The problem is just to gives a smooth manifold structure of all straight lines in $\mathbb{R^2}$ ( not just those which pass through the oringin).Moreover, identify it with a well-known manifold!
My sketch of proof: Regard the straight line as a subset of $\mathbb{R^2}$.
Step1: I try to introduce a transtive Lie group action. Set $G=\mathbb{R^2}\rtimes S^1$ $((v,z)\cdot (v',z')=(v+zv',zz'))$ and it can act on $M$ by$(v,z)\cdot l =v +zl$ (here $l\in M$ as a subset of $\mathbb{R^2}$).
Step2: Consider the isotropy group of $G_m$ of the x-axis $m\in M$. One can easily prove that $$G_m=\{(v,z):v\in m,z=1 or -1\} \approx \mathbb{R} \times S^0$$ is closed in $G$
Step3: Now use Homogeneous Space Characterization (one can see GTM218, Introduction to Smooth Manifold,2nd edited, Page552, theorem 21.18 and Page554 theorem21.20),and then we can get $$M \approx G/G_m$$ where$G/G_m$ is the coset space.
However, I cannot continue. I cannot give a simple representation of $G/G_m$.I cannot identify $M$ with a well-known manifold.
Finally, I state the theorem I have used:(one can see GTM218, Introduction to Smooth Manifold,2nd edited)
Theorem21.20 Suppose$ X$ be a set and we are given a transitive action of a Lie group $G$ on $X$ such that for some point $p\in X$, the isotropy group $G_p$ is closed in G. Then $X$ has a unique smooth manifold structure with respect to which the given action is smooth.