Let ABC be a triangle formed by the lines 7x -- 6y + 3 = 0, x + 2y -- 31 = 0 and 9x -- 2y …

Question

Let \Delta ABC be a triangle formed by the lines 7x -- 6y + 3 = 0, x + 2y -- 31 = 0 and 9x -- 2y -- 19 = 0. Let the point (h, k) be the image of the centroid of \Delta ABC in the line 3x + 6y -- 53 = 0. Then h2 + k2 + hk is equal to :

Accepted Answer

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

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

$$\begin{aligned}
&\begin{aligned}
& \therefore\left(\frac{\mathrm{k}-\frac{28}{3}}{\mathrm{~h}-\frac{17}{3}}\right)\left(-\frac{1}{2}\right)=-1 \
& & 3\left(\frac{\mathrm{~h}+\frac{17}{3}}{2}\right)+6\left(\frac{\mathrm{k}+\frac{28}{3}}{2}\right)=53
\end{aligned}\
\&\text { Solving (1) \& (2) we get } \mathrm{h}=3, \mathrm{k}=4\
\&\therefore \mathrm{h}^2+\mathrm{k}^2+\mathrm{hk}=37
\end{aligned}$$

Solutions · 1