Question

Find the in-center of the triangle formed by the straight line
x+1=0, 3x - 4y =5 and 5x+12y = 27.

Answer 1

Step-by-step explanation:

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