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 these
Correct Answer: A
Solution :
\[\because \ AB=BC=CA=4\sqrt{5}\], i.e., given triangle is equilateral. (In centre of a triangle are same as the centriod when triangle is equilateral) Hence, incentre = \[\left( \frac{7-1+3+2\sqrt{3}}{3},\frac{1+5+3+4\sqrt{3}}{3} \right)\] \[=\left( 3+\frac{2}{\sqrt{3}},3+\frac{4}{\sqrt{3}} \right)\].You need to login to perform this action.
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