Question

Find the coordinates of the centre of the circle inscribed in a triangle whose vertices are $(-36,7),(20,7)$ and $(0,-8)$ ?

Answer 1

The vertices of the triangle are $A(-36,7),B(20,7)andC(0,-8)$

The vertices of the triangle are $A(-36,7),B(20,7)andC(0,-8)$

$a=BC=\sqrt{(0-20)²+(-8-7)²}$

$=\sqrt{(-20)²+(-15)²}$

$=\sqrt{400+225}$

$=\sqrt{625}$

$=\sqrt{25\times 25}$

$=25$

$b=CA=\sqrt{(36-0)²+(7-(-8)²}$

$=\sqrt{\left(36\right)²+(7+8)²}$

$=\sqrt{1296+15²}$

$=\sqrt{1296+225}$

$=\sqrt{1521}$

$=\sqrt{39\times 39}$

$=39$

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

$=\sqrt{(20+36)²+\left(0\right)²}$

$=\sqrt{56²}$

$=\sqrt{56\times 56}$

$=56$

In centre of the circle is

$x₁=-36y₁=7x₂=20y₂=7x₃=0y₃=-8$

$a=25b=39c=56$

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

$=[\frac{(-900+780+0)}{\left(120\right)},\frac{(175+273-448)}{\left(120\right)}]$

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

$=(-1,0)$