The incentre of the triangle with vertices A( | vedclass

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(B) Let the vertices be $A(x_1, y_1) = (1, \sqrt{3})$,$B(x_2, y_2) = (0, 0)$,and $C(x_3, y_3) = (2, 0)$.First,calculate the lengths of the sides oppos

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

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(B) Let the vertices be $A(x_1, y_1) = (1, \sqrt{3})$,$B(x_2, y_2) = (0, 0)$,and $C(x_3, y_3) = (2, 0)$.First,calculate the lengths of the sides opposite to the vertices:$a = BC = \sqrt{(2-0)^2 + (0-0)^2} = 2$$b = AC = \sqrt{(2-1)^2 + (0-\sqrt{3})^2} = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$$c = AB = \sqrt{(1-0)^2 + (\sqrt{3}-0)^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2$Since $a=b=c=2$,the triangle is an equilateral triangle.The incentre $(I)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}\right)$.Substituting the values:$I = \left(\frac{2(1) + 2(0) + 2(2)}{2+2+2}, \frac{2(\sqrt{3}) + 2(0) + 2(0)}{2+2+2}\right)$$I = \left(\frac{2+0+4}{6}, \frac{2\sqrt{3}}{6}\right)$$I = \left(\frac{6}{6}, \frac{\sqrt{3}}{3}\right) = \left(1, \frac{1}{\sqrt{3}}\right)$.

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

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