In Richard R. Patterson's "Parametric cubics as algebraic curves", it states in 3.1 that a line (2d) in homogeneous coordinates can be parameterized by the homogenous parameter $[t, s]$ as $[s\langle x_0, y_0, z_0\rangle + t\langle x_1, y_1, z_1\rangle]$. If I followed that correctly converting to Cartesian coordinates would be a two variable parameterization of a line in 2d: $(\frac{s x_0 + t x_1}{s z_0 + t z_1}, \frac{s y_0 + t y_1}{s z_0 + t z_1})$. This indeed seems to be a parameterization of a line, but it is not clear / intuitive to me how you arrive there and how to clearly see from the equations that it is a line and not another rational parameterization (say a conic). Jim Blinn mentions something related in "Notation, Notation Notation" (Chapter 10, p. 149), that while each $x$ and $y$ is separately a hyperbola, when plotted together they are a line, the key being that $x$ and $y$ share the same denominator.
Could anyone shed some more light on this and the original homogeneous parameterization? How could one have arrived at that parameterization, for example is it possible to start with a 1 variable Cartesian parameterization $(x(t), y(t))$ and extend it to the above homogeneous one somehow?
This parameterization of a line has a fairly straightforward motivation in one of the standard models of the projective plane $\mathbb{RP}^2$.
Consider the plane $z=1$ in $\mathbb R^3$. (Other planes serve as well, but this is a common and simple choice.) Each point on this plane defines a unique line through the origin. Conversely, each line through the origin that is not parallel to this plane intersects it in a unique point. Thus, points on the plane $z=1$ can be identified with lines through the origin. The coordinates of the points on the line through $(x,y,1)$ are its scalar multiples $(wx,wy,w)$, which for $w\ne0$ you should recognize as the set of equivalent homogeneous coordinates of $(x,y)$.
Three noncolinear points in $\mathbb R^3$ define a plane, so we can similarly identify the line through a pair of points $\mathbf p$ and $\mathbf q$ in the plane $z=1$ with the plane that contains those points and the origin: the line through $\mathbf p$ and $\mathbf q$ is the intersection of this plane with $z=1$. The plane is spanned by $\mathbf p$ and $\mathbf q$, so every point on the plane is of the form $s\mathbf p+t\mathbf q$, $s,t\in\mathbb R$. Obviously, we can replace $\mathbf p$ and $\mathbf q$ by any nonzero scalar multiples and still have the same plane, and thus the same line of intersection with $z=1$.
Two nonparallel planes in $\mathbb R^3$ intersect in a line. If that line intersects $z=1$, then it clearly corresponds to the point of intersection of the two lines in $z=1$ represented by the planes. What if their intersection is parallel to $z=1$, i.e., lies in the plane $z=0$? If each plane intersects $z=1$, this means that those intersections are parallel, too. In the projective plane, parallel lines intersect at a point at infinity, so this leads us to identify lines in the plane $z=0$ with points at infinity in $\mathbb{RP}^2$. The plane $z=0$, then, must correspond to the line at infinity. Its intersection with any other plane is a line in $z=0$, i.e., a point at infinity, which is just what we want.
With these additional pairings, the identification with $\mathbb{RP}^2$ is complete: every point in $\mathbb R^3\setminus\{0\}$ corresponds to a point in $\mathbb{RP}^2$, and its coordinates are homogeneous coordinates of the corresponding point on the projective plane. The span of any two linearly independent vectors in $\mathbb R^3$ is a plane through the origin, which maps to the line through the corresponding points in the projective plane. Finally, a line through a pair of distinct points $\mathbf p$ and $\mathbf q$ in $\mathbb{RP}^2$ consists of all linear combinations $s\mathbf p+t\mathbf q$ of those points (excluding zero, if working in homogeneous coordinates).
Expanding this in terms of coordinates, the line through the points with homogeneous coordinates $[x_0:y_0:w_0]$ and $[x_1:y_1:w_1]$ is $$s[x_0:y_0:w_0]+t[x_1:y_1:w_1]=[sx_0+tx_1:sy_0+ty_1:sw_0+tw_1].$$ For the finite points on this line, $sw_0+tw_1\ne0$, and we can convert to Cartesian coordinates: $$\left({sx_0+tx_1\over sw_0+tw_1},{sy_0+ty_1\over sw_0+tw_1}\right).$$
Incidentally, in projective geometry the set of nonzero linear combinations of a collection of objects is called their join: the join of a pair of points is a line.