Concerning the axioms for Incidence geometry; see : Francis Borceux, An Axiomatic Approach to Geometry. Geometric Trilogy I (2014), page 306 :
Ax-I.1 Two distinct points are incident to exactly one line.
Ax-I.2 Each line is incident to at least two distinct points.
Ax-I.3 There exist three points not incident to the same line.
For Ax-I.1, let $P,Q$ such that $P \ne Q$, and suppose that exists $l,l'$ such that $l \ne l'$ and : $P,Q \in l \cap l'$. Then, $l$ and $l'$ are two distinct lines that are incident with two distinct points, contrary to axiom 2).
For Ax-I.2, consider a line $l$ and assume that there is at most one point $P$ such that $P \in l$. By axiom 3) there is onother line $l'$ such that $P \in l \cap l'$.
Consider a third line $l''$ (it exists by axiom 1)) : by axiom 2) there are two points $Q \in l' \cap l''$ and $R \in l'' \cap l$.
We have that $P \ne Q$, otherwise $P \in l, l', l''$, contrary to axiom 3).For the same reason, $P \ne R$ and $Q \ne R$.
Thus $R \in l$, contrary to assumption that $P$ is the only point on $l$.
Finally, the construction of the above proof gives us three distinct points that are not on the same line, i.e.Ax-I-3.
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
Let us assume that incidence structure is a triple $M=(P,L,\epsilon)$, where $P$ is the set of points, $L$ is the set of lines, and $\epsilon$ is the incidence relation between points and lines i.e. $\epsilon\subseteq P\times L$.
Dual model $M^*$ to structure $M=(P,L,\epsilon)$ arises by interchanging the roles of point and lines i.e. $M^*=(L,P,\epsilon^{-1})$ which means that points become lines and vice versa.
As for question 2, it is self-evident that the dual model $M^*$ of some incidence structure $M$ is a model of incidence geometry (axioms A1-A3) if and only if axioms A*1-A*3 are satisfied in $M$.
Example: Any projective plane is a model for incidence geometry such that its dual model is as well a model for incidence geometry.