dfrac{16}{5} - 1 = \dfrac{11}{5}$ So, $P' = \left(\dfrac{3}{5}, \dfrac{11}{5}\right)$. The reflected rays pass through $P'$ and their respective points of incidence. Equation of Reflected Ray 1 (passing through $A$ and $P'$): Slope $m_1' = \dfrac{\dfrac{11}{5} - (-1)}{\dfrac{3}{5} - 3} = \dfrac{\dfrac{16}{5}}{-\dfrac{12}{5}} = -\dfrac{4}{3}$ $y - (-1) = -\dfrac{4}{3}(x - 3) \Rightarrow 3y + 3 = -4x + 12 \Rightarrow 4x + 3y = 9$ Comparing with $ax + by = 9$, we get $a = 4, b = 3$. Equation of Reflected Ray 2 (passing through $B$ and $P'$): Slope $m_2' = \dfrac{\dfrac{11}{5} - 1}{\dfrac{3}{5} - (-1)} = \dfrac{\dfrac{6}{5}}{\dfrac{8}{5}} = \dfrac{3}{4}$ $y - 1 = \dfrac{3}{4}(x - (-1)) \Rightarrow 4y - 4 = 3x + 3 \Rightarrow 3x - 4y = -7 \Rightarrow -3x + 4y = 7$ Comparing with $cx + dy = 7$, we get $c = -3, d = 4$. Finally, we calculate $ad + bc$: $ad + bc = (4)(4) + (3)(-3) = 16 - 9 = 7$ Answer: $7$
2. Q2JEE Main 2025 (02 Apr, Shift 2)Coordinate System
   Let $\mathrm{A}(4,-2), \mathrm{B}(1,1)$ and $\mathrm{C}(9,-3)$ be the vertices of a triangle $A B C$. Then the maximum area of the parallelogram AFDE , formed with vertices $\mathrm{D}, \mathrm{E}$ and F on the sides $\mathrm{BC}, \mathrm{CA}$ and AB of the triangle ABC respectively, is ______ .
   Show answer & solution

   Answer: 3

   Area of $\triangle \mathrm{ABC}=\frac{1}{2}\left|\begin{array}{ccc}4 & -2 & 1 \\ 1 & 1 & 1 \\ 9 & -3 & 1\end{array}\right|$ $=6$ square units Maximum area of $\operatorname{AFDE}=\frac{1}{2} \times 6=3$ sq. units
3. Q3JEE Main 2024 (05 Apr, Shift 1)Basic Forms of Straight Line
   If $\mathrm{A}(1,-1,2), \mathrm{B}(5,7,-6), \mathrm{C}(3,4,-10)$ and $\mathrm{D}(-1,-4,-2)$ are the vertices of a quadrilateral $A B C D$, then its area is :
   1. A.$48 \sqrt{7}$
   2. B.$12 \sqrt{29}$
   3. C.$24 \sqrt{7}$
   4. D.$24 \sqrt{29}$
   Show answer & solution

   Answer: (B)

   $\begin{aligned} & \mathrm{A}(1,-1,2) \\ & \mathrm{B}(5,7,-6) \\ & \mathrm{C}(3,4,-10) \\ & \mathrm{D}(-1,-4,-2) \\ & \text { Area }=\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}|=\frac{1}{2}|(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}) \times(6 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})| \\ & =\frac{1}{2}|112 \hat{\mathrm{i}}-64 \hat{\mathrm{j}}-8 \hat{\mathrm{k}}| \\ & =4|14 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}-\hat{\mathrm{k}}| \\ & =4 \sqrt{196+64+1} \\ & =4 \sqrt{261} \\ & =12 \sqrt{29}\end{aligned}$
4. Q4JEE Main 2022 (25 Jul, Shift 1)Locus
   A line, with the slope greater than one, passes through the point $A\left(4,3\right)$ and intersects the line $x-y-2=0$ at the point $B$. If the length of the line segment $AB$ is $\dfrac{\sqrt{29}}{3}$, then $B$ also lies on the line
   1. A.$2x+y=9$
   2. B.$3x-2y=7$
   3. C.$x+2y=6$
   4. D.$2x-3y=3$
   Show answer & solution

   Answer: (C)

   Given point $B$ lie on the line $x-y-2=0$, so the point will be $B\left({x}_{1},{x}_{1}-2\right)$ Now given distance between $AB$ is $\dfrac{\sqrt{29}}{3}$, so by distance formula, $\sqrt{{\left({x}_{1}-4\right)}^{2}+{\left({x}_{1}-2-3\right)}^{2}}=\dfrac{\sqrt{29}}{3}$ Now on squaring on both side we get, $18{x}_{1}^{2}-162{x}_{1}+340=0$ On solving we get, ${x}_{1}=\dfrac{51}{9}$ or ${x}_{1}=\dfrac{10}{3}$ So, ${y}_{1}=\dfrac{33}{9}$ or ${y}_{1}=\dfrac{4}{3}$ Now we can see only $x+2y=6$ will satisfy the point $\left(\dfrac{10}{3},\dfrac{4}{3}\right)$
5. Q5JEE Main 2021 (31 Aug, Shift 1)Point and Line
   If $p$ and $q$ are the lengths of the perpendiculars from the origin on the lines, $xcosec\alpha -y\sec \alpha =k\cot 2\alpha$ and $x\sin \alpha +y\cos \alpha =k\sin 2\alpha$ respectively, then ${k}^{2}$ is equal to :
   1. A.$2{p}^{2}+{q}^{2}$
   2. B.${p}^{2}+2{q}^{2}$
   3. C.$4{q}^{2}+{p}^{2}$
   4. D.$4{p}^{2}+{q}^{2}$
   Show answer & solution

   Answer: (D)

   ${L}_{1}:xcosec\alpha -y\sec \alpha =k\cot 2\alpha$ $\dfrac{x}{\sin \alpha }-\dfrac{y}{\cos \alpha }=k\dfrac{\cos 2\alpha }{\sin 2\alpha }$ $\dfrac{x\cos \alpha -y\sin \alpha }{\sin \alpha \cos \alpha }=k\dfrac{\cos 2\alpha }{2\sin \alpha \cos \alpha }$ $\Rightarrow x\cos \alpha -y\sin \alpha =\dfrac{k}{2}\cos 2\alpha$ We know that Perpendicular distance from a point $\left({x}_{1},{y}_{1}\right)$to $ax+by+c=0$ is $\left|\dfrac{a{x}_{1}+b{y}_{1}+c}{\sqrt{{a}^{2}+{b}^{2}}}\right|$ Perpendicular distance from $(0,0)$ is $p=\left|\dfrac{0-0-\dfrac{k}{2}\cos 2\alpha }{\sqrt{{\cos }^{2}\alpha +{\sin }^{2}\alpha }}\right|$ $\Rightarrow {p}^{2}=\dfrac{{k}^{2}}{4}{\cos }^{2}2\alpha$ ${L}_{2}:x\sin \alpha +y\cos \alpha -k\sin 2\alpha =0$ Perpendicular distance from $(0,0)$ is $q=\left|\dfrac{0+0-k\sin 2\alpha }{\sqrt{{\sin }^{2}\alpha +{\cos }^{2}\alpha }}\right|$ $\Rightarrow {q}^{2}={k}^{2}{\sin }^{2}2\alpha$ $4{p}^{2}+{q}^{2}={k}^{2}\left({\sin }^{2}2\alpha +{\cos }^{2}2\alpha \right)$ We know that ${\sin }^{2}t+{\cos }^{2}t=1.$ $4{p}^{2}+{q}^{2}={k}^{2}$