Moulton plane

Example

The points of the Moulton plane are the points of $\R^2$. Its lines are of three kinds:

1. the vertical lines $x = c$;
2. the lines $y = m x + b$ with slope $m \ge 0$;
3. for each slope $m < 0$ and each $b \in \R$, the bent line made of the points $(x, m x + b)$ with $x \le 0$ and the points $(x, 2 m x + b)$ with $x \ge 0$.

So lines of negative slope are bent where they cross the $y$-axis, and their slope doubles on the right. With these lines it is an affine plane: any two points lie on exactly one line. For a fixed $m < 0$, the bent lines with parameter $m$ are pairwise parallel, as are the lines of each other slope. Some authors use another factor than $2$; any positive factor other than $1$ gives a plane with the same properties.

The Moulton plane does not satisfy the affine form of Desargues's theorem. Let $O = (-10, -10)$, $A = (-8, -6)$, $B = (-6, 2)$, $C = (-2, -2)$, and $A' = (-7, -4)$, $B' = (-4, 8)$, $C' = (2, 2)$. Check:

• $A' = O + \tfrac{3}{2}(A - O)$, and likewise for $B'$ and $C'$.
• The lines $OA$, $OB$, $OC$ have slopes $2$, $3$, $1$, so they are ordinary lines; they contain $A'$, $B'$, $C'$.
• $AB$ and $A'B'$ both have slope $4$, and $AC$ and $A'C'$ both have slope $2/3$; these are ordinary distinct lines, so $AB \parallel A'B'$ and $AC \parallel A'C'$.
• $B$ and $C$ lie in $x < 0$ on $y = -x - 4$, so $BC$ is the bent line with $m = -1$, $b = -4$.
• The line through $B'$ and $C'$ is bent with $8 = -4m + b$ and $2 = 4m + b$, so $m = -3/4$ and $b = 5$.

Since $-3/4 \ne -1$, the line $B'C'$ is not parallel to $BC$. In the Euclidean plane the same data would force $BC \parallel B'C'$.