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

Step $1:$ Simplification of given data
Let the given points be $A(-2,-1),B(4,0),C(3,3),D(-3,2)$


Let’s calculate slope of $AB,BC,CD$ and $AD$
We know that slope of a line through the points $(x_1,y_1)\&(x_2,y_2)ism=\frac{y_2-y_1}{x_2-x_1}$

Slope of $AB$:
Points are $A(-2,-1),B(4,0)$
$\therefore$ Slope of $AB=\frac{0-(-1)}{4-(-2)}=\frac{1}{4+2}=\frac{1}{6}$

Slope of $CD$:
Points are $C(3,3),D(-3,2)$
$\therefore$ Slope of $CD=\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac{1}{6}$

Slope of $AD$:
Points are $A(-2,-1),D(-3,2)$
$\therefore$ Slope of $AD=\frac{2-(-1)}{-3-(-2)}=\frac{2+1}{-3+2}=\frac{3}{-1}=-3$
Slope of $BC$:
Points are $B(4,0),C(3,3)$
$\therefore$ Slope of $BC=\frac{3-0}{3-4}=\frac{3}{-1}=-3$

Step $2:$ Solve for proving
Slope of $AB$= Slope of $CD$
$\therefore AB||CD$
Slope of $AD$ = Slope of $BC$
$\therefore AD||BC$
Hence, $AB||CD$ and $AD||BC$
i.e., Both pairs of opposite sides of $ABCD$ are parallel.
Hence, $ABCD$ is a parallelogram.