Notation : $\Delta xyz$ is a 2-dimensional triangle. And $[xy]$ is a
line segment between $x,\ y$
Def : $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ is collinear if for
$x,\ y,\ z$ that are distinct points in a line, then $f(x),\ f(y),\
f(z)$ are in a line.
EXE : If $f$ is collinear and bijective, then $f$ sends a line into a line.
EXE : If $f$ is collinear and bijective, then so is $f^{-1}$
Proof : For a line $l$, assume that $f(l)\subset L$ where $L$ is a
line.
Hence $f^{-1}(L)$ contains $l$. Assume that $x\in f^{-1}(L)-l$.
Consider another line $l'$ through $x$ s.t. $l'$ is not parallel to
$l$.
Hence $f(l')$ is in $L$. This implies that if $l''$ is a parallel
line wrt $l$ and $l''$ passes through $x$, then $$ f(X)\subset L,\
X:=\mathbb{R}^2-(l'' -\{x\}) $$
For $y\in l''$, there is $a,\ b\in X$ s.t. $a,\ b,\ y$ are
collinear. Hence $f(\mathbb{R}^2)$ is in $L$, which is a contradiction.
Hence $f:l\rightarrow L$ is a bijection.
Coro : In the above proof, $l,\ l'$ are parallel
lines iff their images are also parallel.
Coro : If $x_i$ is vertexes in parallelogram, then so is $f(x_i)$.
In further, here $f$ preserves order of $x_i$.
Coro : Hence if $z$ is a mid point in $[xy]$, then $f(z)$ is still a
mid point in $[f(x)f(y)]$. This implies that $$
f(tx+(1-t)y)=tf(x)+(1-t)f(y),\ x\neq y,\ t\in \mathbb{R} $$
