The incentre of a triangle with vertices (7, | vedclass

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(A) Let the vertices be $A(7, 1)$,$B(-1, 5)$,and $C(3 + 2\sqrt{3}, 3 + 4\sqrt{3})$.Calculate the side lengths:$AB = \sqrt{(-1 - 7)^2 + (5 - 1)^2} = \s

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$\left( 3 + \frac{2}{\sqrt{3}}, 3 + \frac{4}{\sqrt{3}} \right)$

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(A) Let the vertices be $A(7, 1)$,$B(-1, 5)$,and $C(3 + 2\sqrt{3}, 3 + 4\sqrt{3})$.Calculate the side lengths:$AB = \sqrt{(-1 - 7)^2 + (5 - 1)^2} = \sqrt{(-8)^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$.$BC = \sqrt{(3 + 2\sqrt{3} - (-1))^2 + (3 + 4\sqrt{3} - 5)^2} = \sqrt{(4 + 2\sqrt{3})^2 + (4\sqrt{3} - 2)^2} = \sqrt{(16 + 12 + 16\sqrt{3}) + (48 + 4 - 16\sqrt{3})} = \sqrt{28 + 52} = \sqrt{80} = 4\sqrt{5}$.$CA = \sqrt{(7 - (3 + 2\sqrt{3}))^2 + (1 - (3 + 4\sqrt{3}))^2} = \sqrt{(4 - 2\sqrt{3})^2 + (-2 - 4\sqrt{3})^2} = \sqrt{(16 + 12 - 16\sqrt{3}) + (4 + 48 + 16\sqrt{3})} = \sqrt{28 + 52} = \sqrt{80} = 4\sqrt{5}$.Since $AB = BC = CA$,the triangle is equilateral.For an equilateral triangle,the incentre coincides with the centroid.Centroid = $\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) = \left( \frac{7 - 1 + 3 + 2\sqrt{3}}{3}, \frac{1 + 5 + 3 + 4\sqrt{3}}{3} \right) = \left( \frac{9 + 2\sqrt{3}}{3}, \frac{9 + 4\sqrt{3}}{3} \right) = \left( 3 + \frac{2\sqrt{3}}{3}, 3 + \frac{4\sqrt{3}}{3} \right) = \left( 3 + \frac{2}{\sqrt{3}}, 3 + \frac{4}{\sqrt{3}} \right)$.

The incentre of a triangle with vertices $(7, 1)$,$(-1, 5)$,and $(3 + 2\sqrt{3}, 3 + 4\sqrt{3})$ is

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