2 3 = 7 7 6 = 2 4 = 8 5 = 1 8 1 ∆X ∆Y 1 < m < ∞ ∆X ∆Y 1 < m < ∞ ∆X ∆Y 0 ≤ m ≤ 1 ∆X ∆Y 0 ≤ m ≤ 1 ∆X ∆Y 1< m < ∞ ∆X ∆Y 1< m < ∞ ∆X ∆Y 0 ≤ m ≤ 1 negative positive ∆X: x2 < x1 : change the orientation swap starting and ending points: x2 with x1 , y2 with y1, ∆X = abs(∆X) This will cause the ∆X to become ∆X, 1st condition. 1 < m < ∞: make it 0 ≤ m ≤ 1 Invert the coordinates: x2 with y2 , x1 with y1, Store original m, because remembering it was > 1, will help swapping back the pixels when they are drawn This will cause the placement of the points to represent a line with the slope in the acceptable range, fulﬁlling the 3rd condition Octant conversions: 3 to 7, 4 to 8, 5 to 1, 6 to 2 Octant conversions: 2 to 1 m: calculate with original, unchanged ∆Y and ∆X and never recalculate it again, leave as it is here ∆Y: ∆Y = abs(∆Y) This will cause the ∆Y to become ∆Y, 2nd condition. Octant conversions: 7 to 2, 8 to 1 Now all lines are as in the 1st octant.