The incentre of the triangle with vertices (1 | vedclass

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(D) Let the vertices be $A(1, \sqrt{3})$,$B(0, 0)$,and $C(2, 0)$.Calculate the side lengths:$AB = \sqrt{(1-0)^2 + (\sqrt{3}-0)^2} = \sqrt{1 + 3} = 2$$

Vedclass · Accepted Answer

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

Vedclass · Answer

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

The incentre of the triangle with vertices $(1, \sqrt{3}), (0, 0)$ and $(2, 0)$ is:

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