Prove that General Affine Transformations preserve ratios of lengths Let $A$ be a matrix with determinant 1. Then we call a general affine transformation, a transformation of the form
\begin{align*}
\begin{bmatrix}x'\\y'\end{bmatrix}=A\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}r\\s\end{bmatrix}
\end{align*}
Let $p_1,p_2,p_3$ be colinear points. Prove that
\begin{align*}
\frac{||p_2-p_1||}{||p_2-p_3||}
\end{align*}
Is preserved under these transformations.
I have been working this proof for some time now. I can break these transformations down into pure translations, rotations, and scalings. Everything works out quite nicely for the translations and the rotations but if I try to prove this using a scaling matrix i.e. a matrix of the form
\begin{align*}
\begin{bmatrix}a&0\\0&1/a\end{bmatrix}
\end{align*}.
Any help is greatly appreciated.
 A: Since $p_1, p_2$ and $p_3$ are colinear, we can write $\vec{v} = p_2 - p_3$
and $p2 - p_1 = \lambda \vec{v}$
whence 
$$
\frac{||p_2-p_1||}{||p_2-p_3||} = \lambda
$$
Now write the components of $v$ as $(v_x, v_y)$; we have 
$$
||p_2-p_3|| = (v_x, v_y) \\ ||p_2-p_1|| = (\lambda v_x, \lambda v_y)
$$ 
Now apply (multiply by) a scaling transformation.  Although it is adequate to just use your example, it might be illuminating to use a general matrix 
$$S = \pmatrix{a & b \\ c & d}$$with $ad-bc = 1$.
$$ S (p_2-p_3)= \pmatrix{ a  \,v_x +  b \,v_y \\ c \,v_x + d  \,v_y}\\
S (p_2-p_1) = \pmatrix{ a \lambda  \,v_x +  b \lambda\,v_y \\ c \lambda\,v_x + d \lambda \,v_y} = \lambda S(p2-p3)\\
$$
A: You don’t really need to break the transformation down into cases. This is a straightforward consequence of the linearity of $A$ and homogeneity of norms.  
Note first that the translation part of the transformation—the addition of $[r,s]^T$—is irrelevant since it cancels when you subtract the images of the points from each other.  
The line through $p_2$ and $p_3$ can be parameterized as $(1-\lambda)p_2+\lambda p_3$. Since $p_1$ is colinear with these two points, there’s some real $\lambda$ for which $p_1 = (1-\lambda)p_2+\lambda p_3$, from which $$p_2-p_1 = p_2-((1-\lambda)p_2+\lambda p_3) = \lambda(p_2-p_3)$$ and so $${\|p_2-p_1\|\over\|p_2-p_3\|} = |\lambda|.$$ 
Now, by linearity, $Ap_1 = (1-\lambda)Ap_2+\lambda Ap_3$, therefore $$p_2'-p_1' = Ap_2-((1-\lambda)Ap_2+\lambda Ap_3) = \lambda(Ap_2-Ap_3)=\lambda(p_2'-p_3').$$ If $\det A\ne0$, then $p_2\ne p_3$ implies $p_2'\ne p_3'$ and the result follows immediately.
