Let (X, 0) and (0, Y) be the points at which such a line intercepts the x and y axes, respectively. Then the area of the triangle formed is (1/2)XY= 27 we we must have XY= 54.

Any line through (4,3) can be written y= m(x- 4)+ 3. That means we must have y= m(0-4)+ 3= -4m+ 3= Y and 0= m(X- 4)+ 3 so that mX- 1= 0, mX= 1, and X= 1/m. XY= (-4m+ 3)(1/m)= 54. Solve that for m.