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

Question

Given three points $P, Q, R$ with $P(5,3)$ 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 $	riangle PQR$ lies on the line:

Accepted Answer

$R$ on $x-2y=2$ with $y=0$: $x=2$, so $R=(2,0)$.

$Q$ has the same $y$-coordinate as $P$ (since $PQ\parallel x$-axis), so $y_Q=3$; on $x-2y=2$: $x-6=2\implies x=8$, so $Q=(8,3)$.

$$\text{Centroid} = \left(\frac{5+8+2}3,\frac{3+3+0}3\right) = (5,2)$$

Check: $2(5)-5(2)=0$  --- lies on $2x-5y=0$.

Solutions · 1