Question

# The incentre of the triangle with vertices (1,$\sqrt 3$ ),(0,0) and (2,0) is:A.$\left( {1,\frac{{\sqrt 3 }}{2}} \right)$ B.$\left( {\frac{2}{3},\frac{1}{{\sqrt 3 }}} \right)$ C.$\left( {\frac{2}{3},\frac{{\sqrt 3 }}{2}} \right)$ D.$\left( {1,\frac{1}{{\sqrt 3 }}} \right)$

Hint-Using the vertices first find out the length of the sides of the triangle and proceed
Let us consider the three vertices to be A=(1,$\sqrt 3$ ),B=(0,0),C=(2,0) which in turn forms a $\vartriangle ABC$
So, to find out the length make use of the distance formula and solve it
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{({y_2} - {y_1})}^2}}$
So, we will get the length of the side AB=$\sqrt {{{\left( {0 - 1} \right)}^2} + {{\left( {0 - \sqrt 3 } \right)}^2}} = \sqrt {1 + 3} = \sqrt 4 = 2$ =c
The length of the side CA=$\sqrt {{{\left( {2 - 1} \right)}^2} + {{\left( {0 - \sqrt 3 } \right)}^2}} = \sqrt {1 + 3} = \sqrt 4 = 2$=b
The length of the side BC=$\sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( {0 - 0} \right)}^2}} = \sqrt {4 + 0} = \sqrt 4 = 2$ =a
So, from this , since the length of sides AB=BC=CA=2, we can conclude that it is an equilateral triangle
Since it is an equilateral triangle, the incentre is nothing but equal to the centroid
So, we can write the coordinates of the incentre are
$\begin{gathered} \left( {\frac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\frac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right) = \left( {\frac{{2 \times 1 + 2 \times 0 + 2 \times 2}}{{2 + 2 + 2}},\frac{{2 \times \sqrt 3 + 2 \times 0 + 2 \times 0}}{{2 + 2 + 2}}} \right) \\ = \left( {\frac{6}{6},\frac{{2\sqrt 3 }}{6}} \right) = \left( {1,\frac{1}{{\sqrt 3 }}} \right) \\ \end{gathered}$
So, option D is the correct answer to this question.

Note: Whenever we are given with these kind of questions, first find out what type of
triangle is formed from these sides and then apply the formula with respect to the type of
triangle formed and solve