The fourth vertex $D$ of a parallelogram $ABCD$ whose three vertices are $A(-2, 3)$, $B(6, 7)$ and $C(8, 3)$ is:

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Step 1: In parallelogram, diagonals bisect each other. Midpoint of $AC = \left(\frac{-2+8}{2}, \frac{3+3}{2}\right) = (3, 3)$. Midpoint of $BD = \left(\frac{6+x}{2}, \frac{7+y}{2}\right) = (3, 3) \Rightarrow x = 0, y = -1$

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