Question

B) Show that the points (1,1), (-1,-1) and ( -root3, root 3 ) are the vertices of an equilateral triangle.

Answer 1

Given:

The vertices of a triangle are $(1,1),(-1,-1),(-\sqrt{3},\sqrt{3})$.

To prove:

The given vertices are the vertices of an equilateral triangle.

Solution:

Distance formula:

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

Let the vertices of the triangle are $A(1,1),B(-1,-1),C(-\sqrt{3},\sqrt{3})$. Then, by using the distance formula, we get

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

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

$AB=\sqrt{4+4}$

$AB=\sqrt{8}$

Similarly,

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

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

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

$BC=\sqrt{1+3+1+3}$

$BC=\sqrt{8}$

And,

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

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

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

$CA=\sqrt{1+3+1+3}$

$CA=\sqrt{8}$

Clearly, $AB=BC=CA$.

Since all sides of the given triangle are equal, therefore the given vertices are the vertices of an equilateral triangle.

Hence proved.