There are various approaches, but it is common to go to homogeneous coordinates. So the projective lines have homogeneous equations $2x+y=0$ and $4x+2y+z=0$. They meet where $z=0$ and $2x+y=0$, so at $(1,-2,0)$.
Because we are using homogeneous coordinates, each component of $(1,-2,0)$ can be multiplied by the same non-zero constant.
Remark: I do not know what notation is used in your course, so used standard old-fashioned notation. Your version may use equivalence classes. If so, it should not be difficult to translate.
Suppose that your eye is at the origin, and the "canvas" on which you draw is in front of your eye, and is the $z = 1$ plane. Then the point $(2, 3, 5)$ in space will project, in your "drawing" or "seeing" of the world, to the point $(2/5, 3/5, 1)$ in the $z = 1$ plane. In general, any point $(x, y, z)$ will project to $(x/z, y/z, 1)$.
Now consider two parallel lines; the first one consisting of all points of the form $(1, t, 4)$ and the second consisting of points $(3, t, 2)$. The projections of these to the drawing plane will consist of points of the form $(1/4, t/4, 1)$ and $(3/2, t/2, 1)$, respectively, i.e., they'll still be parallel.
Now look at the lines $(-1, 0, t)$ and $(1, 0, t)$. These project to
$(-1/t, 0, 1)$ and $(1/t, 0, 1)$, which "meet" at the point $(0, 0, 1)$ when $t$ goes to $\infty$.
Does that help at all?
Best Answer
$\newcommand{\Reals}{\mathbf{R}}$Here's a "parametric" way to think of it: When you write "last coordinate $0$", presumably you're thinking of $\Reals^{3}$ embedded in $\Reals^{4}$ as $(x, y, z, 1)$. Take a non-zero direction $v = (a, b, c)$ in $\Reals^{3}$. A pair of parallel lines in $\Reals^{3}$ can be parametrized by \begin{align*} \ell_{1}: &\quad (x_{1} + at, y_{1} + bt, z_{1} + ct, 1) \sim \tfrac{1}{t}(x_{1} + at, y_{1} + bt, z_{1} + ct, 1), \\ \ell_{2}: &\quad (x_{2} + at, y_{2} + bt, z_{2} + ct, 1) \sim \tfrac{1}{t}(x_{2} + at, y_{2} + bt, z_{2} + ct, 1). \end{align*} Distribute the division by $t$, and let $|t| \to \infty$.