Find the incenter of the triangle with vertic | vedclass

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Question

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

Find the incenter of the triangle with vertices $(1, \sqrt{3})$,$(0, 0)$,and $(2, 0)$.

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