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) A≡(3,2),B≡(6,7),andC≡(5,8)represent the vertices of the triangle ABC
Angular bisector atA:
A-=(3,2)
B-=(6,7)
C-=(5,8)A≡(3,2)B≡(6,7)C≡(5,8)
Slope of line AB is m_(AB)=(7-2)/(6-3)=5/3mAB=7−26−3=53
Slope of line AB is m_(AC)=(8-2)/(5-3)=6/2=3mAC=8−25−3=62=3
Let the slope of the bisector be m
(m-5/3)/(1+mxx5/3)=(3-m)/(1+3xxm)m−531+m×53=3−m1+3×m
=(3m-5)/(3+5m)=(3-m)/(1+3m)=3m−53+5m=3−m1+3m
Cross multiplying
(1+3m)(3m-5)=(3-m)(3+5m)(1+3m)(3m−5)=(3−m)(3+5m)
3m-5+9m^2-15m=9-3m+15m-5m^23m−5+9m2−15m=9−3m+15m−5m2
9m^2+5m^2+3m-15m+3m-15m-5-9=09m2+5m2+3m−15m+3m−15m−5−9=0
14m^2-24m-14=014m2−24m−14=0
Dividing by 14
m^2-12/7m-1=0m2−127m−1=0
m=2.17, m=-0.46m=2.17,m=−0.46
Equation of the bisector through A(3,2) is
(y-2)/(x-3)=2.17y−2x−3=2.17
Simplifying
y-2=2.17(x-3)y−2=2.17(x−3)
y-2=2.17x-6.51y−2=2.17x−6.51
y=2.17x-4.51y=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)=5/3mAB=7−26−3=53
Slope of line BC is m_(BC)=(8-7)/(5-6)=1/-1=-1mBC=8−75−6=1−1=−1
Let the slope of the bisector be m
(m-5/3)/(1+mxx5/3)=(-1-m)/(1+(-1)xxm)m−531+m×53=−1−m1+(−1)×m
=(3m-5)/(3+5m)=(-1-m)/(1-m)=3m−53+5m=−1−m1−m
Cross multiplying
(1-m)(3m-5)=(-1-m)(3+5m)(1−m)(3m−5)=(−1−m)(3+5m)
3m-5-3m^2+5m=-3-5m-3m-5m^23m−5−3m2+5m=−3−5m−3m−5m2
5m^2-3m^2+5m+3m+5m+3m+3-5=05m2−3m2+5m+3m+5m+3m+3−5=0
2m^2+16m-2=02m2+16m−2=0
Dividing by 2
m^2+8m-1=0m2+8m−1=0
m=0.12, m=-8.12m=0.12,m=−8.12
Equation of the bisector through B(6,7) is
(y-7)/(x-6)=0.12y−7x−6=0.12
Simplifying
y-7=0.12(x-6)y−7=0.12(x−6)
y-7=0.12x-0.75y−7=0.12x−0.75
y=0.12x+6.25y=0.12x+6.25
The lines y=2.17x-4.51y=2.17x−4.51 and y=0.12x+6.25y=0.12x+6.25 intersect at the center of the incircle
Equating rhs
2.17x-4.51=0.12x+6.252.17x−4.51=0.12x+6.25
2.17x-0.12x=6.25+4.512.17x−0.12x=6.25+4.51
2.05x=10.762.05x=10.76
x=10.76/2.05x=10.762.05
x=5.25x=5.25
y=2.17x-4.51y=2.17x−4.51
y=2.17xx5.25-4.51y=2.17×5.25−4.51
y=11.39-4.51y=11.39−4.51
y=6.88y=6.88
y=0.12x+6.25y=0.12x+6.25
y=0.12xx5.25+6.25=0.63+6.25y=0.12×5.25+6.25=0.63+6.25
y=6.88y=6.88
Verified
The coordinates of the centre of incircle is
O-=(5.25,6.88)O≡(5.25,6.88)
Slope of line AB
m_(AB)=5/3mAB=53
Point A is A-=(3,2)A≡(3,2)
Equation of the line AB is
(y-2)/(x-3)=5/3y−2x−3=53
3(y-2)=5(x-3)3(y−2)=5(x−3)
3y-6=5x-153y−6=5x−15
5x-3y+6-15=05x−3y+6−15=0
5x-3y-9=05x−3y−9=0
Center is O-=(5.25,6.88)O≡(5.25,6.88)
Tangent is 5x-3y-9=05x−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∣∣
∣
∣∣5×5.25−3×6.88−9√52+(−3)2∣∣
∣
∣∣=∣∣∣−3.395.83∣∣∣=0.581
The radius of the circle is 0.581
