Question

Prove that the points $(4,5),(7,6),(6,3),(3,2)$ are the vertices of a parallelogram. Is it a rectangle?

Answer 1

let the vertices of a parallelogram be $A(4,5),B(7,6),C(6,3)D(3,2)$

So by distance formula we have,

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

$AB=\sqrt{(7-4{)}^2+(6-5{)}^2}=\sqrt{9+1}=\sqrt{10}$

$BC=\sqrt{(6-7{)}^2+(3-6{)}^2}=\sqrt{1+9}=\sqrt{10}$

$CD=\sqrt{(3-6{)}^2+(2-3{)}^2}=\sqrt{9+1}=\sqrt{10}$

$AD=\sqrt{(3-4{)}^2+(2-5{)}^2}=\sqrt{1+9}=\sqrt{10}$

$\therefore$ $AB=CD$ and $BC=AD$

$\therefore$ Opposite sides are equal

Hence,$ABCD$ is a parallelogram (Proved)

Also,$AB=BC=CD=AD=$$\sqrt{10}$ i.e. all the sides are

equal.

Therefore,it is not a rectangle.