Understanding the connection between the projective space and the affine plane Suppose we have a point $P=[x,y,z]\in \mathbb P^2$. 
Then at least one of the coordinates is not zero. Suppose $z\neq 0$. So we have write $P$ as $[x/z,y/z,1]$ and this point belongs to $(x/z,y/z)$ in the affine plane. 
But the every point on the line at infinity is of the form $[x,y,0]$, right? What are the images of this points?
Vice versa a point $Q=(a,b)$ in the affine plane belongs to the projective point $[a,b,1]$. Which points will be send to points on the line at infinity ?
 A: 
I want to add to the good answer of Ashwin a representation of the projective plane that I consider very nice and intuitive  that is given embedding the projective plane $\mathbb{P}^2$ in $\mathbb{R}^3$.
The figure illustrates the situation.
The points on a plane $z=k$ have coordinates $P=(a,b,k)$ and correspond to points of $\mathbb{P}^2$ as $(a,b,k) \rightarrow (a/k,b/k,1)$ (usually we chose $k=1$ ), but we can identify these points also with the straight lines from the origin that contain these points.
Now, for any such line, we have a point at infinity $(a,b,0)$ that correspond to a line in the plane $z=0$ passing through the point $(a,b,0)$ that gives the ''direction'' at infinity, in the sense that this line is the ''limit position'' of the line passing through $(a,b,k)$ when $k \rightarrow 0$.
A: Note that $\mathbb{P}^2$ contains many copies of $\mathbb{A}^2$; specifically, if you take the complement of a line in $\mathbb{P}^2$, you get the affine plane (in higher dimensions, you take the complement of a hyperplane). In your case, you are taking the complement of the line defined by $z=0$. So in this case, you do consider the "line at infinity" to be the line defined by $z=0$.
These points on the infinity line do not have well defined realizations in the affine plane, since the affine plane is explicitly defined as the complement of that line. Conversely, when you embed the affine plane into the projective plane, you have to make a choice of how to embed it (equivalently, choose a curve in $\mathbb{P}^2$ and take its complement). Therefore, no points in the affine plane will be sent to the line at infinity; you can think of it as being "added on" to the affine plane.
To see this explicitly in your case, note that the map $\mathbb{P}^2 \to \mathbb{A}^2$ given by $[x,y,z] \mapsto [x/z,y/z]$ is not defined (in the usual sense of division by $0$ being meaningless) when $z=0$; this is basically by construction.
