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 :
Key Idea: If triangle is an equilateral, then orthocentre, incentre and centroid will be same. Let \[A(1,\,\,\sqrt{3}),\,\,B(0,\,\,0),\,\,C(0,\,\,1)\] be the given points. \[\therefore \]\[a=BC\] \[=\sqrt{{{(2-0)}^{2}}+{{(0-0)}^{2}}}=2\] \[\therefore \]\[b=CA\] \[=\sqrt{{{(2-1)}^{2}}+{{(0-\sqrt{3})}^{2}}}=2\] \[c=AB\] \[=\sqrt{{{(0-1)}^{2}}+{{(0-\sqrt{3})}^{2}}}=2\] 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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