Question

Show that ${\displaystyle \square }$ ABCD formed by the vertices A(-4,-7), B(-1,2), C(8,5) and D(5,-4) is a rhombus.

Answer 1

${\displaystyle AD=\sqrt{{(X_2-X_1)}^2+{(Y_2-Y_1)}^2}}$

${\displaystyle =\sqrt{({(5+4)}^2+{(-4+7)}^2}}$

${\displaystyle =\sqrt{81+9}}$

=${\displaystyle \sqrt{90}}$

=${\displaystyle 3\sqrt{10} .................\left(1\right)}$

BC =${\displaystyle \sqrt{{(8+1)}^2+{(5-2)}^2}}$

${\displaystyle =\sqrt{81+9}}$

=${\displaystyle \sqrt{90}}$

${\displaystyle =3\sqrt{10} ............\left(2\right)}$

${\displaystyle AB={\sqrt{-1+4}}^2+{(2+7)}^2}$

${\displaystyle =\sqrt{9}+81}$

${\displaystyle =\sqrt{90}}$

${\displaystyle =3\sqrt{10}}$      ...............(3)

${\displaystyle CD=\sqrt{{(8-5)}^2+{(5+4)}^2}}$

${\displaystyle =\sqrt{9+81}}$

${\displaystyle =\sqrt{90}}$

${\displaystyle =3\sqrt{10}}$      ...............(4)

From (1), (2), (3) and (4); AB = BC = CD = DA
∴ ${\displaystyle \square }$ ABCD is a rhombus.