Coordinates of Fourth Vertex of Parallelogram

In a parallelogram $ABCD$, the diagonals $AC$ and $BD$ bisect each other. Thus, the midpoint of $AC$ is equal to the midpoint of $BD$.

Midpoint of $AC = \left(\frac{1 + (-2)}{2}, \frac{1 + 8}{2}\right) = \left(-\frac{1}{2}, \frac{9}{2}\right)$

Let $D = (x, y)$. Midpoint of $BD = \left(\frac{3 + x}{2}, \frac{4 + y}{2}\right)$

Equating coordinates: $\frac{3 + x}{2} = -\frac{1}{2} \implies 3 + x = -1 \implies x = -4$ $\frac{4 + y}{2} = \frac{9}{2} \implies 4 + y = 9 \implies y = 5$

So, the coordinates of $D$ are $(-4, 5)$.