The $(x, y)$ coordinates of vertices P, Q and R of a parallelogram PQRS are $(-3, -2)$, $(1, -5)$ and $(9, 1)$, respectively. If the diagonal SQ intersects the $x$-axis at $(a, 0)$, then the value of $a$ is

In a parallelogram, diagonals bisect each other — meaning the midpoint of diagonal $PR$ equals the midpoint of diagonal $QS$.

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Midpoint of $PR$:

$\left(\dfrac{-3 + 9}{2},\ \dfrac{-2 + 1}{2}\right) = \left(3,\ -\dfrac{1}{2}\right)$

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Let $S = (x_S,\ y_S)$. Setting the midpoint of $QS$ equal to $\left(3,\ -\dfrac{1}{2}\right)$:

$\dfrac{1 + x_S}{2} = 3 \implies x_S = 5$

$\dfrac{-5 + y_S}{2} = -\dfrac{1}{2} \implies y_S = 4$

So $S = (5,\ 4)$.

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Slope of line $SQ$ through $S(5,\ 4)$ and $Q(1,\ -5)$:

$m = \dfrac{-5 - 4}{1 - 5} = \dfrac{-9}{-4} = \dfrac{9}{4}$

Using point-slope form with $Q(1,\ -5)$:

$y + 5 = \dfrac{9}{4}(x - 1)$

$y = \dfrac{9}{4}x - \dfrac{9}{4} - 5$

$y = \dfrac{9}{4}x - \dfrac{29}{4}$

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The line crosses the $x$-axis where $y = 0$:

$0 = \dfrac{9}{4}x - \dfrac{29}{4}$

$\dfrac{9}{4}x = \dfrac{29}{4}$

$x = \dfrac{29}{9}$

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$a = \dfrac{29}{9}$