Let A = (a,b) and B = (c,d). The midpoint M of AB is (2, 3) = ((a + c)/2, (b + d)/2), so
a +c = 4 ....(1)
b +d = 6. ....(2)
Since the point A(a,b) lies on the line 2x + y −3 = 0 and the point B(c,d) lies on the line 3x − 2y + 1 = 0, we also have
2a + b = 3 .....(3)
3c −2d = −1. ....(4)
From (1) and (2), we have c = 4− a and d = 6−b.
Substituting these into (4) gives
3a − 2b = 1. (5)
We can now solve (3) and (5) to obtain a = 1 and b = 1.
We have found that A is (a,b) = (1, 1). So the gradient of the line l is m = (1−3)/(1−2) = 2. Thus l has equation y −1 = 2(x −1) or, equivalently, y = 2x −1.