Determine the area in square units of parallelogram $ABCD$ with vertices $A (-6, 2)$, $B (12, 2)$, $C (12, -2)$ and $D (-6, -2)$.
A(-6,2) B (12,2)
D(-6, -2) C (12, -2)
AB = 12 - -6 = 18
BC = 2 - - 2 = 4
AB * BC = 18 * 4 = 72 = the area
+36 
