Let $A B C$ 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:

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Aicharya · Accepted Answer

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

Let \(A B C\) 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…

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