To find the coordinates of the incentre of a triangle whose vertices are given
Definition:
The point of concurrence of the internal bisectors of the angles of a triangle is called the incentre and is generally denoted by I.
Let A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) be the vertices of the triangle. Let AD and BE, the bisectors of the angles A and B meet in I.
AD bisects angleBAC of triangle ABC
Hence BD/DC = AB/BC = c/b
D divides BC internally in the ratio c:b
Hence the coordinates of D are [(cx_{3}+bx_{2})/(c+b), (cy_{3} + by_{2})/(c+b)]
Now, BI bisects angle ABD of triangle ABD
Hence AI/ID = AB/BD = c/BDequation 1
But BD/DC = c/b(Proved above)
Hence BD/(BD+ DC) = c/(c+b)
BD/BC = c/(c+b)
Hence BD = ca/(c+b)(because BC = a)
From equation 1
AI/ID = c/ca/(c+b) = (c+b)/a
This shows that I divides AD internally in the ratio c+b:a
The coordinates of I are
x = {(c+b)[(cx_{3}+bx_{2})/(c+b)]+ax_{1}}/(c+b)+a = (ax_{1}+bx_{2}+cx3)/(a+b+c)
y = {(c+b)[(cy_{3}+by_{2})/(c+b)] + ay_{1}}/(c+b) + a = (ay_{1}+by_{2}+cy_{3})/(a+b+c)
Directions: Solve the following.
