Question

Show that the points $(-4,-4),(-2,-4),(4,0)$ and $(2,3)$ are not the vertices points of a rectangle.

Answer 1

Let, rectangle be ABCD, where $A=(-4,-4),B=(-2,-4),C=(4,0)$ and $D=(2,3)$

In Rectangle, opposite sides are equal and diagonals are equal.

Distance between two points $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$AB=\sqrt{{(-4+2)}^2+{\left(0\right)}^2}=2BC=\sqrt{{(-2-4)}^2+{(-4-0)}^2}=\sqrt{52}CD=\sqrt{{(4-2)}^2+{(3-0)}^2}=\sqrt{13}DA=\sqrt{{(-4-2)}^2+{(-4-3)}^2}=\sqrt{85}AC=\sqrt{{(-4-4)}^2+{(-4)}^2}=\sqrt{80}BD=\sqrt{{(-2-2)}^2+{(-4-3)}^2}=\sqrt{65}$

No side is equal,

Hence, It is not a rectangle.