Straight Lines JEE Main previous year questions with solutions

4 solved JEE Main questions on Straight Lines, free to read — no sign-in needed. The full chapter has 177 questions; sign in to attempt the remaining 173 in the exam simulator.

Answers checked against the official answer keys.

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 (27 Jan, Shift 2)Concurrency & Family of Lines
   If the sum of squares of all real values of $\alpha$, for which the lines $2x-y+3=0,6x+3y+1=0$ and $\alpha x+2y-2=0$ do not form a triangle is $p$, then the greatest integer less than or equal to $p$ is ________.
   Show answer & solution

   Answer: 32

   Given lines are $2x-y+3=0$, $6x+3y+1=0$ and $\alpha x+2y-2=0$. These lines will not form a triangle if $\alpha x+2y-2=0$ is concurrent with $2x-y+3=0$ and $6x+3y+1=0$ or parallel to either of them. Case-$1$: Concurrent lines $\Rightarrow \left|\begin{matrix}2 & -1 & 3 \\ 6 & 3 & 1 \\ \alpha & 2 & -2\end{matrix}\right|=0$ $\Rightarrow 2\left(-6-2\right)+1\left(-12-\alpha \right)+3\left(12-3\alpha \right)=0$ $\Rightarrow -16-12-\alpha +36-9\alpha =0$ $\Rightarrow -10\alpha +8=0$ $\Rightarrow \alpha =\dfrac{4}{5}$ Case-$2$: Parallel lines $\dfrac{-\alpha }{6}=\dfrac{-2}{3}$, $-\dfrac{\alpha }{2}=\dfrac{-2}{-1}$ $\Rightarrow \alpha =4$, $\alpha =-4$ $\Rightarrow P=16+16+\dfrac{16}{25}$ $[P]=\left[32+\dfrac{16}{25}\right]=32$
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 $\left(\dfrac{10}{3},\dfrac{4}{3}\right)$