Ans: A
Since $AP$ is a median of $\Delta ABC$, then $P$ is the mid-point of $BC$. The coordinates of $P$
$\begin{array}{cl}
= & \left(\dfrac{5+9}{2}, \dfrac{8+2}{2} \right) \\
= & (7, 5)
\end{array}$
Therefore, the equation of the straight line passes through $A$ and $P$ is
$\begin{array}{rcl}
\dfrac{y-3}{x-3} & = & \dfrac{5-3}{7-3} \\
2y-6 & = & x-3 \\
x -2y +3 & = & 0
\end{array}$
