Question

Show that the points $\left(2,-2\right),\left(8,4\right),\left(5,7\right)$ and $\left(-1,1\right)$ taken in order constitute the vertices of a Rectangle.

Answer 1

Let these A(2,-2) ,B(8,4) C(5,7) and D(-1,1) vertices of quadrilateral.

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

$=\sqrt{6^2+6^2}$

$=\sqrt{72}$

$=6\sqrt{2}$

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

$=\sqrt{3^2+3^2}$

$=\sqrt{18}$

$=3\sqrt{2}$

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

$=\sqrt{6^2+6^2}$

$=\sqrt{72}$

$=6\sqrt{2}$

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

$=\sqrt{3^2+3^2}$

$=\sqrt{18}$

$=3\sqrt{2}$

now ,

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

$=\sqrt{3^2+9^2}$

$=\sqrt{90}$

$=3\sqrt{10}$

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

$=\sqrt{9^2+3^2}$

$=\sqrt{90}$

$=3\sqrt{10}$

Since , $AB=CD,BC=DA,AC=BD$

so , it is $Rectangle$