This is from a practice problem set:
Consider the space of all lines in $\mathbb R^2$ (not necessarily through the origin). Prove that it is homeomorphic to $\mathbb RP^3 \setminus \{\mathrm{pt}\}$.
I know I can think of $\mathbb RP^3$ as $S^3$ with its antipodal points identified. However, it's not clear to me how to construct a homomorphism from $\mathbb RP^3 \setminus \{\mathrm{pt}\}$ to the set of all lines in $\mathbb R^2$. Per this answer, it should help to think of the set of all lines in $\mathbb R^2$ as $[a : b : c]$ s.t. $a, b$ are not simultaneously $0$ s.t. $ax + by + c = 0$ describes a line in $\mathbb R^2$.
How do I construct a homeomorphism?
The linked answer gives you a nice map $f: \Bbb{R}P^2 - \{[0 : 0 : 1]\} \to X$, where $X$ is your space of lines. Namely, $f([a : b : c]) = \{(x, y): ax + by + c = 0\}$. $f$ is well defined because $ax + by + c = 0$ is equivalent to $\lambda ax + \lambda by + \lambda c = 0$ for $\lambda \neq 0$. $X$ excludes only those triples $(a, b, c)$ of the form $(0, 0, c)$, which is exactly the point removed from the punctured $\Bbb{R}P^2$. This means $f$ has an obvious inverse. Then you have to argue that $f$ and $f^{-1}$ are continuous in the topology that you want on $X$ (and you have to define that topology).
(Intuitively, if you want to wiggle a line slightly, there are two different ways to do it: shift it to a parallel line, or rotate it around one of its points. This suggests that $X$ is locally two-dimensional, unlike the punctured $\Bbb{R}P^3$ which is three-dimensional. So that's a hint that the original statement is wrong.)