Since there have been excellent answers on why one should work in projective space, let me add some comments on how to visualize it.
You probably know affine space $A^n$ sits in $P^n$ naturally: the complement of any hyperplane in $P^n$ is isomorphic to $A^n$. Let us denote affine coordinates by $x_1, \ldots, x_n$, and projective ones by $z_0, \ldots, z_n$, and embed $A^n$ in $P^n$ by sending $(x_1,\ldots, x_n)$ to $[1:x_1:,\ldots,x_n]$. i.e. $A^n = [x_0 \neq 0]$.
Now it is easy to see that $[x_0 = 0] \subset P^n$ is isomorphic to $P^{n-1}$ so we derive $ P^n = A^n \cup P^{n-1} $ as sets.
We can think of this $P^{n-1}$ as the "points at infinity", i.e. the different "directions" on how to travel to infinity in the affine space. For example any line in $A^n$ meets the points at infinity at exactly one point corresponding to his slope (thats why parallel lines in $A^n$ which have the same slope, meet at the same point at infinity). Another example would be a parabola $y=x^2$, when we go to infinity, the tangent direction of the graph will eventually become almost vertical, thats why its point at infinity is $[0:1:0]$ (for this homogenize to $yz = x^2$ and set $z=0$).
Lastly, i find it easy to visualize the whole $P^n$ as the quotient of the n-sphere with anitpodal points identified, more specifically picking a set of representatives as the lower hemispehere plus the equator (where of course we still need to identify on the equator). So if n=2 and our field is the reals, we can pick the 2-sphere which is easily visualizable. Now if we restrict to the lower hemisphere and the equator, we found one representative for each line through the lower hemisphere, and on the equator we still need to identify points. You can see that the lower hemisphere (excluding the equator) is $A^2$, and the points on the equator are the points at infinity: $P^1$. Its also clear why parallel lines meet, they converge to the same point on the equator.
Hopefully this was helpful.
Joachim