Pasch's axiom and plane separation can be shown equivalent. But both depend on the line ordering (or betweenness) axiom for meaning. Does the latter imply those two axioms? I read not, but I seem to have proved it does:
Draw all possible lines from a given point P. They will all intersect a given line L, except for the line thru P parallel to L. Every point in the plane is on one of the lines we drew. If P’ is on the parallel line we drew, or if it is on one of the lines that cuts L at a point P”, but P” is not between P and P’, we say P’ is in Region 1 (on the same side of L as P). We can get from P’ to P, and then from P to any other point in this region, without passing P”, because of the order axioms. Thus this region is connected.
The remaining part of the plane consists of the points P’ such that P” is on L between P and P’. We call this region ‘the other side’ of L from P. To show this other side of L is connected, take two points on it, P’1 and P’2. Each of these is on a line from P. Call them L’1 and L’2. Draw a line thru P’1, parallel to L. Since L’2 cuts L, it is not parallel to L, and so it is not parallel to L’2. Let it cut L’2 at P’3. We can now get from P’1 to P’2 via P’3, provided we can show that the segment from P’3 to P’2 does not contain P’ (the intersection of L’2 with L).
Assume P’ were on this segment. Since P’ divides L2 into just two connected parts (by the order axioms), P’2 and P’3 would be on opposite parts of L1. Since we took P’2 to be on the part of L2 not on the P side of L, P’3 must be on the P side of L. But this contradicts the fact that it is connected to P’1 by a line parallel to L, and so must be on the same side of L as P’1. Thus P’ cannot be on the segment from P’3 to P’2. Thus the ‘other side’ of L is connected.