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:

• A

  47
• Correct answer
  B

  37
• C

  36
• D

  40

Solution

$\begin{aligned} \therefore \text { centroid of } \triangle \mathrm{ABC} & =\left(\frac{9+3+5}{3}, \frac{11+4+13}{3}\right) \\ & =\left(\frac{17}{3}, \frac{28}{3}\right)\end{aligned}$
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 $\mathrm{h}=3, \mathrm{k}=4$ $\therefore \mathrm{h}^2+\mathrm{k}^2+\mathrm{hk}=37$

Hence, the correct answer is option: 'B'.