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Q.
The incentre of a triangle with vertices $(7,1),(-1,5)$ and $(3+2 \sqrt{3}, 3+4 \sqrt{3})$ is
Straight Lines
Solution:
$\because AB = BC = CA = 4\sqrt{5}$,
i.e., given triangle is equilateral. (Incentre of a 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) $
$=\left(3+\frac{2}{\sqrt{3}}, 3+\frac{4}{\sqrt{3}}\right)$
