Coordinate Geometry – Straight Lines – JEE Main 29 Jan 2025 Shift 1

Solution:

First, find the vertices of the triangle by solving the equations in pairs.

Solving $7x-6y+3=0$ and $x+2y-31=0 \Rightarrow A(9, 11)$.

Solving $7x-6y+3=0$ and $9x-2y-19=0 \Rightarrow B(3, 4)$.

Solving $x+2y-31=0$ and $9x-2y-19=0 \Rightarrow C(5, 13)$.


The centroid $G(x_1, y_1)$ of $\triangle ABC$ is:

$G = \left(\frac{9+3+5}{3}, \frac{11+4+13}{3}\right) = \left(\frac{17}{3}, \frac{28}{3}\right)$.

Let $I(h, k)$ be the image of $G$ in the line $3x+6y-53=0$ ($a=3, b=6, c=-53$).


Using the image formula $\frac{h-x_1}{a} = \frac{k-y_1}{b} = -2 \left( \frac{ax_1+by_1+c}{a^2+b^2} \right)$:

$\frac{h – \frac{17}{3}}{3} = \frac{k – \frac{28}{3}}{6} = -2 \left( \frac{3(\frac{17}{3}) + 6(\frac{28}{3}) – 53}{3^2 + 6^2} \right) $

$\frac{h – \frac{17}{3}}{3} = \frac{k – \frac{28}{3}}{6} = -\frac{8}{9}$

$\frac{h – \frac{17}{3}}{3} = -\frac{8}{9} \Rightarrow h – \frac{17}{3} = -\frac{8}{3} \Rightarrow h = \frac{9}{3} = 3$.

$\frac{k – \frac{28}{3}}{6} = -\frac{8}{9} \Rightarrow k – \frac{28}{3} = -\frac{16}{3} \Rightarrow k = \frac{12}{3} = 4$.

We need to find $h^{2}+k^{2}+hk$:

$= (3)^{2} + (4)^{2} + (3)(4)$

$= 9 + 16 + 12$

$= 37$

Ans. (1)