Question

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

(1, 0), (5, 3), (2, 7) and (-2, 4)

Answer 1

Consider the given points.

$A(1,0),B(5,3),C(2,7),D(-2,4)$

So,

$AB=\sqrt{(5-1{)}^2+(3-0{)}^2}$

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

$AB=\sqrt{25}=5$

Similarly,

$BC=\sqrt{(2-5{)}^2+(7-3{)}^2}$

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

$BC=\sqrt{25}=5$

Similarly,

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

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

$CD=\sqrt{25}=5$

Similarly,

$DA=\sqrt{(1+2{)}^2+(0-4{)}^2}$

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

$DA=\sqrt{25}=5$

Therefore,

$AB=BC=CD=AD$

Now,

$AC=\sqrt{(2-1{)}^2+(7-0{)}^2}$

$AC=\sqrt{(1{)}^2+(7{)}^2}$

$AC=\sqrt{1+49}=\sqrt{50}$

Similarly,

$BD=\sqrt{(-2-5{)}^2+(4-3{)}^2}$

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

$BD=\sqrt{49+1}=\sqrt{50}$

So,

$AC=BD$

As all the sides are equal and diagonals are equal.

Its shows that the following vertices are of rhombus.

Hence, this is the answer.