Question

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

Answer 1

Here AB $=\sqrt{(3+1{)}^2+(1-0{)}^2}$

$=\sqrt{17}$

BC$=\sqrt{(2-3{)}^2+(2-1{)}^2}$

$=\sqrt{2}$

CD$=\sqrt{(-2-2{)}^2+(1-2{)}^2}$

$=\sqrt{17}$

AD$=\sqrt{(-2+1{)}^2+(1-0{)}^2}$

$=\sqrt{2}$

Here,$AB=CD$ and $BC=AD$

Therefore its a parallelogram