Question

Find the coordinates of the incenter of the triangle whose vertices are $(7, - 36),(7,20)$ and $( - 8,0)$

Answer 1

Hint: Write down the formula for finding the incenter of the triangle. Find the required variables for the formula which are the distance between two vertices and the coordinates. Substitute the values and evaluate to get the coordinates for the incenter of the triangle.

Formula used: The formula for finding the distance between any two coordinates $({x_1},{y_1}),({x_2},{y_2})$ is given by, $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$
The formula for finding the incenter of a triangle is given that $a,b,c\;$ are the lengths of the sides of the triangle. $(x,y) = \left[ {\left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)} \right]$

Complete step-by-step solution:
To find the incenter we need the lengths of the sides and their coordinates.
$A = (7, - 36),B = (7,20),C = ( - 8,0)$
So, let’s first find out the distance between the coordinates AB
Using the distance formula which is,
$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$
Substitute in the formula.
$\Rightarrow c = \sqrt {{{(7 - 7)}^2} + {{(20 - ( - 36))}^2}}$
$\Rightarrow c = \sqrt {{{(0)}^2} + {{(20 + 36)}^2}}$
Now simplify further to get the distance $c$
$\Rightarrow c = \sqrt {{{(56)}^2}}$
$\Rightarrow c = 56$
Now let’s find the distance between BC
Using the distance formula which is,
$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$
Substitute in the formula.
$\Rightarrow a = \sqrt {{{( - 8 - 7)}^2} + {{(0 - 20)}^2}}$
$\Rightarrow a = \sqrt {{{( - 15)}^2} + {{(20)}^2}}$
Now simplify further to get the distance $a$
$\Rightarrow a = \sqrt {225 + 400}$
$\Rightarrow a = \sqrt {625} = 25$
Now let’s find the distance between CA
Using the distance formula which is,
$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$
Substitute in the formula.
$\Rightarrow b = \sqrt {{{(7 - ( - 8))}^2} + {{( - 36 - 0)}^2}}$
$\Rightarrow b = \sqrt {{{(15)}^2} + {{( - 36)}^2}}$
Now simplify further to get the distance $b$
$\Rightarrow b = \sqrt {225 + 1296}$
$\Rightarrow b = \sqrt {1521} = 39$
Now substitute these in the incenter formula which is,
$(x,y) = \left[ {\left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)} \right]$
Substituting the values, we get,
$\Rightarrow (x,y) = \left[ {\left( {\dfrac{{25{x_1} + 39{x_2} + 56{x_3}}}{{56 + 25 + 39}},\dfrac{{25{y_1} + 39{y_2} + 56{y_3}}}{{56 + 25 + 39}}} \right)} \right]$
Since we have substituted the lengths of the triangle now, we place the values of the coordinates which are, $({x_1},{y_1}) = (7, - 36),({x_2},{y_2}) = (7,20),({x_3},{y_3}) = ( - 8,0)$
$\Rightarrow (x,y) = \left[ {\left( {\dfrac{{(25 \times 7) + (39 \times 7) + (56 \times - 8)}}{{120}},\dfrac{{(25 \times - 36) + (39 \times 20) + (56 \times 0)}}{{120}}} \right)} \right]$
Perform the multiplication operations in the numerator part.
$\Rightarrow (x,y) = \left[ {\left( {\dfrac{{175 + 273 - 448}}{{120}},\dfrac{{ - 900 + 780}}{{120}}} \right)} \right]$
Perform addition operation in the numerator as well
$\Rightarrow (x,y) = \left[ {\left( {\dfrac{0}{{120}},\dfrac{{ - 120}}{{120}}} \right)} \right]$
$\Rightarrow (x,y) = (0, - 1)$

$\therefore$ The incenter for the triangle whose three vertices are given is $(0, - 1)$.

Note: The incenter of a triangle is said to be a point of intersection of all the angular bisectors. While taking the lengths, always take the length of the side opposite to the other angle i.e., if we take the side BC then we consider the angle A which is opposite to them. This is also the reason why the length of BC is named $a$ .