Mathematics · Straight lines

JEE Main 2025 — 22 January, Morning Shift — Question 3

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

Option A:
24
Option B:
19
Option C:
21
Option D:
22
Correct

Answer: D

Step-by-step solution

Let ' $G$ ' be the centroid of $\Delta$ formed by $(1,3)(3,1)$ & $(2,4)$

$\mathrm{G} \cong\left(2, \frac{8}{3}\right)$

Image of $G$ w.r.t. $x+2 y-2=0$

$\frac{\alpha-2}{1}=\frac{\beta-\frac{8}{3}}{2}=-2 \frac{\left(2+\frac{16}{3}-2\right)}{1+4}$

$=\frac{-2}{5}\left(\frac{16}{3}\right)$

$\Rightarrow \alpha=\frac{-32}{15}+2=\frac{-2}{15}, \beta=\frac{-32 \times 2}{15}+\frac{8}{3}=\frac{-24}{15}$

$15(\alpha-\beta)=-2+24=22$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Morning Shift
Subject: Mathematics
Chapter: Straight lines
Topic: Angle between lines, perpendicular distance & distance between parallel lines, foot, image