You can start with the form ∆y(x -x1) -∆x(y -y1) = 0 Here, we have ∆y = 11-(-3) = 14 ∆x = -3-1 = -4 and we can choose (x1, y1) = (1, -3). This gives 14(x -1) -(-4)(y -(-3)) = 0 14x +4y -2 = 0 All these terms have a common factor of 2 that we can remove. Adding 1 to the result puts it in standard form: