Question

Show that the following point taken in order form the vertices of a rhombus.

(2, -3), (6, 5), (-2, 1) and (-6, -7)

Answer 1

Consider the given points.

$A(2,-3),B(6,5),C(-2,1),D(-6,-7)$

So,

$AB=\sqrt{(6-2{)}^2+(5+3{)}^2}$

$AB=\sqrt{(4{)}^2+(8{)}^2}$

$AB=\sqrt{16+64}=\sqrt{80}$

Similarly,

$BC=\sqrt{(-2-6{)}^2+(1-5{)}^2}$

$BC=\sqrt{(-8{)}^2+(-4{)}^2}$

$BC=\sqrt{64+16}=\sqrt{80}$

Similarly,

$CD=\sqrt{(-6+2{)}^2+(-7-1{)}^2}$

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

$CD=\sqrt{16+64}=\sqrt{80}$

Similarly,

$DA=\sqrt{(2+6{)}^2+(-3+7{)}^2}$

$DA=\sqrt{(8{)}^2+(4{)}^2}$

$DA=\sqrt{64+16}=\sqrt{80}$

Therefore,

$AB=BC=CD=AD$

Now,

$AC=\sqrt{(-2-2{)}^2+(1+3{)}^2}$

$AC=\sqrt{(-4{)}^2+(4{)}^2}$

$AC=\sqrt{16+16}=\sqrt{32}$

Similarly,

$BD=\sqrt{(-6-6{)}^2+(-7-5{)}^2}$

$BD=\sqrt{(-12{)}^2+(-12{)}^2}$

$BD=\sqrt{144+144}=\sqrt{288}$

So,

$AC\ne BD$

As all the sides are equal and diagonals are not equal.

Its shows that the following vertices are of rhombus.

Hence, this is the answer.