Question

The vertices of a triangle are \[\left( {0,0} \right),\left( {\sqrt 3 ,3} \right),\left( { - \sqrt 3 ,3} \right)\] then the in-centre is
\[\left( {0,2} \right)\]
\[\left( {2,0} \right)\]
\[\left( {1,1} \right)\]
\[\left( {1,2} \right)\]

Answer 1

Hint:
Here, we have to find the Incentre of the triangle with the given vertices of a triangle. We will use the distance formula to find the length of the sides of a triangle. We will then substitute the length of the sides in the Incentre formula and solve it further to find the Incentre of the triangle.

Formula Used:
We will use the following formula:
1) Distance between two points is given by \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \] where \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)\] are the two points respectively.
2) Incentre of a triangle is given by the formula \[I = \left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)\] where \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\] are the vertices of the triangle and \[a,b,c\] are the length of the sides of the triangle respectively.
3) Rules of surds: \[a\sqrt c \pm b\sqrt c = \left( {a \pm b} \right)\sqrt c \]

Complete Step by step Solution:
We are given the vertices of a triangle \[\left( {0,0} \right),\left( {\sqrt 3 ,3} \right),\left( { - \sqrt 3 ,3} \right)\]. We will draw the diagram using the given information

Now, we will find the lengths of each side of the triangle from the vertices of a triangle using the distance formula.
First, we will find the length of the side AB of a triangle .
Substituting \[{x_1} = 0,{y_1} = 0,{x_2} = \sqrt 3 \] and \[{y_2} = 3\] in the formula \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \], we get
Length of AB \[ = \sqrt {{{\left( {\sqrt 3 - 0} \right)}^2} + {{\left( {3 - 0} \right)}^2}} = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( 3 \right)}^2}} \]
Applying the exponent on the terms, we get
\[ \Rightarrow \] Length of AB \[ = \sqrt {3 + 9} = \sqrt {12} \]
Computing the square root, we get
\[ \Rightarrow \] Length of AB \[ = 2\sqrt 3 \]
Now, we will find the length of the side BC of a triangle
Substituting \[{x_1} = - \sqrt 3 ,{y_1} = 3,{x_2} = \sqrt 3 \] and \[{y_2} = 3\] in the formula \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \], we get
Length of BC \[ = \sqrt {{{\left( {\sqrt 3 - \left( { - \sqrt 3 } \right)} \right)}^2} + {{\left( {3 - 3} \right)}^2}} = \sqrt {{{\left( {\sqrt 3 + \sqrt 3 } \right)}^2}} \]
Adding the terms, we get
\[ \Rightarrow \] Length of BC \[ = \sqrt {{{\left( {2\sqrt 3 } \right)}^2}} = \sqrt {4 \times 3} \]
Computing the square root, we get
\[ \Rightarrow \] Length of BC \[ = 2\sqrt 3 \]
Now, we will find the length of the side CA of a triangle
Substituting \[{x_1} = 0,{y_1} = 0,{x_2} = - \sqrt 3 \] and \[{y_2} = 3\] in the formula \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \], we get
Length of CA \[ = \sqrt {{{\left( {\left( { - \sqrt 3 } \right) - 0} \right)}^2} + {{\left( {3 - 0} \right)}^2}} = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( 3 \right)}^2}} \]
Applying the exponent on the terms, we get
\[ \Rightarrow \] Length of CA \[ = \sqrt {3 + 9} = \sqrt {12} \]
Computing the square root, we get
\[ \Rightarrow \] Length of CA \[ = 2\sqrt 3 \]
So, we get \[AB = BC = CA = 2\sqrt 3 \]
Since all the sides of a triangle are equal, then the triangle formed with the given vertices is an equilateral triangle.
Substituting the values in the formula Incentre of the triangle \[I = \left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)\], we get
\[ \Rightarrow I = \left( {\dfrac{{2\sqrt 3 \left( 0 \right) + 2\sqrt 3 \left( {\sqrt 3 } \right) + 2\sqrt 3 \left( { - \sqrt 3 } \right)}}{{2\sqrt 3 + 2\sqrt 3 + 2\sqrt 3 }},\dfrac{{2\sqrt 3 \left( 0 \right) + 2\sqrt 3 \left( 3 \right) + 2\sqrt 3 \left( 3 \right)}}{{2\sqrt 3 + 2\sqrt 3 + 2\sqrt 3 }}} \right)\]
By using the rules of surds,
\[ \Rightarrow I = \left( {\dfrac{{2\left( 3 \right) + 2\left( { - 3} \right)}}{{6\sqrt 3 }},\dfrac{{2\sqrt 3 \left( 3 \right) + 2\sqrt 3 \left( 3 \right)}}{{6\sqrt 3 }}} \right)\]
\[ \Rightarrow I = \left( {\dfrac{{6 - 6}}{{6\sqrt 3 }},\dfrac{{6\sqrt 3 + 6\sqrt 3 }}{{6\sqrt 3 }}} \right)\]
Adding and subtracting the terms, we get
\[ \Rightarrow I = \left( {\dfrac{0}{{6\sqrt 3 }},\dfrac{{12\sqrt 3 }}{{6\sqrt 3 }}} \right)\]
By simplifying the terms, we get
\[ \Rightarrow I = \left( {0,2} \right)\]

Therefore, the Incentre of a triangle is \[\left( {0,2} \right)\]. Thus Option(A) is the correct answer.

Note:
We know that the distance between two points can be used to find the length of the sides of the triangle, since the length and the distance are equal. The Incentre of a triangle is defined as the point of Intersection of the angle bisectors of the angles of a triangle. It is also the centre of the circle which touches all the vertices of a triangle.