The lines are

          y      = 0     (1)

    4x + 3y - 50 = 0     (2)

    3x - 4y      = 0     (3)



    


    Lines  y=0 (red), 4x+3y-50 = 0 (green), 3x-4y = 0 (blue)


The triangle is in QI. Its vertices are

    P1 = (0,0)        (the intersection points of lines (1) and (3) )

    P2 = (12.5,0)     (the intersection point of lines  (1) and (2) )

    P3 = (8,6)        (the intersection point of lines  (2) and (3) )


Let the point (p,q) be the center of the inscribed circle into the triangle.



        Then the point (p,q) is EQUALLY REMOTED from the triangle sides.



So, I will write two equations. One equation will say that the point (p,q) is equally remoted from lines (1) and (2).

Seqond equation will say that the point (p,q) is equally remoted from lines (1) and (3).


The formula for the distance of the point (p,q) on a coordinate plane from the line given by its general equation Ax + By - C = 0 is


    d = .        (4)


So, my first equation is


     = ,

or

     = .      (5)


The center of the triangle lies, obviously, in QI  --  therefore, I can take off the modulus sign in the left part at "q"
   (since q is positive in QI (!) )


The center of the triangle lies, obviously, BELOW line (2) -- therefore, 4p+3q-50 < 0 (!).  Therefore, I can take off
the modulus sign in the right part, replacing it by the sign " - " (minus).

Then equation (5) takes the form

    q = -(4p + 3q -50)/5,  

or, in standard equivalent form

    4p + 8q = 50.     (6)



My second equation is


     = ,

or

     = .      (7)


Again, similarly to what was said above, I can take off the modulus sign in the left part at "q"
   (since q is positive in QI (!) )


The center of the triangle lies, obviously, BELOW line (3) -- therefore, 3p-4q > 0 (!).  Therefore, I can take off
the modulus sign in the right part, replacing it by the sign " + " (plus).

Then equation (7) takes the form

    q = (3p - 4q)/5,  

or, in standard equivalent form

    3p - 9q = 0.      (8)


Now I have two equations (6) and (8) to find p and q. 

    4p + 8q = 50,     (6)

    3p - 9q =  0.     (8)


Simplify them

    2p + 4q = 25      (6')

     p - 3q =  0      (8')


From (8'), express p = 3q and substitute it into (6'). You will get

    2*(3q) + 4q = 25,

    6q     + 4q = 25

    10q      = 25

      q =  = 2.5.


Then p = 3q = 7.5.


The radius of the circle r is, obviously, equal to q:  r = q = 2.5.


Therefore, the equation of the circle is


     +  = ,   or


     +  = .     ANSWER


You can transform it further as you want.