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 $\Delta PQR$ lies on the line
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 $\Delta PQR$ lies on the line
- A
$x -2y + 1 = 0$
- B
$2x + y -9 = 0$
- C
$2x -5y = 0$
- D
$5x -2y = 0$
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