SOLUTION: please help me solve this problem " Find the equation of the circle inscribed in a triangle, if the triangle has its sides on the lines ; x-y-3=0 , x+y-11=0 , and 7x+y-5=0.

The equations of lines are

 x - y =  3,    (1)
 x + y = 11,    (2)
7x + y =  5.    (3)


1.  Notice that the lines (1) and (2) are perpendicular. So the triangle is right-angled.


2.  The intersection point of lines (1) and (2)  is  C = (7,4)  (you can find it mentally by adding equations (1) and (2) ).

    It is the right-angle vertex.

    The intersection point of lines (1) and (3)  is  A = (1,-2)  (you can find it mentally by adding equations (1) and (3) ).

    The intersection point of lines (2) and (3)  is  B = (-1,12)  (you can find it mentally by subtracting equations (2) from equation (3) ).


3.  The side CA is the vector (-6,-6) of the length .  It is the leg "b" of the triangle.

    The side CB is the vector (-8,8)  of the length . It is the leg "a" of the triangle.

    The side AB is the vector (-2,14) of the length .  It is the hypotenuse "c" of the triangle.

        (classic 3-4-5 right-angled triangle).


4.  It is well known fact that the radius of a right-angled triangle with the legs "a" and "b" and the hypotenuse "c" is equal to  ,

    which is in this case   = .


5.  The center of the circle lies on the angle bisector of the right angle at the vertex C.
    This angle bisector is the horizontal line y = 4.


6.  The last thing to find is the x-coordinate of the center of the inscribed circle.

    This x-coordinate is equal to  7 -  = 7 -  = 7 -  = 7 - 4 = 3.


7.  Now we have everything to write the equation of the inscribed circle. It is 

     = 8.    (8 =  =  ).


    The center is at (3,4); the radius is .