As far as I can see, a line with two points satisfies this system of axioms.
None of these axioms postulate the existence of noncollinear points, but that is normally a feature of axioms for the projective plane and projective $3$-space.
Perhaps the author has given these axioms in addition to some others that occurred earlier?
One possible example is coordinate geometry over the dyadic rationals: numbers of the form $\frac{a}{2^b}$,where $a,b \in \mathbb Z$.
We give this the inherited structure of the usual geometry on $\mathbb R^2$, but throw away all points whose $x$- and $y$-coordinates don't have this form, and throw away all lines not containing at least two of the remaining points.
Since all of your axioms were satisfied before we threw away points, the only ones we need to check are the axioms that claim something exists. (We need to check we didn't throw it away.) So:
- There is still a line through any two points. We only threw away lines that only contained at most one dyadic rational point, such as the line $y = \pi x$.
- Any line contains at least two points. If it didn't, then we threw it away!
- If $P,Q$ are distinct points, there is a point $R$ such that $B(P,Q,R)$. If $P = (x_1,y_1)$ and $Q = (x_2,y_2)$, then we can take $R = (2x_2 - x_1, 2y_2 - y_1)$.
- If $P,Q$ are distinct points, there is a point $R$ such that $B(P,R,Q)$. If $P=(x_1,y_1)$ and $Q = (x_2,y_2)$, then we can take $R = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$.
Finally, Pasch's axiom is violated because some lines no longer have the intersection points they should. For example, if we take the triangle with vertices $A = (1,0)$, $B = (-1,0)$, and $C = (0,1)$, then the line $y=2x$ intersects side $AB$ at $(0,0)$ but doesn't intersect $AC$ or $BC$. (It used to intersect $AC$ at $(\frac13,\frac23)$, but then we threw away that point.)
The other common axioms this violates are:
- Completeness: we can extend this plane to a bigger one that contains it by returning to $\mathbb R^2$. (We don't even have any of the circle intersection properties, which are weaker than completeness.)
- The parallel postulate: lots of lines become parallel because we threw away their intersection points.
- Segment copying: we can't, for example, copy a segment of length $1$ onto the line $y=x$.
Also, depending on how some of your angle axioms are stated, they might be iffy here since this geometry doesn't have plane separation. (This will always be an issue, though, since plane separation is equivalent to Pasch's axiom!)
Another kind of example is the missing strip plane. Here, we also throw away some points: we start with $\mathbb R^2$ and throw away the points $(x,y) : 0<x \le 1$, along with vertical lines with equation $x=c$ where $0<c\le1$. We modify the definition of the length of a line segment: if a line segment $AB$ lies on a line with slope $m$, and $A$ and $B$ are on opposite sides of the missing strip, then we decrease the length of $AB$ by $\sqrt{1+m^2}$ from the usual length.
This plane actually satisfies some notions of completeness: for example, it satisfies the line completeness axiom, since every line in the missing strip plane looks exactly like a line in the usual Cartesian plane, on its own.
Apart from Pasch's axiom (which is violated because we can draw a diagram where an intersection point ought to be in the missing strip), it only violates the parallel postulate (for similar reasons) and SAS triangle congruence (which Hilbert takes as an axiom).
Best Answer
1) There are two points $P$ and $Q$ and one line $\{PQ\}$. Axioms 1 and 2 are satisfied but 3 is not.
2) There are three points $P$, $Q$, $R$ and one line $\{PQ\}$. Axioms 2 and 3 are satisfied but 1 is not.
1) There are two points $P$ and $Q$ and two lines $\{PQ\}$ and $\{P\}$. Axioms 1 and 3 are satisfied but 2 is not.