Does the model $\mathbb{R}^2$ (The usual Cartesian plane.) with the distance defined as $d(P,Q) = | x_1-x_2| + |y_1-y_2|$ satisfy the ruler postulate?
The ruler postulate is defined as the following
For any line $l$ and any two distinct points $O$ and $P$ on $l$, there exists a bijection $c : l \rightarrow \mathbb{R}$ such that the following holds:
- $c(O)=0$ and $c(P) \geq 0$.
- $d(A,B)=|c(A)-c(B)|$, for all points $A$ and $B$ on $l$
Now the distance is defined in this space
Symmetry: If $P(x_1,y_1)$ and $Q(x_2,y_2)$, then $$d(Q,P)=|x_2-x_1|+|y_2-y_1|=|x_1-x_2|+|y_1-y_2|=d(P,Q)$$
- Also $d(P,Q)$ is clearly nonnegative. If $d(P,Q)=|x_1-x_2|+|y_1-y_2|=0$, then $|x_1-x_2|=|y_1-y_2|=0$. This implies that $x_1=x_2$ and $y_1=y_2$, that is $P=Q$.
- Triangular inequality: If $P(x_1,y_1)$, $Q(x_2,y_2)$ and $R(x_3,y_3)$, then $$ d(P,R)=|x_1-x_3|+|y_1-y_3| \leqslant |x_1-x_2|+|x_2-x_3|+|y_1-y_2|+|y_2-y_3| = |x_1-x_2|+|y_1-y_2|+|x_2-x_3|+|y_2-y_3| = d(P,Q)+d(Q,R) $$
We can conclude that $d$ is indeed a distance.
How can I show that the distance exists therefore the ruler postulate will hold? It seems that since a distance exists there exists a way to measure it.
Edit: Would this mean a line in the usual sense of the solutions $(x,y)$ to some equation $ax+by = d$, where $(a,b) \not = (0,0)$ ?