Q. Given three points $P, Q, R$ with $P(5, 3)$ and $R$ lies on the $x$-axis. If equation of $RQ$ is $x — 2y = 2$ and $PQ$ is parallel to the $x$-axis, then the centroid of $\Delta PQR$ lies on the line :
Solution:
equation of $RQ\equiv x-2y=2$
$\Rightarrow R \left(2, 0\right)$
equation of $PQ = y = 3$
point of intersection of $PQ$ and $RQ$
$x - 2\left(3\right) = 2$
$x=8$
$\Rightarrow R\left(8, 3\right)$
Centroid $\left(\frac{2+8+5}{3}, \frac{0+3+3}{3}\right)$
$\equiv \left(5, 2\right)$ as is simplified by $2x - 5y = 0$
