Question

show that the points (3,2), (0,5), (-3,2) and (0,-1) are the vertices of a square .

Answer 1

apply the distance formula...and make all the 4sides equal....

Answer 2

Given:

A(3, 2) = (x₁, y₁)

B(0,5) = (x₂, y₂)

C(-3, 2) = (x₃, y₃)

D(0, -1) = (x₄, y₄)

To Prove:

Given points (ABCD) are the vertices of square.

Solution:

By applying the distance formula,

Distance of AB = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

On substituting the values, we get

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

                          = $\sqrt{9+9}$

                          = $\sqrt{18}$

Distance of BC = $\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$

On substituting the value, we get

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

                          = $\sqrt{9+9}$

                          = $\sqrt{18}$

Distance of CD = $\sqrt{(x_4-x_3)^2+(y_4-y_3)^2}$

On substituting the values, we get

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

                          = $\sqrt{9+9}$

                          = $\sqrt{18}$

Distance of DA = $\sqrt{(x_4-x_1)^2+(y_4-y_1)^2}$

On substituting the values, we get

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

                          = $\sqrt{9+9}$

                          = $\sqrt{18}$

All the four sides are equal. Thus, the given points are the vertices of a square.