If the points (1, 1),(-1, -1),and (-sqrt{3}, | vedclass

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(A) Let the vertices be $A(1, 1)$,$B(-1, -1)$,and $C(-\sqrt{3}, \sqrt{3})$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB =

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Equilateral

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(A) Let the vertices be $A(1, 1)$,$B(-1, -1)$,and $C(-\sqrt{3}, \sqrt{3})$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(-1 - 1)^2 + (-1 - 1)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.$BC = \sqrt{(-\sqrt{3} - (-1))^2 + (\sqrt{3} - (-1))^2} = \sqrt{(1 - \sqrt{3})^2 + (\sqrt{3} + 1)^2} = \sqrt{(1 - 2\sqrt{3} + 3) + (3 + 2\sqrt{3} + 1)} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.$CA = \sqrt{(1 - (-\sqrt{3}))^2 + (1 - \sqrt{3})^2} = \sqrt{(1 + \sqrt{3})^2 + (1 - \sqrt{3})^2} = \sqrt{(1 + 2\sqrt{3} + 3) + (1 - 2\sqrt{3} + 3)} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.Since $AB = BC = CA = 2\sqrt{2}$,all sides are equal.Therefore,the triangle is an equilateral triangle.

If the points $(1, 1)$,$(-1, -1)$,and $(-\sqrt{3}, \sqrt{3})$ are the vertices of a triangle,then this triangle is:

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