Question

Show that the points $A(1,0),B(5,3),C(2,7)$ and $D(-2,4)$ are the vertices of a parallelogram.

Answer 1

We have, $A(1,0),B(5,3),C(2,7)andD(-2,4)$

So by distance formula we have,

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

$AB=\sqrt{{(5-1)}^2+{\left(3\right)}^2}=5BC=\sqrt{{(5-2)}^2+{(3-7)}^2}=5CD=\sqrt{{(-2-2)}^2+{(7-4)}^2}=5DA=\sqrt{{(1+2)}^2+{\left(4\right)}^2}=5$

Opposite sides of a parallelogram are equal. Here, Also opposite sides are equal.

$\therefore ABCD$ is a parallelogram. ( HENCE PROVED)