A) \[\left( 1,\frac{\sqrt{3}}{2} \right)\]
B) \[\left( \frac{2}{3},\frac{1}{\sqrt{3}} \right)\]
C) \[\left( \frac{2}{3},\frac{\sqrt{3}}{2} \right)\]
D) \[\left( 1,\frac{1}{\sqrt{3}} \right)\]
Correct Answer: D
Solution :
Let\[A(1,\sqrt{3}),B(0,0),C(2,0)\]be the given points. \[\therefore \] \[a=BC\] \[=\sqrt{{{(2-0)}^{2}}+{{(0-0)}^{2}}}\] \[=2\] \[b=CA\] \[=\sqrt{{{(2-1)}^{2}}+(0-\sqrt{3})}\] \[c=AB\] \[=\sqrt{1+3}=2\] \[\therefore \]Triangle is equilateral. \[\therefore \]In centre is the same as centroid of the triangle \[\therefore \]In centre is\[\left( \frac{1+0+2}{3},\frac{\sqrt{3}+0+0}{3} \right),\]i.e. \[\left( 1,\frac{1}{\sqrt{3}} \right)\]
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