Parallelogram Diagonal Intersection

In parallelogram $PQRS$, the diagonals $PR$ and $QS$ bisect each other. Midpoint of $PR = \left(\frac{-3+9}{2}, \frac{-2+1}{2}\right) = \left(3, -\frac{1}{2}\right)$. Since this is also the midpoint of $QS$ and $Q = (1, -5)$: $\frac{1 + x_S}{2} = 3 \implies x_S = 5$ $\frac{-5 + y_S}{2} = -\frac{1}{2} \implies y_S = 4$ So $S = (5, 4)$.

The line $SQ$ passes through $(1, -5)$ and $(5, 4)$. Slope of $SQ = \frac{4 - (-5)}{5 - 1} = \frac{9}{4}$. Equation of $SQ$: $y - 4 = \frac{9}{4}(x - 5) \implies 9x - 4y = 29$. Since $SQ$ intersects the x-axis at $(a, 0)$, substitute $y = 0$: $9a = 29 \implies a = \frac{29}{9}$.