Given three points P, Q, R with P(5, 3) and R lies on the x - axis....

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

$⇒ R ≡ \left(2 , 0\right)$ lies


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

$∴ Q ≡ \left(8, 3\right)</p><p>Centroid </p><p>$≡ \left(\dfrac{5 + 8 + 2}{3},\dfrac{3 + 3 + 0}{3}\right) ≡ \left(5 , 2\right)

\left(∵ \dfrac{{x}_{1} + {x}_{2} + {x}_{3}}{3}, \dfrac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right)

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