Given three points P , Q , R with P 5 , 3 and R lies on the x - axis. If the equation of R Q is x - 2 y = 2…

Given, $P\left(5,3\right)$ and equation of $RQ$ is $x-2y=2$

$\Rightarrow R\equiv \left(2,\; 0\right)$ lies


on $x-$axis and $Q$ lies on a line parallel to $x-$axis through $P$ which is $y=3$. Hence, on solving $y=3$ and $x-2y=2\Rightarrow x=8$

$\therefore Q\equiv \left(8,3\right)$

Centroid 

$\equiv \left(\frac{5+8+2}{3},\frac{3+3+0}{3}\right)\equiv \left(5,\; 2\right)$

$\left(\because \frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

$\left(5,\; 2\right)$ lies on$2x-5y=0$