Question

Prove that the points $(3,-2),(4,0),(6,-3)$ and $(5,-5)$ are the vertices of a parallelogram.

Answer 1

Let the vertices of a parallelogram be $A(-3,2)$, $B(4,0)$, $C(6,-3)$ and $D(5,-5)$

So by distance formula we have,

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

$AB=\sqrt{{(4+3)}^2+{(0+2)}^2}=\sqrt{5}CD=\sqrt{{(5-6)}^2+{(-5+3)}^2}=\sqrt{5}BC=\sqrt{{(6-4)}^2+{(-3)}^2}=\sqrt{13}AD=\sqrt{{(5-3)}^2+{(-5+2)}^2}=\sqrt{13}$

Opposite site of a parallelogram are equal.

Therefore $AB=CD=\sqrt{5}$ and $BC=AD=\sqrt{13}$

$ABCD$ is a parallelogram (PROVED)