Fourth vertex of a parallelogram

Find the fourth vertex $D$ of the parallelogram $ABCD$ with $A(0, 0)$, $B(3, 0)$, and $C(4, 2)$. In a parallelogram the two diagonals $AC$ and $BD$ bisect each other, so they share a midpoint.

The midpoint of $AC$ is $\left( \dfrac{0 + 4}{2}, \dfrac{0 + 2}{2} \right) = (2, 1)$. Writing $D(x, y)$, the midpoint of $BD$ is $\left( \dfrac{3 + x}{2}, \dfrac{0 + y}{2} \right)$, which must equal $(2, 1)$; hence $x = 1$, $y = 2$, that is $D(1, 2)$. The large dots mark the four vertices, with $AB \parallel DC$ and $AD \parallel BC$.