Question

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

(-4, -7), (-1, 2), (8, 5) and (5, -4)

Answer 1

Consider the given points.

$A(-4,-7),B(-1,2),C(8,5),D(5,-4)$

Distance of $AB$,

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

$AB=\sqrt{(3{)}^2+(9{)}^2}$

$AB=\sqrt{9+81}=\sqrt{90}$

Similarly,

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

$BC=\sqrt{(9{)}^2+(3{)}^2}$

$BC=\sqrt{81+9}=\sqrt{90}$

Similarly,

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

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

$CD=\sqrt{9+81}=\sqrt{90}$

Similarly,

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

$DA=\sqrt{(-9{)}^2+(-3{)}^2}$

$DA=\sqrt{81+9}=\sqrt{90}$

Therefore,

$AB=BC=CD=AD$

Now,

$AC=\sqrt{(8+4{)}^2+(5+7{)}^2}$

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

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

Similarly,

$BD=\sqrt{(5+1{)}^2+(-4-2{)}^2}$

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

$BD=\sqrt{36+36}=\sqrt{72}$

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.