Given:
The vertices of a triangle are [tex](1,1),(-1,-1),(-\sqrt{3},\sqrt{3})[/tex].
To prove:
The given vertices are the vertices of an equilateral triangle.
Solution:
Distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Let the vertices of the triangle are [tex]A(1,1),B(-1,-1),C(-\sqrt{3},\sqrt{3})[/tex]. Then, by using the distance formula, we get
[tex]AB=\sqrt{(-1-1)^2+(-1-1)^2}[/tex]
[tex]AB=\sqrt{(-2)^2+(-2)^2}[/tex]
[tex]AB=\sqrt{4+4}[/tex]
[tex]AB=\sqrt{8}[/tex]
Similarly,
[tex]BC=\sqrt{(-\sqrt{3}-(-1))^2+(\sqrt{3}-(-1))^2}[/tex]
[tex]BC=\sqrt{(1-\sqrt{3})^2+(1+\sqrt{3})^2}[/tex]
[tex]BC=\sqrt{(1)^2+(\sqrt{3})^2-2\sqrt{3}+(1)^2+(\sqrt{3})^2+2\sqrt{3}}[/tex]
[tex]BC=\sqrt{1+3+1+3}[/tex]
[tex]BC=\sqrt{8}[/tex]
And,
[tex]CA=\sqrt{(1-(-\sqrt{3}))^2+(1-\sqrt{3})^2}[/tex]
[tex]CA=\sqrt{(1+\sqrt{3}))^2+(1-\sqrt{3})^2}[/tex]
[tex]CA=\sqrt{(1)^2+(\sqrt{3})^2+2\sqrt{3}+(1)^2+(\sqrt{3})^2-2\sqrt{3}}[/tex]
[tex]CA=\sqrt{1+3+1+3}[/tex]
[tex]CA=\sqrt{8}[/tex]
Clearly, [tex]AB=BC=CA[/tex].
Since all sides of the given triangle are equal, therefore the given vertices are the vertices of an equilateral triangle.
Hence proved.
