Incentre of a Triangle





In this page incentre of a triangle we are going to see how to find incentre,

Definition:

The internal bisectors of the three vertical angle of a triangle are concurrent.This point of concurrency is called the incentre of the triangle..The incentre is denoted by I.

To find the coordinates of the incentre of the triangle formed by the points (x₁,y₁) (x₂,y₂) and (x₃,y₃)

Let ABC be a triangle whose vertices are (x₁,y₁) (x₂,y₂) and (x₃,y₃). Let AD,BE and CF be the internal bisectors of the angles of the triangle ABC.The incentre I of the triangle ABC is the point of intersection of AD,BE and CF.

Incentre of the triangle =

   [(ax₁ + bx₂ + cx₃)/(a+b+c),(ay₁ + by₂ + cy₃)/(a+b+c)]


Example 1:

Find the coordinates of the incentre of the triangle whose vertices are

A (1,1) ,B (2,1) and C (2,2)

Solution:

The vertices of the triangle are A (1,1) ,B (2,1) and C (2,2)

a = BC = √(2-2)² + (2-1)²

          =  √(0)² + (1)²

          =  √1

          =  1    

b = CA = √(1-2)² + (1-2)²

          =  √(-1)² + (-1)²

          =  √1+1

          =  √2     

c = AB = √(2-1)² + (1-1)²

          =  √(1)² + (0)²

          =  √1+0

          =  √1

          =   1

Incentre I of the triangle is

[(ax₁ + bx₂ + cx₃)/(a+b+c),(ay₁ + by₂ + cy₃)/(a+b+c)]

      x₁ = 1 y₁ = 1 x₂ = 2 y₂ = 1  x₃ = 2 y₃ = 2

       a = 1 b = √2  c = 1

= [1(1) + √2(2) + 1(2)/(1 + √2 + 1),1(1) + √2(2) + 1(2)/(1 + √2 + 1)  

= [(1+2+2√2) / (2+√2),(1+2+2√2) / (2+√2)]

= [(3+2√2)/(2+√2),(3+2√2)/(2+√2)]



Example 2:

Find the coordinates of the incentre of a triangle whose vertices are

A (-36,7) ,B (20,7) and C (0,-8)

Solution:

The vertices of the triangle are A (-36,7) ,B (20,7) and C (0,-8)

a = BC = √(0-20)² + (-8-7)²

          =  √(-20)² + (-15)²

          =  √400+225

          =  √625

          =  √25 x 25

          =  25  

b = CA = √(36-0))² + (7-(-8))²

          =  √(36)² + (7+8)²

          =  √1296 + 15²

          =  √1296+225

          =  √1521

          =  √39 x 39

          =  39

c = AB = √(20-(-36))² + (7-7)²

          =  √(20+36)² + (0)²

          =  √56²

          =   √56 x 56

          =  56

Incentre I of the triangle is

[(ax₁ + bx₂ + cx₃)/(a+b+c),(ay₁ + by₂ + cy₃)/(a+b+c)]

      x₁ = -36 y₁ = 7 x₂ = 20 y₂ = 7  x₃ = 0 y₃ = -8

       a = 25 b = 39  c = 56

= [25(-36)+39(20)+56(0)/(25+39+56),25(7)+39(7)+56(-8)/(25+39+56)]  

= [(-900+780+0)/(120),(175+273-448/(120)]

= [(-120)/120,(448-448)/120]

= (-1,0)


So the incentre is (-1,0) incentre of a triangle

Related Topics






Incentre Of A Triangle to Analytical Geometry