Find the coordinates of the center of the circle inscribed in a triangle whose vertices are (-36,7) , (20,7) and (0,-8).

Explanation

Let `ABC` is the triangle with the vertices `A(-36,7), B(20,7) and C(0,-8)`.

Then, `AB = sqrt((20-(-36))^2 + (7-7)^2) = 56`

`BC = sqrt((0-(-20))^2+((-8)-7)^2 ) )= sqrt(400+225) = sqrt(625) = 25`

`AC = sqrt((-36-0)^2 + (7-(-8))^2)) = sqrt(1521) = 39`

Let, `O(O_x,O_y)` is the center of the circle inscribed in triangle `ABc`.

Then, `O_x = (A_xa+B_xb+C_xc)/P and O_y = (A_ya+B_yb+C_yc)/P`

Here, `P = AB+BC+AC = 56+25+39 = 10` and `a = 25,b = 39, c = 56`

So, `O_x = (-36**25+20**39+0**56)/120 = -120/120 = -1`

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