A) \[(1,\,\sqrt{3}/2)\]
B) \[(2/3,\,\,\,1\,\sqrt{3})\]
C) \[(2/3,\,\,\sqrt{3}/2)\]
D) \[(1,\,\,1/\sqrt{3})\]
Correct Answer: D
Solution :
Let the vertices of a triangle be \[A(1,\sqrt{3}),\,\,\,\,\,B(0,0),\,\,\,\,\,\,\,C(2,0)\] Let \[a=BC=\sqrt{{{(2-0)}^{2}}+{{0}^{2}}}=2\] \[b=AC=\sqrt{{{(2-1)}^{2}}+{{(0-\sqrt{3})}^{2}}}=2\] and \[c=BA=\sqrt{{{(1-0)}^{2}}+{{(\sqrt{3}-0)}^{2}}}=2\] \[\therefore \] Incentre of triangle, \[I=\left[ \frac{a{{x}_{1}}+b{{x}_{2}}+c{{x}_{3}}}{a+b+c},\,\,\frac{a{{y}_{1}}+b{{y}_{2}}+c{{y}_{3}}}{a+b+c} \right]\] \[=\left( \frac{2(1)+2(0)+2(2)}{2+2+2},\,\frac{2(\sqrt{3})+2(0)+2(0)}{2+2+2} \right)\] ie, \[\left( \frac{6}{6},\,\frac{2\sqrt{3}}{6} \right)\] or \[\left( 1,\,\,\frac{1}{\sqrt{3}} \right)\]
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