Describe the orbits of the action. So $L$ denotes the set of oriented straight lines through the origin in $\mathbb{R}^{2}$ (that is, straight lines with a preferred direction, indicated by an arrow). The group
$$(\mathbb{R},+)\cong \left \{ \begin{pmatrix}
1 & x\\ 
0 & 1
\end{pmatrix}\mid x\in \mathbb{R} \right \}\leq SL_{2}(\mathbb{R})$$
is made up of linear transformations of $\mathbb{R}^{2}$, and hence acts on the set $L$ (i.e. lines through the origin are taken to lines through the origin).
How do we describe the orbits of this action?
 A: There are four orbits. Before we look at them let us describe how to represent the lines with direction. Consider the nonzero, vectors $\mathbf{v}=\begin{pmatrix}a\\b\\\end{pmatrix}$ in $\mathbb{R}^2$. We say $\mathbf{v}$ and $\mathbf{w}$ are equivalent if there is $r\in \mathbb{R}$ such that $r>0$ and $\mathbf{v}=r\mathbf{w}$. This is clearly an equivalence relation, and the equivalence classes are in 1-1 correspondence with the directed lines through the origin.
Further the action of the group preserves  the equivalence classes and this action is the same as the action of the group on the lines (I am assuming that this is the action that you have in mind). 
Now we can chose some canonical representatives. Given  $\begin{pmatrix}a^{\prime}\\b^{\prime}\\\end{pmatrix}$ with $b^{\prime}\neq 0$ multiply by the positive real $\frac{1}{|b^{\prime}|}$ to get the vectors $\begin{pmatrix}a\\1\\\end{pmatrix}$ and $\begin{pmatrix}a\\-1\\\end{pmatrix}$ these are canonial representitives. If on the other hand $b^{\prime}=0$ multiply by $\frac{1}{|a^{\prime}|}$ to get the vectors  $\begin{pmatrix}1\\0\\\end{pmatrix}$ and $\begin{pmatrix}-1\\0\\\end{pmatrix}$.
Two of the orbits are singletons, those the positive and negative $x$-axis.
$$\begin{pmatrix}1&x\\0&1\end{pmatrix} \begin{pmatrix}1\\0\\\end{pmatrix}=\begin{pmatrix}1\\0\\\end{pmatrix}$$
$$\begin{pmatrix}1&x\\0&1\end{pmatrix} \begin{pmatrix}-1\\0\\\end{pmatrix}=\begin{pmatrix}-1\\0\\\end{pmatrix}$$
The remaining two orbits consist firstly, of all lines directed above the $x$-axis and secondly, those directed below the $x$-axis.
$$\begin{pmatrix}1&x\\0&1\end{pmatrix} \begin{pmatrix}a\\1\\\end{pmatrix}=\begin{pmatrix}a+x\\1\\\end{pmatrix}$$
$$\begin{pmatrix}1&x\\0&1\end{pmatrix} \begin{pmatrix}a\\-1\\\end{pmatrix}=\begin{pmatrix}a+x\\-1\\\end{pmatrix}$$
Where here we note that as $x\in \mathbb{R}$ the values of $a+x$ also run through the real numbers. 
In summary there are four classes, the positive and negative $x$-axis directions, both with one line in each orbit; and all lines directed above and below the $x$-axis, each with infinitely many lines (the same lines in each actually, but with opposite directions).
A: Hint: Look at the orbits of $\left( \begin{matrix} 0 \\ 1 \end{matrix} \right)$ and $\left( \begin{matrix} 0 \\ -1 \end{matrix} \right)$. 
Which lines are missing? How can you conclude?
