Question

Find the co-ordinates of in center of the triangle whose vertices are $(7,-36),(7,-20)$ and $(-8,0)$.

Answer 1

R.E.F image

Find the co-ordinates of in-center of the triangle whose vertices are

(7,-36)(7,-20) and (-8,0)

Ans-

# co-ordinates of incenter of $△$ ABC is

given by

$I(x,y)=(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$

# Evaluation of a,b,c :

$a=l\left(BC\right)=\sqrt{(x_2-x_3{)}^2+(y_2-y_3{)}^2}$

$=\sqrt{(7-(8{)}^2+(-20-0{)}^2)}=25units$

$c=l\left(BA\right)=\sqrt{\left(x_{2-x_1+(y_2-y_1{)}^2}\right)}$

$=\sqrt{(7-7{)}^2+(-20-\left(36{)}^2\right)}=16$ units

$b=l\left(AC\right)=\sqrt{(x_3-x_1+(y_3-y_1{)}^2)}$

$=\sqrt{(-8-7{)}^2+(0-(-36{)}^2)}=39$ units

$I(x,y)=(\frac{(25\times 7)+(39\times 7)+(16\times -8)}{25+16+39},\frac{(25\times -36)+(39\times -20)+(16\times 0)}{25+16+39})$

=(4,-21)

$\therefore$ co-ordinates of in center of $△$ ABC=(4,-21)