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To find the coordinates of the incenter of a triangle whose vertices are given as $A\left ( -36,15 \right ) B\left ( 0,0 \right ),$ and $C\left ( 20,15 \right )$ .
Option: 1
$\left ( -1,-8 \right )$

Option: 2
$\left ( -1,8 \right )$

Option: 3
$\left ( 8,-1 \right )$

Option: 4
$\left ( -8,-1 \right )$

Answers (1)

best_answer

Given that,

$A\left ( -36,15 \right )=\left ( x_{1},y_{1} \right )$

$B\left ( 0,0 \right )=\left ( x_{2},y_{2} \right )$

$C\left ( 20,15 \right )=\left ( x_{3},y_{3} \right )$

To find the length of the sides by using the distance formula,

$AB=c=\sqrt{\left ( -36-0 \right )^{2}+\left ( 15-0 \right )^{2}}$

$AB=c=39$

$BC=a=\sqrt{\left ( 0-20 \right )^{2}+\left ( 0-15 \right )^{2}}$

$BC=a=25$

$CA=b=\sqrt{\left ( 20+36 \right )^{2}+\left ( 15-15 \right )^{2}}$

$CA=b=56$

Substituting the values in the incenter formula,

$\left( \frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c},\frac{ay_{1}+by_{2}+cy_{3}}{a+b+c} \right )$

$\left( \frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c},\frac{ay_{1}+by_{2}+cy_{3}}{a+b+c} \right )=\left( \frac{-900+0+780}{120},\frac{375+0+585}{120} \right )$

$\left( \frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c},\frac{ay_{1}+by_{2}+cy_{3}}{a+b+c} \right )=\left ( -1,8 \right )$

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