The circumcentre of the triangle formed by th | vedclass

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

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$\left(1, -\frac{1}{\sqrt{3}}\right)$

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(D) Let the vertices be $A(1, \sqrt{3})$,$B(-1, -\sqrt{3})$,and $C(3, -\sqrt{3})$.Calculate the side lengths:$AB = \sqrt{(-1-1)^2 + (-\sqrt{3}-\sqrt{3})^2} = \sqrt{(-2)^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = 4$.$BC = \sqrt{(3 - (-1))^2 + (-\sqrt{3} - (-\sqrt{3}))^2} = \sqrt{4^2 + 0^2} = 4$.$AC = \sqrt{(3-1)^2 + (-\sqrt{3}-\sqrt{3})^2} = \sqrt{2^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = 4$.Since $AB = BC = AC = 4$,the triangle is an equilateral triangle.In an equilateral triangle,the circumcentre coincides with the centroid.The centroid $(G)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.$G = \left(\frac{1-1+3}{3}, \frac{\sqrt{3}-\sqrt{3}-\sqrt{3}}{3}\right) = \left(\frac{3}{3}, \frac{-\sqrt{3}}{3}\right) = \left(1, -\frac{1}{\sqrt{3}}\right)$.

The circumcentre of the triangle formed by the points $A(1, \sqrt{3})$,$B(-1, -\sqrt{3})$,and $C(3, -\sqrt{3})$ is

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