A linear invertible plane transformation preserve lines. I need verification in my attempt of proof.
Let $B:\mathbb{R}^2 \to \mathbb{R}^2$ a linear invertible transformation.
Since it's linear, then $B(0)=B(x-x)=B(x)-B(x)=0$.
Let $p \in l$ some line, then $ax_p +by_p +c=0$. Then $aB(x_p)+bB(y_p)+cB(1)=0$, then $B(p)$ is in a line of equation $ax+by+cB(1)=0$, then $B$ does preserve lines.
I'm not sure if I did it right, specially because I can't see where invertible property was used.
Thanks.

Note that $B$ "eats" vectors (elements of $\mathbb{R}^2$) while $x_p,y_p$ are real numbers (the coordinates of the point $(x_p,y_p)$ that lies on the line $l$ whose equation is given by $ax + by + c = 0$) and so you can't apply $B$ to $x_p,y_p$!
Instead of describing lines in $\mathbb{R}^2$ as a solution to a non-homogeneous linear equations, it is better to describe them using a single parameter and vectors and then you can apply $B$ to the points on the line and see what you get.
Namely, a line in $\mathbb{R}^2$ can be described by
$$ l = \{ p_0 + t \cdot v \, | \, t \in \mathbb{R} \} $$
where $p_0 \in \mathbb{R}^2$ is some point on the line $l$ and $0 \neq v \in \mathbb{R}^2$ is the direction vector of the line. Then
$$ B(l) = \{ B(p_0 + t \cdot v) \, | \, t \in \mathbb{R} \} = \{ B(p_0) + t B(v) \, | \, t \in \mathbb{R} \} $$
and since $B$ is invertible, $B(v)$ is non-zero and so $B(l)$ is a line that passes through the point $B(p_0)$ with direction $B(v)$.
