# Find incentre of triangle x+1=0,3x-4y=5,5x+2y=27

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vertices: A (-1, -2) , B(-1, 16) and C(59/13, 28/13)
Inradius r = 5/2 ;;; Incenter (xin, yin) = (3/2, 3) ;;;
sides are: a=90/13 ;; b = 36 sqrt(29)/13 ;; c = 18
s = 19.918 ;;; Area = 49.846

2015-08-26T18:01:07+05:30

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If we want to find the Incircle radius "r", then we can find vertices A (x1,y1), B(x2, y2), C(x3, y3), lengths of sides BC = a, CA = b, AB = c, and then area  Δ of the triangle ABC.  Then
r = Δ / s      where s = (a+b+c)/2

If we want to find the Incenter I (xin, yin) then
x_in = [a x1 + b x2 + c x3] / (a + b + c)
y_in = [a y1 + b y2 + c y3) / (a + b + c)

The sides of the triangle are given by the equations of straight lines:
AB:  x + 1 = 0         =>  x = -1
BC:  3 x - 4y = 5     =>  y = 3/4 x - 5/4
CA :  5x + 2y = 27  =>  y = -5/2  x - 27/2

Solve the above three equations in pairs to get:
Intersection of AB and CA :  A (x1, y1) = (-1, 16)
Intersection of  AB and BC :   B (x2 , y2) = (-1, -2)
Intersection of  BC and CA :  C (x3 , y3) = (59/13, 28/13)

lengths of sides

s=(a+b+c)/2 = [18 + 2√29]*9/13 = 19.918....

x_in = (a * x1 + b * x2 + c * x3)  / (a+b+c) = 3/2
y_in  = (a * y1 + b * y2 + c * y3) / (a+b+c) = 3

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