Using slopes. Prove that the points A

The property of parallelogram states that opposite sides are equal. 

We have 4 sides as AB,BC,CD,DA

Given points are A(-2,-1),B(1,0),C(4,3) and D(1,2)

AB and CD are opposite sides, and BC and DA are the other two opposite sides.

So slopes of AB = CD and slopes BC = DA

slope $=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$

Slope of $A B=\left(\frac{0+1}{1+2}\right)=\frac{1}{3}$

The slope of $B C=\left(\frac{3-0}{4-1}\right)=\frac{3}{3}=1$

The slope of $C D=\left(\frac{2-3}{1-4}\right)=\frac{-1}{-3}=\frac{1}{3}$

Slope of $D A=\left(\frac{2+1}{1+2}\right)=\frac{3}{3}=1$

Therefore the Slope of $A B=$ Slope of $C D$ and

The slope of $B C=$ Slope of $D A$

Also, the product of slope of two adjacent sides is not equal to $-1$, therefore it is not a rectangle.

Hence ABCD is a parallelogram.