The figure below represents the triangle and lines bisectors of its sides.
The equation of AB is 3x−2y+c=0 As it passes through A(3,2), the value of c=−5 Thus, 3x−2y−5=0 The coordinates of D is (1,-1) as D is the mid-point of AB Consider the coordinates of B are (α,β)
3+α
2
=1,
2+β
2
=−1 α=−1,β=−4 B(−1,−4) Similarly, the equation of AC is 2x−y+c=0 As it passes through (3,2), the value of c=−4 Thus, 2x−y−4=0 E is a point of intersection of x+2y−2=0 and 2x−y−4=0 E=(2,0) and E is the midpoint of AC The coordinates of C are,