Question

Prove that the points $A(-2,-1),B(1,0),C(4,3)$ and $D(1,2)$ are the vertices of a parallelogram $ABCD$.

Answer 1

Given points are

$A(-2,-1),B(1,0),C(4,3)$ and $D(1,2)$

A quadrilateral is a parallelogram if the opposite sides are equal.

$\therefore A{B}^2=(-2-1{)}^2+(-1-0{)}^2$

$=9+1$

$=10$

$B{C}^2=(4-1{)}^2+(3-0{)}^2$

$=9+9$

$=18$

$C{D}^2=(1-4{)}^2+(2-3{)}^2$

$9+1$

$=10$

$A{D}^2=(-2-1{)}^2+(-1-2{)}^2$

$=9+9$

$=18$

$\therefore AB=CD=\sqrt{10}$ and $BC=AD=\sqrt{18}$

$\therefore$ The opposite sides of the quadrilateral ABCD are equal, the four points from a parallelogram.