Given three points $P , Q , R$ with $P ( 5,3 )$ and $R$ lies on the $x$-axis. If the equation of $RQ$ is $x - 2 y = 2$ and $PQ$ is parallel to the $x$-axis, then the centroid of $\triangle PQR$ lies on the line
(1) $x - 2 y + 1 = 0$
(2) $2 x + y - 9 = 0$
(3) $2 x - 5 y = 0$
(4) $5 x - 2 y = 0$
Given three points $P , Q , R$ with $P ( 5,3 )$ and $R$ lies on the $x$-axis. If the equation of $RQ$ is $x - 2 y = 2$ and $PQ$ is parallel to the $x$-axis, then the centroid of $\triangle PQR$ lies on the line\\
(1) $x - 2 y + 1 = 0$\\
(2) $2 x + y - 9 = 0$\\
(3) $2 x - 5 y = 0$\\
(4) $5 x - 2 y = 0$