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High School Mathematics - 2
9.17 Incentre of a Triangle

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(x1, y1), B(x2, y2) and C(x3, y3) 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 [(cx3+bx2)/(c+b), (cy3 + by2)/(c+b)]
Now, BI bisects angle ABD of triangle ABD
Hence AI/ID = AB/BD = c/BD----equation 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)[(cx3+bx2)/(c+b)]+ax1}/(c+b)+a = (ax1+bx2+cx3)/(a+b+c)
y = {(c+b)[(cy3+by2)/(c+b)] + ay1}/(c+b) + a = (ay1+by2+cy3)/(a+b+c)


Directions: Solve the following.

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High School Mathematics - 2
9.17 Incentre of a Triangle

Q 1: Find the coordinates of the incentre of the triangle whose vertices are (4,-2), (-2,4) and (5,5)
1,-3
2,-1
5,5
5/2, 5/2

Q 2: Find the incentre of the triangle formed by (-4,1), (-1,2) and (0,5)
(2,-1)
(1,2)
(1,-2)
(-2,1)

Q 3: The mid points of the sides of a triangle are at (1/2,0), (1/2, 1/2) and (0,1/2).Find the co-ordinates of the incentre.
(1,-2)
(0,1)
(1- 2/2, 1-/2)
(1,2)

Q 4: If (0,1), (1,1)and (1,0) are the mid points of the sides of the triangle, find the incentre.
(1,2)
(2-22, 2-22)
(2,-1)
(2,2)

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