Straight Lines JEE Main PYQs (2020–2026)

Solved Straight Lines PYQs

3 recent questions with answers. The full set of 115 is free to practise.

1. JEE Main 2025EasyConcurrency of Lines & Family of Lines
   Q1

   Let ABC be the triangle such that the equations of lines AB and AC be $3y-x=2$ and $x+y=2$ , respectively, and the points B and C lie on $x$ -axis. If P is the orthocentre of the triangle ABC , then the area of the triangle PBC is equal to

   • A8
   • B4
   • C10
   • D6

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   Correct answer: (D)

   Equation of line $AB$ is $3y-x=2$
   
   And $AC$ is $x+y=2$
   
   In line $AB$ ,
   
   When $y=0,x=-2$
   
   $$\therefore B(-2,0)$$
   
   In line $AC$ ,
   
   When $y=0,x=2$
   
   $$\therefore C(2x,0)$$
   
   Equation of altitude of $BC$ ,
   
   $$Y=x+2$$
   
   Similarly, equation of altitude of $AB$ ,
   
   $$y=-3x+6$$
   
   $\therefore$ On solving, orthocentre $P(1,3)$
   
   $$\therefore ar(△PBC)=6$$
2. JEE Main 2025HardSlope, Collinearity & Basic Coordinate Geometry
   Q2

   If A and B are the points of intersection of the circle $x^2+y^2-8x=0$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ and a point P moves on the line $2x-3y+4=0$ , then the centroid of $\Delta PAB$ lies on the line :

   • A$x+9y=36$
   • B$9x-9y=32$
   • C$4x-9y=12$
   • D$6x-9y=20$

   Show answer & solution

   Correct answer: (D)

   $$\begin{array}{rlr}& x^2+y^2-8x=0,\frac{x^2}{9}-\frac{y^2}{4}=1\text{.... (1)}& \\ & 4x^2-9y^2=36\text{.... (2)}& \\ & \text{ Solve }(1)& (2)\\ & 4x^2-9\left(8x-x^2\right)=36& \\ & 13x^2-72x-36=0& \\ & (13x+6)(x=6)=0& \\ & x=\frac{-6}{13},x=6& \\ & x=\frac{-6}{13}(\text{ rejected })& \\ & y\to \text{ Imaginary }& \\ & n=6,\frac{36}{9}-\frac{y^2}{4}=1& \\ & y^2=12,y=I\sqrt{12}& \\ & A(6,\sqrt{12}),B(6,-\sqrt{12})& \\ & p\left(\alpha ,\frac{2\alpha +4}{3}\right)P\text{ lies on }& \end{array}$$
   
   $$\begin{array}{rl}& \text{ centroid }(h,k)2x-3y+y=0\\ & h=\frac{12+\alpha }{3},\alpha =3h-12\\ & k=\frac{\frac{2\alpha -3y}{3}}{3}\Rightarrow 2\alpha +4=9y\\ & \alpha =\frac{9k-4}{2}\\ & 6h-2y=9k-4\\ & 6x-9y=20\end{array}$$
3. JEE Main 2025EasyFoot of Perpendicular, Image of a Point & Reflection
   Q3

   Let the triangle PQR be the image of the triangle with vertices $(1,3),(3,1)$ and $(2,4)$ in the line $x+2y=2$ . If the centroid of $△PQR$ is the point $(\alpha ,\beta )$ , then $15(\alpha -\beta )$ is equal to :

   • A21
   • B19
   • C22
   • D24

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   Correct answer: (C)

   Let ’ $G$ ’ be the centroid of $\Delta$ formed by $(1,3)(3,1)$ & $(2,4)$
   
   $$G\cong \left(2,\frac{8}{3}\right)$$
   
   Image of G w.r.t. $x+2y-2=0$
   
   $$\begin{array}{rl}& \frac{\alpha -2}{1}=\frac{\beta -\frac{8}{3}}{2}=-2\frac{\left(2+\frac{16}{3}-2\right)}{1+4}\\ & =\frac{-2}{5}\left(\frac{16}{3}\right)\\ & \Rightarrow \alpha =\frac{-32}{15}+2=\frac{-2}{15},\beta =\frac{-32\times 2}{15}+\frac{8}{3}=\frac{-24}{15}\\ & 15(\alpha -\beta )=-2+24=22\end{array}$$

Topic-wise question count

Where JEE Main has asked from Straight Lines, most frequent first.

• Slope, Collinearity & Basic Coordinate Geometry37
• Foot of Perpendicular, Image of a Point & Reflection19
• Concurrency of Lines & Family of Lines17
• Angle between Lines & Angle Bisectors14
• Distance of a Point from a Line & Distance between Parallel Lines14
• Various Forms of Equations of a Line13
• General Equation of Second Degree & Homogenization1

Year-wise Straight Lines questions in JEE Main

Number of questions from this chapter in each year's paper(s).