Question

Show that the points
A(2,-2), B(8,4) , C(5,7) ,D(-1,1) are the vertices of a rectangle .

Answer 1

We know the distance between the two points is given by,

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Here $AB=\sqrt{(8-2{)}^2+(4+2{)}^2}$

$=\sqrt{36+36}$

$=6\sqrt{2},$

$BC=\sqrt{(5-8{)}^2+(7-4{)}^2}=\sqrt{9+9}$

$=3\sqrt{2},$

$CD=\sqrt{(-1-5{)}^2+(1-7{)}^2}=\sqrt{36+36}$

$=6\sqrt{2}$ and

$AD=\sqrt{(2+1{)}^2+(-2-1{)}^2}=\sqrt{9+9}$

$=3\sqrt{2}$

$AC=\sqrt{(5-2{)}^2+(7+2{)}^2}=\sqrt{90}$ and

$BD=\sqrt{(-1-8{)}^2+(1-4{)}^2}=\sqrt{90}$

Here$AB=CD,BC=AD,AC=BD$

i.e.,opposite sides are equal and diagonal are equal.

Hence it is a rectangle