Question

The incentre 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

Answer 1

Answer: (A)

$\left( 3 + \frac{2}{\sqrt{3}},3 + \frac{4}{\sqrt{3}} \right)$

Answer 2

$\because A B = B C = C A = 4 \sqrt { 5 }$ ,

i.e., given triangle is equilateral.

(In centre of a triangle are same as the centriod when triangle is equilateral)

Hence, incentre

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

$= \left( 3 + \frac { 2 } { \sqrt { 3 } } , 3 + \frac { 4 } { \sqrt { 3 } } \right)$.