Show that the following points form an equilateral triangle. (- $\sqrt{3}$ , 1), (2 $\sqrt{3}$ , -2) and (2 $\sqrt{3}$ , 4)

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Solution

Given $A (- \sqrt{3}, 1) B (2 \sqrt{3}, - 2), C (2 \sqrt{3}, 4)$

$A B = \sqrt{(3 \sqrt{3})^{2} + (3)^{2}}$ $= 6$

$B C = \sqrt{(0)^{2} + 6^{2}} = 6$

$A C = \sqrt{(3 \sqrt{3})^{2} + (3)^{2}} = 6$

$∴ A B = B C = A C$

Hence it is an equilateral triangle.

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