A) \[\left( 3+\frac{2}{\sqrt{3}},\,3+\frac{4}{\sqrt{3}} \right)\]
B) \[\left( 1+\frac{2}{3\sqrt{3}},1+\frac{4}{3\sqrt{3}} \right)\]
C) \[(7,1)\]
D) None of the above
Correct Answer: A
Solution :
Let the vertices of the triangle are \[A(7,1),\,\,\,B(-1,5)\]and \[C(3+2\sqrt{3},\,3+4\sqrt{3})\]. Now, \[AB=\sqrt{{{(7+1)}^{2}}+{{(1-5)}^{2}}}=\sqrt{64+16}=\sqrt{80}\] \[BC=\sqrt{{{(-1-3-2\sqrt{3})}^{2}}+{{(5-3-4\sqrt{3})}^{2}}}\] \[=\sqrt{16+12+16\sqrt{3}+4+48-16\sqrt{3}}\] \[=\sqrt{80}\] and \[CA=\sqrt{{{(3+2\sqrt{3}-7)}^{2}}+{{(3+4\sqrt{3}-1)}^{2}}}\] \[=\sqrt{16+12-16\sqrt{3}+4+48+16\sqrt{3}}\] \[=\sqrt{80}\] Here, \[AB=BC=CA=\sqrt{80}\] \[\therefore \] Given, triangle is an equilateral triangle. So, in centre \[=\left( \frac{7-1+3+2\sqrt{3}}{3},\frac{1+5+3+4\sqrt{3}}{3} \right)\] \[=\left( \frac{9+2\sqrt{3}}{3},\frac{9+4\sqrt{3}}{3} \right)\] \[=\left( 3+\frac{2}{\sqrt{3}},3+\frac{4}{\sqrt{3}} \right)\]
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