Mathematics · Straight lines

JEE Main 2025 — 29 January, Morning Shift — Question 48

Let ABC be a triangle formed by the lines $7 x-6 y+3=0, x+2 y-31=0$ and $9 x-2 y-19=0$ , Let the point $(h, k)$

be the image of the centroid of $\Delta A B C$ in the line $3 x+6 y-53=0$ . Then $h^{2}+k^{2}+h k$ is equal to

Answer: A

$\therefore$ centroid of $\triangle \mathrm{ABC}=\left(\frac{9+3+5}{3}, \frac{11+4+13}{3}\right)$ $=\left(\frac{17}{3}, \frac{28}{3}\right)$

Let image of centroid with respect to line mirror is (h,k)

\begin{aligned} & \therefore\left(\frac{\mathrm{k}-\frac{28}{3}}{\mathrm{~h}-\frac{17}{3}}\right)\left(-\frac{1}{2}\right)=-1 \\& 3 \cdot\left(\frac{\mathrm{h}+\frac{17}{3}}{2}\right)+6 \cdot\left(\frac{\mathrm{k}+\frac{28}{3}}{2}\right)=53 \end{aligned}

Solving (1) & (2) we get $\mathrm{h}=3, \mathrm{k}=4$

$\therefore \mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{hk}=37$

Answer key and solution verified before publishing.

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