I have the group $$G = \left\{ \begin{bmatrix}a&b\\0&1\end{bmatrix} : a,b \in R, a \neq 0\right\}$$ and the map $\wedge: G \times R^2 \to R^2$ defined by $$\begin{bmatrix}a&b\\0&1\end{bmatrix} \wedge(x,y) = (ax+by, y).$$
I have already shown that $\wedge$ defines a group action on the plane. All I have left to do is determine and describe its orbits:
When $y = 0$ we have $$Orb(x,0) = \{(ax,0): a \in R,a \neq 0 \} .$$ This is just the $x$ axis excluding the origin.
When $x = 0$ we have $$Orb(0,y) = \{(by,y): b \in R \}$$ If I set $x = by \implies y = x/b$ so this is the set of lines through the origin with gradient $1/b$. Not including the origin itself.
Finally with both $x$ and $y$ not equal to $0$, $$Orb(x,y) = \{(ax+by,y): a,b \in R, a \neq 0 \}.$$ Again setting $x = ax+by$ we have $y = \frac{x(1-a)}{b}$. This is the set of lines through by not including the origin with gradient $\frac{(1-a)}{b}$.
Finally with both $x,y$ equal to $0$, we have the origin.
I don't feel as though this is correct as the orbits of a group action partition the plane into distinct sets. Could someone give me a little help here please?
The decomposition given in the question is not the orbit decomposition.
Hint Note that the given action, $$\pmatrix{a & b \\ 0 & 1} \wedge (x, y) := (ax + by, y)$$ (which is just the restriction of the standard action of $GL(2, \Bbb R)$ on $\Bbb R^2$, namely the one given by left matrix multiplication, to our group $G$), fixes the second component of every element in $\Bbb R^2$. So, whatever the orbit decomposition is, each orbit is a subset of a horiztonal line $\{y = y_0\}$ for some $y_0$, and henceforth we need only understand the behavior of the first component.