Question

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

Answer 1

A] Given A(7, -36) B(7,20) C(-8,0)

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

$=\sqrt{400+625}=25$

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

$=39$

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

$=56$

a = 25, b = 39, c = 56

In center $=\left(\frac{a{x}_1+b{x}_2+c{x}_3}{x+b}\right),\left(\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right)$

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

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

$\therefore$ In center is (0,-1)