Question:
Show that the points $\mathbf{A}(1,1), \mathbf{B}(-1,-1)$ and $\mathbf{C}(-\sqrt{3}, \sqrt{3})$ are the vertices of an equilateral triangle each of whose sides is 22 units.
Solution:
Given: The 3 points are $A(1,1), B(-1,-1)$ and $C(-\sqrt{3}, \sqrt{3})$.
$A B=\sqrt{(-1-1)^{2}+(-1-1)^{2}}$
$=\sqrt{4+4}$
$=2 \sqrt{2}$ units …..(1)
$B C=\sqrt{(-\sqrt{3}+1)^{2}+(\sqrt{3}+1)^{2}}$
$=\sqrt{3-2 \sqrt{3}+1+3+2 \sqrt{3}+1}$
$=2 \sqrt{2}$ units …..(2)
$A C=\sqrt{(-\sqrt{3}-1)^{2}+(\sqrt{3}-1)^{2}}$
$=\sqrt{3+2 \sqrt{3}+1+3-2 \sqrt{3}+1}$
$=2 \sqrt{2}$ units .....(3)
From equations 1,2 and 3, we have
$A B=B C=A C=2 \sqrt{2}$ units.
Therefore, $\triangle \mathrm{ABC}$ is an equilateral triangle each of whose sides is $2 \sqrt{2}$ units.
Hence, proved.