Start pixel and End pixel determine the smallest place you can reference

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   Pixel 1 (0,0)
           Pixel 2 (3,1)

   y = mx+b    Slope-intercept

   m = slope
   b = y-intercept
   y = deltaY / deltaX

Let's look at:
     - |delatX| > |deltaY|
       Right now 0 <= m <= 1
 />NE
/
--->E

Problem 1: (0,0) (3,1)

deltaY / deltaX = 1-0 / 3-0 = 1/3

Now we have: y = 1/3x

current pixel - right side of the line
    draw(0,0)
                                     DDA algorithm - Digital Differential Analyzer
    x++;
    y = y + m;

    draw(x,y)---> 1,1/3 => (1,0)     | | | | |
                                     | | | | |
    x++;                             | | |#|#|   # = filled in pixels
    y = y + m;                       |#|#| | |

    draw(x,y)---> 2,2/3 => (2,1)     "Jaggies" or Aliasing

    MSAA = opposite of Aliasing, performance is hit but presentation is up
    Temporal Artifact

-1 <= m <= 0
              x++
--> E         y = mx + b
\
 \>SE

IF (-10, -20)
                    y = mx + b
^N                  x = y - b / m
| />NW              x1 = y1 - b / m
|/                  y2 = y1 + 1
                    x2 in terms of x1

x2 = y1+1-b / m = (y1-b / m) + (1 / m) = x1 + 1/m


y++;
x = x + 1/m;     = x + deltaX/deltaY


NaN - Not a Number ex. 1/0


 1.000000 * 10^0
+ .0000009
------------
 Accumulation Errors

m is slobe--slope!