How does this version of an equation of a line work: $E(x, y)=aX+bY+C$

Question

I'm reading this article to understand hardware triangle rasterization, but I'm getting stuck in a particular part that defines line equations (that make up a triangle) as follows:

Note that when we move e.g. one pixel to the right, we add one to X and leave Y the same. Our edge equations have the form $E(X,Y)=aX+bY+c$ with a, b, c being per-triangle constants, so for $X+1$ it will be $E(X+1,Y)=a(X+1)+bY+c=E(X,Y)+a$. In other words, once you have the values of the edge equations at a given point, the values of the edge equations for adjacent pixels are just a few adds away.

I'm used to equations of a line like $y=mx+b$, the point slope form of this, or $Ax+Bx+C=0$, where the equation holds true if and only if a point is on a line. The equations of a line in this article are very confusing coming from this background, though:

• They look like they're functions $f: \mathbf{R}^2 \rightarrow \mathbf{R}$, but this doesn't define a line in any meaningful way that I can think of.
• Somehow we need separate equations for the same line at different points, which also does not make sense.
• The claim that $a$, $b$, and $c$ are per-triangle constants, combined with the fact that these numbers, taken together, identify a unique equation would seem to imply that all three equations of a line to make up a triangle have the same equation which is just, what?

What's going on here? How does this style of equation of a line work?

Answer 1

The function $\, E(X,Y)=aX+bY+c \,$ is used to define a one dimensional curve in two dimensions. If you look at the graph of this function using $\, Z = E(X,Y)\,$ you can see that it is a plane. The intersection of this plane with the plane $\, Z = 0\,$ is a straight line whose equation is $\, E(X,Y) = 0.\,$ Another way to see this is to solve the equation to get $\, Y = -(aX+c)/b\,$ which of the form $\, y = mx+b\,$ which you mentioned.

This generalizes to any number of dimensions. For example $\, E(X,Y,Z)=aX+bY+cZ+d\,$ can be used to define a plane in three dimensions in the same way. The equation of a plane being $\,E(X,Y,Z) = 0.\,$ In the same way, you can solve the equation to get $\, Z = -(aX+bY+d)/c\,$ which is the equation of a plane as mentioned before.

The article linked to states:

The general approach can be summarized in 2 sentences: the signed distance to a line can be computed with a 2D dot product (plus an add) – just as a signed distance to a plane can be compute with a 3D dot product (plus add). And the interior of a triangle can be defined as the set of all points that are on the correct side of all three edges So… just loop over all candidate pixels and test whether they're actually inside the triangle. That's it. That's the basic algorithm.

just before giving the function $\, E(X,Y)=aX+bY+c. \,$ In this context $\, E(X,Y)\,$ gives a measure of the signed distance from a point $\,(X,Y)\,$ to a line $\, E(X,Y)=0 \,$ which contains one of the edges of a triangle, and where, obviously, the distance of a point on the line to the line itself is zero.