Action of $GL_3(\mathbb{R})$ on pairs $(p, l)$ where $p$ is point and $l$ is a line in $\mathbb{R}P^2$ Consider action of $GL_3(\mathbb{R})$ on pairs $(p, l)$ where $p$ is point and $l$ is a line in $\mathbb{R}P^2$ by projective transforms. The question is:

How to describe orbits of such action, their dimensions and Lie algebras of stabilizers of such action as subalgebras of $\mathfrak{gl}_3(\mathbb{R})$?

I have understood that $GL_3(\mathbb{R})$ acts transitivly on $\mathbb{R}P^2$ because for any $v, w \in \mathbb{R}^3$ exits $g\in GL_3(\mathbb{R})$ such that $g(v) = w$. Also I've computed that stabilizer at point $p = (1, 0, 0)$.It is matrices of the following form:
$$ 
\begin{pmatrix}
 x & y & z \\
 0 & g_{11} & g_{12} \\
 0 & g_{21} & g_{22}
\end{pmatrix} 
$$
Where $x, y, z \in \mathbb{R}$ and x is nonzero, and $g \in GL_2(\mathbb{R})$. But I'm stuck how to use this.
 A: In fact , you have two orbits : the first is the set $\{(p,l), p\in l\}$, the second $\{(p,l), p\notin l\}$.
To check that the second is an orbit, let us remind that a point (line) in $\bf RP^2$ is a line (plane) in $\bf R^3$, through the origin. Now if you choose a non zero vector $e$ on the line, and a base $(f,g)$ of the plane you get a base $(e,f,g)$ of $\bf R^3$. This prove that the (equivariant) map from the set of bases of $\bf R^3$, which send the base $(e,f,g)$ to the pair : teh line  ${\bf R}e$,  the plane  ${\bf R}f+{\bf R}g$  t is onto.
To check that the first  one is an orbit, we can argue the same way : if $p$ is a point in $RP^2$ and $l$ a projective line through $p$ there exists a base $(e,f,g)$ of $\bf R^3$ such that $e$ is a base of the line $p$ and $(e,f)$ a base of the plane $l$. Therefore the map from the et of bases to this set which send $(e,f,g)$ to the pair ${\bf R}e$, ${\bf R}e+{\bf R}f$ is onto.
As the sources of these two maps is an orbit, and as these maps are equivariant, their images are orbits
