element by performing the action of all group elements: O(x) = {f(x)|f ∈ G}.

The importance of this notion is the following:
Qualitatively different configurations are orbits of the automorphism group.

Here, we start with configurations consisting of three lines, i.e. we consider the orbits for all sets {l1, l2, l3} of (at most) three lines18 in Euclidean space with respect to the group of all affine bijections (and not just the orientation preserving ones - orientations will come in at a later stage). This group is usually called the affine group of R
2 and denoted by GA(R
2).

A line in Euclidean space is given by the set of all points (*x, y*) for which y =
mx+b. Given three lines y = mix+bi (i = 1, 2, 3), we list their orbits by giving a defining property. In each case, it is fairly obvious that the defining property is preserved by affine bijections. Moreover, in each case, we show a transformation property, namely that given two instances of the defining properties, the first can be transformed into the second by an affine bijection. Together, this means that the defining property exactly specifies an orbit. The transformation property often follows from the following basic facts about affine bijections, see [55]:

1. An affine bijection is uniquely determined by its action on an affine basis, the result of which is given by another affine basis. Since an affine basis of the Euclidean plane is a point triple in general position, given any two point triples in general position, there is a unique affine bijection mapping the first point triple to the second.
2. Affine maps transform lines into lines.

3. Affine maps preserve parallelism of lines.

That is, it suffices to show that an instance of the defining property is determined by three points in general position and drawing lines and parallel lines.

We will consider the intersection of line i with line j (i 6= j ∈ {1, 2, 3}). This is given by the system of equations:
{y = mix + bi, y = mjx + bj}.

For mi 6= mj , this has a unique solution:

$$x=-{\frac{b_{i}-b_{j}}{m_{i}-m_{j}}},\ y={\frac{m_{i}b_{j}-m_{j}b_{i}}{m_{i}-m_{j}}}.$$

For mi = mj , there is either is no solution (bi 6= bj ; the lines are parallel), or there are infinitely many solutions (bi = bj ; the lines are identical).

We can now distinguish seven cases:
18We do not require that l1, l2 and l3 are distinct; hence, the set {l1, l2, l3} may also consist of two elements or be a singleton.