Consider a most general affine transformation which transforms point $(x,y)$ to $(X,Y)$ such that $$X=ax+by+c$$ and $$Y=dx+ey+f$$ $a,b,c,d,e$ and $f$ won't be entirely independent of each other.
I think $a^2 + d^2 = b^2 + e^2$ and $ab=ed$. Is that correct? How can I prove that?
Actually, I concluded that should hold true. Because I was taking ratio $(AB/BC)$ between three collinear points (say $A, B, C$) and equating it to $A'B'/B'C'$ when these points become $A', B', C'$ after affine transformation.
Thanks.
Every affine transformation can be decomposed into a linear transformation followed by a translation. Your transformation can be expressed in matrix form as $$\begin{bmatrix}a&b\\d&e\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}c\\f\end{bmatrix}.\tag{*}$$ Translations are isometries so have no effect on colinearity, lengths of line segments or angles between lines. We can therefore safely ignore this part of the affine transformation, i.e., assume that $c=f=0$.
A line through points $\mathbf p$ and $\mathbf q$ can be represented parametrically as $(1-\lambda)\mathbf p+\lambda \mathbf q$. This is essentially linear interpolation: if $\mathbf r$ lies between these two points, it divides the segment $\overline{\mathbf p\mathbf q}$ in the proportion $\lambda:1-\lambda$. If you apply a linear transformation $L$ to this line, you get $$L[(1-\lambda)\mathbf p+\lambda \mathbf q] = (1-\lambda)L(\mathbf p)+\lambda L(\mathbf q).$$ As long as $L(\mathbf p)\ne L(\mathbf q)$, the image of the line is also a straight line, and it should be fairly clear (and is not difficult to prove formally) that $L$ preserves ratios of segments on the line. We therefore require that $L$ is injective, which is equivalent to its matrix being nonsingular, i.e., $ae-bd\ne0$.
A similar computation shows that a nonsingular linear transformation preserves parallelism. If $\mathbf p$ and $\mathbf q$ are distinct points on a line, and $\mathbf p'$ and $\mathbf q'$ distinct points on a parallel line, then $\mathbf p-\mathbf q = \mu(\mathbf p'-\mathbf q')$ for some $\mu\ne 0$. By linearity again, $$L(\mathbf p)-L(\mathbf q) = L(\mathbf p-\mathbf q) = \mu(L(\mathbf p')-L(\mathbf q')),$$ therefore the images of the two lines are also parallel.