Given, P(5,3) and equation of RQ is x−2y=2
⇒R≡(2,0) 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≡(8,3)
Centroid
≡(35+8+2,33+3+0)≡(5,2)
(∵3x1+x2+x3,3y1+y2+y3)
(5,2) lies on2x−5y=0