Question

Prove that the points $(2,-2),(8,4),(5,7),$ and $(-1,1)$ are the angular points of a rectangle.

Answer 1

Given points
$A(2,-2),B(8,4),C(5,7)$ and $D(-1,1)$
$AC=\sqrt{(2-5{)}^2+(-2-7{)}^2}=\sqrt{90}$
$BD=\sqrt{(8-(-1){)}^2+(4-1{)}^2}=\sqrt{90}$
$\Rightarrow$ Diagonals $AC$ and $BD$ are equal. (property of rectangle)
$AB=\sqrt{(2-8{)}^2+(-2-4{)}^2}=\sqrt{72}$
$CD=\sqrt{(5-(-1){)}^2+(7-1{)}^2}=\sqrt{72}$
$AB=CD$
Similarly by distance formula
$BC=AD$
$\Rightarrow$ $AB=CD$ & $BC=AD$ and diagonals are also equal.
$\Rightarrow$ ABCD is a rectangle