The triangle with vertices A(2, 4),B(2, 6),an | vedclass

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(C) The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$AB = \sqrt{(2 - 2)^2 + (6 -

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Equilateral

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(C) The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$AB = \sqrt{(2 - 2)^2 + (6 - 4)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$.$BC = \sqrt{(2 + \sqrt{3} - 2)^2 + (5 - 6)^2} = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.$CA = \sqrt{(2 + \sqrt{3} - 2)^2 + (5 - 4)^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.Since $AB = BC = CA = 2$,all sides are equal.Therefore,the triangle is an equilateral triangle.

The triangle with vertices $A(2, 4)$,$B(2, 6)$,and $C(2 + \sqrt{3}, 5)$ is a . . . .

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