Question

Given three points $P,Q,R$ with $P\equiv (5,3)$ 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 :

• A. $2x+y-9=0$
• B. $x-2y+1=0$
• C. $5x-2y=0$
• D. $2x-5y=0$

Answer 1

The correct option is D $2x-5y=0$

Solution:

Equation of $RQ$ : $x-2y=2$

Since $R$ lies on x-axis$⟹R(2,0)$

Equation of $PQ$ $=y=3$

Point of intersection of $PQ$ and $RQ$ is

$x-(2\times 3)=2$

or, $x=8$

$⟹Q(8,3)$

Centroid: $(\frac{5+8+2}{3},\frac{3+3+0}{3})$

or, Centroid $(5,2)$

$\equiv (5,2)$ as is simplified by $2x-5y=0.$