The incentre of the triangle with vertices (1 | vedclass

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

Vedclass · Accepted Answer

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

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

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

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