Question

Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinatesof the vertices of a parallelogram and find the angle between its diagonals.

Answer 1

Let ABCD be a quadrilateral

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

Using distance formula

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

$=\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13}$

$BC=\sqrt{(2-0{)}^2+(3-2{)}^2}=\sqrt{4+1}=\sqrt{5}$

$CD=\sqrt{(4-2{)}^2+(0-3{)}^2}=\sqrt{4+9}=\sqrt{13}$

$DA=\sqrt{(4-2{)}^2+(0+1{)}^2}=\sqrt{4+1}=\sqrt{5}$

Since opposite sides (AB and CD) and (BC and DA) are equal.

$\therefore$ The given quadrilateral is a parallelogram.