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

Solution: Please refer to video for the figure.Here, slope of $AB=\frac{0-\left(-1\right)}{4-\left(-2\right)}=\frac{1}{6}$
Slope of $BC=\frac{3-0}{3-4}=-3$
Slope of $CD=\frac{2-3}{-3-3}=\frac{1}{6}$
Slope of $DA=\frac{2-\left(-1\right)}{-3-\left(-2\right)}=-3$
As, slope of $AB$ = slope of $CD$
and slope of $BC$ = slope of $AD$
It means, $AB\mid \mid CD$ and $BC\mid \mid AD$
So, ABCD is a parallelogram.