How can I see a visual representation of the difference between Euclidean and Affine geometry? I have just started studying different types of geometry with the first to I have covered being Euclidean geometry and Affine geometry. I am aware that Euclidean geometry is what we have used the most up until degree-level maths and Affine geometry is to do with the preservation of parallel lines but beyond this I am struggling to visualise the differences between these geometries. I was wondering if anybody knew a good way of explaining this or some resources to aid in my understanding.

First of all, it can be very useful to know that there is an even more general geometry including Euclidean and affine geometry: Projective geometry (shortly said: the geometry of perspective).
This said, I think that there at least two ways to consider this issue.
First point of view attached to figures :

*

*Euclidean geometry: the image of a square can be a/any square.


*Affine geometry: the image of a square is a/any parallelogram.


*Projective geometry: the image of a square is a/any quadrilateral.
Second point of view attached to group of transformations (The "modern" point of view developed by Klein in the so-called Erlangen program):

*

*Euclidean geometry is characterized by the group of transformations generated by rotations, symmetries, translations, all of them preserving (Eucledean) norm:

$$\begin{cases}x'&= &a x - \varepsilon b y + e\\y'&=& b x + \varepsilon a y +f\ \end{cases} \ \text{with} \ a^2+b^2=1 \ \ \text{or} \ \ \begin{cases}x'&=&x+ e\\y'&=&y +f\ \end{cases}$$
(with $\varepsilon = +1$ for a rotation, $\varepsilon = -1$ for a symmetry with respect to a straight line ; the second case deals with pure translations).

*

*Affine geometry is characterized by any transformation of the form :

$$\begin{cases}x'&=&ax+by+e\\y'&=&cx+dy+f \end{cases}$$

*

*Projective geometry is characterized by any transformation of the form :

$$\begin{cases}x'&=&(ax+by+e)/(gx+hy+i)\\y'&=&(cx+dy+f)/(gx+hy+i) \end{cases}$$
(please note that if, in the common denominator, we take $f=h=0$ and $i=1$, we are back to an affine transformation).
