map colinear triple of points to another triple of points in $\mathbb{R}^2$ Given two triples of pw different colinear points in $\mathbb{R}^2$ so $(x_1,x_2,x_3),(y_1,y_2,y_3) \in (\mathbb{R}^2)^3$.
There is a map of the form $T:\mathbb{R}^2\to\mathbb{R}^2,x\mapsto Ax+b$, where $A \in Gl(\mathbb{R},2), b\in \mathbb{R}^2$ with $T(x_k)=y_k,k=1,2,3$.
$$\iff$$
$$\frac{||x_1-x_2||}{||x_2-x_3||} = \frac{||y_1-y_2||}{||y_2-y_3||}$$

Intuitively it is somewhat clear, but how to pin that down?
 A: If all three of $x_1,x_2,x_3$ are collinear, then they can be expressed as $x_1 = (x_1 - x_2) + x_2$ and $x_3 = (x_3 - x_2) + x_2 = s(x_1 - x_2) + x_2$. 
 Likewise, $y_1 = (y_1 - y_2) + y_2$ and $y_3 = (y_3 - y_2) + y_2 = t(y_1 - y_2) + y_2$.
For convenience, put $v = x_1 - x_2$ and $w = y_1 - y_2$. The vector $v$ is the direction of the line on which $x_1,x_2,x_3$ lie, and the vector $x_2$ is the offset of the line from $0$. Likewise, but $w = y_1 - y_2$.
Now we can set about creating the affine map. Let's first assume that $x_1 = y_1 = 0$ (we'll construct $b$ for the general case in a moment), so that the two lines pass through the origin. Clearly the linear map $A$ will have to take the span of $v$ to the span of $w$. In order to carry $x_2$ to $y_2$, it will have to take $v$ exactly to $w$. But it must also carry $x_3$ to $y_3$, which gives the condition
$$A(sv) = sAv = sw = tw.$$
This implies that $s = t$, i.e., 
$$\frac{\|x_3 - x_2\|}{\|x_1 - x_2\|} = \frac{\|y_3 - y_2\|}{\|y_1 - y_2\|}.$$
Likewise, if $s=t$, then $A$ is given by any map that takes $v$ to $w$.
Now what if $x_1$ or $y_1$ is not $0$? Then the map $A$ will take the $x$-line to a line parallel to the $y$-line; we can correct by adding $b = y_1 - Ax_1$.
A: This is wrong as stated. Consider in ${\mathbb R}^2$ the two triples $$x_1=(0,0), \quad x_2=(1,0),\quad x_3=(3,0)$$ and 
$$y_1=(0,1),\quad y_2=(1,1),\quad y_3=(-1,1)\ .$$
