Question

Prove that the points (0, 5), B(-2, -2), C(5, 1) and D(7, 7) are the vertices of a rhombus.

Answer 1

Let the given points be A(0, 5), B(-2, -2), C(5, 0) and D(7, 7).

Then ABCD is a quadrilateral in which:

AB = $\sqrt{(-2-0{)}^2+(-2-5{)}^2}=\sqrt{(-2{)}^2+(-7{)}^2}=\sqrt{4+49}=\sqrt{53}\text{units}$

BC = $\sqrt{(5+2{)}^2+(0+2{)}^2}=\sqrt{7^2+2^2}=\sqrt{49+4}=\sqrt{53}\text{units}$

CD = $\sqrt{(7-5{)}^2+(7-0{)}^2}=\sqrt{2^2+7^2}=\sqrt{4+49}=\sqrt{53}\text{units}$

DA = $\sqrt{(7-0{)}^2+(7-5{)}^2}=\sqrt{7^2+2^2}=\sqrt{49+4}=\sqrt{53}\text{units}$

Thus, AB = BC = CD = DA = $\sqrt{53}\text{units}$

Also, AC = $\sqrt{(5-0{)}^2+(0-5{)}^2}=\sqrt{5^2+(-5{)}^2}=\sqrt{50}=5\sqrt{2}\text{units}$

And, BD = $\sqrt{(7+2{)}^2+(7+2{)}^2}=\sqrt{9^2+9^2}=\sqrt{81+81}$

=$\sqrt{162}=9\sqrt{2}\text{units}$

$\therefore AC\ne BD.$

Thus, ABCD is a quadrilateral all of whose sides are equal but its diagonals are not equal.

Hence, ABCD is a rhombus.