Without using the distance formula,show that the points $(-2,-1), (4,0), (3,3),$ and $(-3,2)$ are the vertices of a parallelogram.

(N/A) Let the points $(-2,-1), (4,0), (3,3),$ and $(-3,2)$ be denoted by $A, B, C,$ and $D$ respectively.
The slope of $AB = \frac{0 - (-1)}{4 - (-2)} = \frac{1}{6}$.
The slope of $CD = \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6}$.
Since the slope of $AB = \text{slope of } CD$,$AB$ is parallel to $CD$.
Now,the slope of $BC = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3$.
The slope of $AD = \frac{2 - (-1)}{-3 - (-2)} = \frac{3}{-1} = -3$.
Since the slope of $BC = \text{slope of } AD$,$BC$ is parallel to $AD$.
Since both pairs of opposite sides are parallel,the quadrilateral $ABCD$ is a parallelogram.