JEE Main 2025 — Straight Lines Question with Solution

$\begin{array}{rl}\therefore \text{ centroid of }△\mathrm{ABC}& =\left(\frac{9+3+5}{3},\frac{11+4+13}{3}\right)\\ & =\left(\frac{17}{3},\frac{28}{3}\right)\end{array}$
Let image of centroid with respect to line mirror is $\begin{aligned} & (\mathrm{h}, \mathrm{k}) \\ & \therefore\left(\frac{\mathrm{k}-\frac{28}{3}}{\mathrm{~h}-\frac{17}{3}}\right)\left(-\frac{1}{2}\right)=-1 \\ & \& 3\left(\frac{\mathrm{~h}+\frac{17}{3}}{2}\right)+6 \cdot\left(\frac{\frac{\mathrm{k}+28}{3}}{2}\right)=53 \end{aligned}$ Solving (1) \& (2) we get $\mathrm{h}=3,\mathrm{k}=4$ $\therefore {\mathrm{h}}^2+{\mathrm{k}}^2+\mathrm{hk}=37$