Question

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

(0, 0), (3, 4), (0, 8) and (-3, 4)

Answer 1

Consider the given points.

$A(0,0),B(3,4),C(0,8),D(-3,4)$

Distance of $AB$,

$AB=\sqrt{(3-0{)}^2+(4-0{)}^2}$

$AB=\sqrt{(3{)}^2+(4{)}^2}$

$AB=\sqrt{9+16}=\sqrt{25}=5$

Similarly,

$BC=\sqrt{(0-3{)}^2+(8-4{)}^2}$

$BC=\sqrt{(-3{)}^2+(4{)}^2}$

$BC=\sqrt{9+16}=5$

Similarly,

$CD=\sqrt{(-3-0{)}^2+(4-8{)}^2}$

$CD=\sqrt{(-3{)}^2+(-4{)}^2}$

$CD=\sqrt{9+16}=5$

Similarly,

$DA=\sqrt{(0+3{)}^2+(0-4{)}^2}$

$DA=\sqrt{(3{)}^2+(-4{)}^2}$

$DA=\sqrt{9+16}=5$

Therefore,

$AB=BC=CD=AD$

Now,

$AC=\sqrt{(0-0{)}^2+(8-0{)}^2}$

$AC=\sqrt{(0{)}^2+(8{)}^2}$

$AC=\sqrt{64}=8$

Similarly,

$BD=\sqrt{(-3-0{)}^2+(4-8{)}^2}$

$BD=\sqrt{(-3{)}^2+(-4{)}^2}$

$BD=\sqrt{9+16}=5$

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.