A triangle has corners at `(3 , 2 )`, `(6 ,7 )`, and `(5 ,8 )`. What is the radius of the triangle's inscribed circle?

We determine the coordinates of the center of inscribed circle by investigation of the intersection of the angular bisectors.

Let,`A-=(3,2), B-=(6,7), and C-=(5,8)`represent the vertices of the triangle ABC

Angular bisector atA:
`A-=(3,2) B-=(6,7) C-=(5,8)`

Slope of line AB is `m_(AB)=(7-2)/(6-3)=5/3`

Slope of line AB is `m_(AC)=(8-2)/(5-3)=6/2=3`

Let the slope of the bisector be m

`(m-5/3)/(1+mxx5/3)=(3-m)/(1+3xxm)`

`=(3m-5)/(3+5m)=(3-m)/(1+3m)`

Cross multiplying

`(1+3m)(3m-5)=(3-m)(3+5m)`

`3m-5+9m^2-15m=9-3m+15m-5m^2`

`9m^2+5m^2+3m-15m+3m-15m-5-9=0`

`14m^2-24m-14=0`

Dividing by 14

`m^2-12/7m-1=0`

`m=2.17, m=-0.46`

Equation of the bisector through A(3,2) is

`(y-2)/(x-3)=2.17`

Simplifying

`y-2=2.17(x-3)`

`y-2=2.17x-6.51`

`y=2.17x-4.51`

Angular bisector atB:
A-=(3,2)
B-=(6,7)
C-=(5,8)

Slope of line AB is `m_(AB)=(7-2)/(6-3)=5/3`

Slope of line BC is `m_(BC)=(8-7)/(5-6)=1/-1=-1`

Let the slope of the bisector be m

`(m-5/3)/(1+mxx5/3)=(-1-m)/(1+(-1)xxm)`

`=(3m-5)/(3+5m)=(-1-m)/(1-m)`

Cross multiplying

`(1-m)(3m-5)=(-1-m)(3+5m)`

`3m-5-3m^2+5m=-3-5m-3m-5m^2`

`5m^2-3m^2+5m+3m+5m+3m+3-5=0`

`2m^2+16m-2=0`

Dividing by 2

`m^2+8m-1=0`

`m=0.12, m=-8.12`

Equation of the bisector through B(6,7) is

`(y-7)/(x-6)=0.12`

Simplifying

`y-7=0.12(x-6)`

`y-7=0.12x-0.75`

`y=0.12x+6.25`

The lines `y=2.17x-4.51` and `y=0.12x+6.25` intersect at the center of the incircle

Equating rhs

`2.17x-4.51=0.12x+6.25`

`2.17x-0.12x=6.25+4.51`

`2.05x=10.76`

`x=10.76/2.05`

`x=5.25`

`y=2.17x-4.51`

`y=2.17xx5.25-4.51`

`y=11.39-4.51`

`y=6.88`

`y=0.12x+6.25`

`y=0.12xx5.25+6.25=0.63+6.25`

`y=6.88`

Verified

The coordinates of the centre of incircle is

`O-=(5.25,6.88)`

Slope of line AB

`m_(AB)=5/3`

Point A is `A-=(3,2)`

Equation of the line AB is

`(y-2)/(x-3)=5/3`

`3(y-2)=5(x-3)`

`3y-6=5x-15`

`5x-3y+6-15=0`

`5x-3y-9=0`

Center is `O-=(5.25,6.88)`

Tangent is `5x-3y-9=0`

The distance from O to the line AB is

given by

`|(5xx5.25-3xx6.88-9)/sqrt(5^2+(-3)^2)|=|(-3.39)/5.83|=0.581`

The radius of the circle is 0.581