Question

Find the incentre of the triangle formed by the vertices $(0,0),(5,12),(16,12)$ .
A. $\left( {7,9} \right)$
B. $\left( {9,7} \right)$
C. $\left( { - 9,7} \right)$
D. $\left( { - 7,9} \right)$

Answer 1

Hint: First say that the given points are A, B and C. Now, obtain the distance between A, B; B, C and C, A. Then name the distances as BC as a, CA as b and AB as c. Then apply the formula of incentre to obtain the required value.

Formula Used:
Distance between two points $(a, b), (c, d)$ is $\sqrt {{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}}$.
The formula to obtain the incentre of the triangle ABC is,
$\left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)$ , where $\left( {{x_i},{y_i}} \right),i = 1,2,3$are the vertices and a, b and c are the distance between the vertices BC, CA and AB respectively.

Complete step by step solution:
Suppose that $A(0,0),B(5,12),C(16,12)$.
The distance between BC is (a)=$\sqrt {{{\left( {16 - 12} \right)}^2} + {{\left( {12 - 5} \right)}^2}} $
That is $a = \sqrt {25 + 144} $
$ = 13$
The distance between CA is (B)=$\sqrt {{{\left( {16 - 0} \right)}^2} + {{\left( {12 - 0} \right)}^2}} $
That is $b = \sqrt {156 + 144} $
$ = 20$
The distance between AB is (c)=$\sqrt {{{\left( {5 - 0} \right)}^2} + {{\left( {12 - 0} \right)}^2}} $
That is $c = \sqrt {25 + 144} $
$ = 13$
The incentre is,
$\left( {\dfrac{{13 \times 0 + 20 \times 5 + 13 \times 13}}{{13 + 13 + 20}},\dfrac{{13 \times 0 + 20 \times 12 + 13 \times 12}}{{13 + 13 + 20}}} \right)$
$ = \left( {\dfrac{{308}}{{44}},\dfrac{{396}}{{44}}} \right) = (7,9)$

Option ‘A’ is correct

Note: Sometime students get confused incentre with the centroid and just add the coordinates and divide them by 3 to obtain the correct answer, but that is not correct, to obtain the incentre we need to find the distance between the vertices first and then applying the formula have to obtain the required incentre.