Explanation for why the Moulton plane satisfies the axioms of a projective plane

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These series of short questions on the Moulton plane are taken from Stillwell's The Four Pillars of Geometry. Please note that text defines the Moulton plane a little differently than it is normally defined (Please see the picture for the definition used in the book).

Questions 6.1.1 was fairly straightforward, I got $x=(1,0)$, which I believe is the correct answer. Can someone confirm this? More importantly, I am having trouble making any progress with the other questions. I would really appreciate an intuitive explanations for how to approach them.

One more thing that is not clear to me from the text: is it part of the definition of the moulton plane that lines with negative gradient are unaffected by the 'bend' that hitting the $x$ axis produces?


My work:

Based on Brian's comment:

For $6.1.2$:

Suppose we have two points $A=(x_1, y_1)$ and $B=(x_2, y_2)$, where $y_2 \ge y1$.

If $y_1 \ge 0$ and $y_2 \ge 0$ OR if If $y_1 \le 0$ and $y_2 \le 0$ OR if $x_2 \le x_1$ then the points $A,B$ simply lie on the Euclidean line $y=mx +c$.

For the case where $y_2>0$, $y_1<0$ and $x_2>x_1$, to find the line connecting $A$ and $B$ we need to find $k$ and $m$ such that $y_1=mx_1 + k$ and $y_2=2mx_2 + k$



For $6.1.4$:

The points $(1,1), (-1,1), (-1,-1)$ and $(1,-1)$