If P(0,0), Q(1,0) and Rleft(frac{1}{2}, frac{ | vedclass

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(C) The points $P(0,0), Q(1,0)$ and $R\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ form an equilateral triangle because the side lengths are:$PQ = \s

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$\left(\frac{1}{2}, \frac{1}{2 \sqrt{3}}\right)$

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(C) The points $P(0,0), Q(1,0)$ and $R\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ form an equilateral triangle because the side lengths are:$PQ = \sqrt{(1-0)^2 + (0-0)^2} = 1$$QR = \sqrt{(\frac{1}{2}-1)^2 + (\frac{\sqrt{3}}{2}-0)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$$RP = \sqrt{(0-\frac{1}{2})^2 + (0-\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$Since the triangle is equilateral, the centre of the inscribed circle (incenter) is the same as the centroid.The coordinates of the incenter $I$ are given by $\left(\frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}\right)$, where $a, b, c$ are the side lengths opposite to vertices $P, Q, R$ respectively.Here $a=b=c=1$, so $I = \left(\frac{0+1+\frac{1}{2}}{3}, \frac{0+0+\frac{\sqrt{3}}{2}}{3}\right) = \left(\frac{\frac{3}{2}}{3}, \frac{\frac{\sqrt{3}}{2}}{3}\right) = \left(\frac{1}{2}, \frac{\sqrt{3}}{6}\right) = \left(\frac{1}{2}, \frac{1}{2\sqrt{3}}\right)$.

If $P(0,0), Q(1,0)$ and $R\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ are three given points, then the centre of the circle for which the lines $PQ, QR$ and $RP$ are the tangents is

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