question_answer

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

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 :

[a] \[\because AB=BC=CA=4\sqrt{5},\] i.e., given triangle is equalitateral. (Incentre of triangle are same as the centroid when triangle is equilateral) Hence, incentre \[=\left( \frac{7-1+3+2\sqrt{3}}{3},\frac{1+5+3+4\sqrt{3}}{3} \right)\] \[=\left( 3+\frac{2}{\sqrt{3}},3+\frac{4}{\sqrt{3}} \right)\]