The triangle with the vertices (1, 1), (-1, - | vedclass

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(D) 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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an equilateral

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

The triangle with the vertices $(1, 1)$, $(-1, -1)$, and $(-\sqrt{3}, \sqrt{3})$ is $\ldots \ldots \ldots$ triangle.

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