question_answer

The incentre of a triangle with vertices (7, 1) (-1, 5) and \[(3+2\sqrt{3},\,3+4\sqrt{3})\] is   [J & K 2005]

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 = \[\left( \frac{7-1+3+2\sqrt{3}}{3},\frac{1+5+3+4\sqrt{3}}{3} \right)\]                \[=\left( 3+\frac{2}{\sqrt{3}},3+\frac{4}{\sqrt{3}} \right)\].