The incentre of the triangle with vertices A (1, √3), B (0, 0) and C (2, 0) is

Concept:

Incentre: The incentre is the point where the angle bisectors intersect.

• For a triangle with side lengths a, b, c with vertices at the points (x₁, y₁), (x₂, y₂), (x₃, y₃) the incentre lies at \(\left( {\frac{{{\rm{a}}{{\rm{x}}_1} + {\rm{b}}{{\rm{x}}_2} + {\rm{c}}{{\rm{x}}_3}}}{{{\rm{a}} + {\rm{b}} + {\rm{c}}}},{\rm{\;}}\frac{{{\rm{a}}{{\rm{y}}_1} + {\rm{b}}{{\rm{y}}_2} + {\rm{c}}{{\rm{y}}_3}}}{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}} \right)\)
• For equilateral triangle incentre coincide with the centroid.

Calculation:

OA = 2

\({\rm{OB\;}} = {\rm{\;}}\sqrt {{{\left( {1 - 0} \right)}^2} + {\rm{\;}}{{\left( {\sqrt 3 - 0} \right)}^2}} = 2{\rm{\;}}\)

\({\rm{AB\;}} = {\rm{\;}}\sqrt {{{\left( {1 - 2} \right)}^2} + {\rm{\;}}{{\left( {\sqrt 3 - 0} \right)}^2}} = 2\)

Here OA = OB = AB,

Since, the given triangle is an equilateral triangle.

Therefore, incentre coincide with the centroid.

Incentre of triangle = \(\left( {\frac{{0\; + \;2\; +{\rm{\;}}1}}{3},\frac{{0\; + {\rm{\;}}0\; + {\rm{\;}}\sqrt 3 }}{3}} \right) = {\rm{}}\left( {1,\frac{1}{{\sqrt 3 }}} \right)\)