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Show that the points $(-3, -3)$, $(3, 3)$ and $(-3\sqrt{3}, 3\sqrt{3})$ are the vertices of an equilateral triangle.
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Let $A (-3, -3)$, $B (3, 3)$ and $C (-3\sqrt{3}, 3\sqrt{3})$ be the given points.
Using distance formula
$AB = \sqrt{(3 + 3)^2 + (3 + 3)^2} = 6\sqrt{2}$ units
$BC = \sqrt{(-3\sqrt{3}- 3)^2 + (3\sqrt{3} - 3)^2} = 6\sqrt{2}$ units
$CA = \sqrt{(-3 + 3\sqrt{3})^2 + (-3 - 3\sqrt{3})^2} = 6\sqrt{2}$ units
As $AB = BC = CA$, so the given points are the vertices of an equilateral triangle.
Using distance formula
$AB = \sqrt{(3 + 3)^2 + (3 + 3)^2} = 6\sqrt{2}$ units
$BC = \sqrt{(-3\sqrt{3}- 3)^2 + (3\sqrt{3} - 3)^2} = 6\sqrt{2}$ units
$CA = \sqrt{(-3 + 3\sqrt{3})^2 + (-3 - 3\sqrt{3})^2} = 6\sqrt{2}$ units
As $AB = BC = CA$, so the given points are the vertices of an equilateral triangle.