Given three points P, Q, R with P=(5,3) and R lies on the x-axis. If equation of RQ is x-y=2 and PQ is parallel to the x-axis, then the centroid of ΔPQR lies on the line :

Solution:Equation of RQ: x-y=2
Since R lies on x-axis ⇒R(2,0)
Equation of PQ=y=3
Point of intersection of PQ and RQ is x-(2×3)=2 or, x=8 ⇒Q(8,3)
Centroid: (5+8+2)/3, (3+3+0)/3 or, Centroid(5,2) as is simplified by 2x-y=0.