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
