Question

If $(a,b)$ are the coordinates of the center of the circle inscribed in a triangle whose vertices are $(-36,7),(20,7)$ and $(0,-8)$ then $a+b$ is equal to ____.

Answer 1

The verities of the triangle are $A(-36,7),B(20,7)$ and $C(0,-8)$

$a=BC=\sqrt{(0-20{)}^2+(-8-7{)}^2}=\sqrt{(-20{)}^2+(-15{)}^2}$

$BC=\sqrt{400+225}=\sqrt{625}$

$BC=25$

$b=CA=\sqrt{(36-0{)}^2+(7-(-8){)}^2}$

$=\sqrt{{36}^2+{15}^2}$

$=\sqrt{1296+225}=\sqrt{1521}$

$CA=39$

$c=AB=\sqrt{(20-(-36){)}^2+(7-7{)}^2}$

$c=AB=\sqrt{{51}^2+0^2}$

$AB=56$

Incenter $I$ of the triangle is

$\left[\right(a{x}_1+b{x}_2+c{x}_3\left)/\right(a+b+c),(4y_1+b{y}_2+c{y}_3\left)/\right(a+b+c\left)\right]$

$x_1=-36,y_1=7,x_2=20,y_2=7,x_3=0,y_3=8$

$[\frac{25(-36)+39\left(20\right)+56\left(0\right)}{25+39+56},\frac{25\left(7\right)+39\left(7\right)+56(-8)}{25+39+56}]$

$=[\frac{(-120)}{120},\frac{448-448}{120}]=(-1,0)$

so, the incentre is $(-1,0)$

Comparing with $(a,b)$

We get, $a=-1,b=0$

$a+b=-1+0=-1$