Let points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ be respectively denoted by $A , B , C ,$ and $D$.

Slopes of $AB =\frac{0+1}{4+2}=\frac{1}{6}$

Slopes of $CD =\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac{1}{6}$

$\Rightarrow$ Slope of $AB =$ Slope of $CD$

$\Rightarrow AB$ and $CD$ are parallel to each other.

Now, slope of $BC =\frac{3-0}{3-4}=\frac{3}{-1}=-3$

Slope of $AD =\frac{2+1}{-3+2}=\frac{3}{-1}=-3$

$\Rightarrow$ Slope of $BC =$ Slope of $AD$

$\Rightarrow BC$ and $AD$ are parallel to each other.

Therefore, both pairs of opposite side of quadrilateral $ABCD$ are parallel. Hence, $ABCD$ is a parallelogram.

Thus, points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are the vertices of a parallelogram.