Question

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

Answer 1

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

Now, slope of $AB$ $=\frac{0+1}{4+2}=\frac{1}{6}$

Slope 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

Also, 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 are pairs of opposite sides of quadrilateral $ABCD$ are parallel.

Hence $ABCD$ is parallelogram.