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Show that the points $(a, a)$, $(-a, -a)$ and $(-\sqrt{3}a, \sqrt{3}a)$ are the vertices of an equilateral triangle.
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Let A $(a, a)$, B $(–a, –a)$ and C $(-\sqrt{3}a, \sqrt{3}a)$ be the given points.
AB = $\sqrt{(-a - a)^2 + (-a - a)^2} = \sqrt{8a^2}$ or $2\sqrt{2} a$
BC = $\sqrt{(-\sqrt{3}a + a)^2 + (\sqrt{3}a + a)^2} = \sqrt{8a^2}$ or $2\sqrt{2} a$
CA = $\sqrt{(-\sqrt{3}a – a)^2 + (\sqrt{3}a – a)^2} = \sqrt{8a^2}$ or $2\sqrt{2} a$
Since AB = BC = CA
Therefore, $\triangle$ ABC is an equilateral triangle.
AB = $\sqrt{(-a - a)^2 + (-a - a)^2} = \sqrt{8a^2}$ or $2\sqrt{2} a$
BC = $\sqrt{(-\sqrt{3}a + a)^2 + (\sqrt{3}a + a)^2} = \sqrt{8a^2}$ or $2\sqrt{2} a$
CA = $\sqrt{(-\sqrt{3}a – a)^2 + (\sqrt{3}a – a)^2} = \sqrt{8a^2}$ or $2\sqrt{2} a$
Since AB = BC = CA
Therefore, $\triangle$ ABC is an equilateral triangle.