Find a ruler in Taxicab plane 
In the Taxicab Plane, find a ruler $f$ with $f(P)=0$ and $f(Q)>0$ for
  the given pair $P$ and $Q$:
$1. \ P = (2,3), Q=(2,-5)$
$2. \ P= (2,3), Q = (4,0).$

The definition of ruler is
Ruler: Let $l$ be a line in an incidence geometry. Assume that there is a distance function $d$. A function $f: l \to \mathbb{R}$ is a ruler for $l$ if $f$ is a bijection and for each pair of points $P$ and $Q$ on $l$ we have $|f(P) - f(Q)| = d(P,Q)$. Where $f(P)$ is called the coordinate of $P$ with respect to $f$.
I am not sure how to do this question.
 A: *

*In the taxicab metric, $d(P,Q)=|2-2|+|3-(-5)|=8$. The line connecting these two points is $$\ell\equiv\{(x,y)\in\mathbb R^2\,|\,x=2\}.$$ We have the the requirement that $$|f(P)-f(Q)|=|0-f(Q)|=|f(Q)|=d(P,Q)=8.$$ Therefore, either $f(Q)=8$ or $f(Q)=-8$. But we need $f(Q)>0$, so $f(Q)=8$. We can then extend $f(\cdot)$ onto the whole line $\ell$ as follows:
\begin{align*}
f(x,y)=3-y\quad\forall (x,y)\in\ell.
\end{align*}
You can check that $f$ is a bijection between the line $\ell$ and $\mathbb R$, and the definition of $f(\cdot)$ makes it apparent that $f(P)=f(2,3)=3-3=0$ and $f(Q)=f(2,-5)=3-(-5)=8$, as desired. Finally, if $P'=(p_1',p_2')$ and $Q'=(q_1',q_2')$ are another pair of points on the line $\ell$, then we have $p_1'=q_1'=2$ (with $q_1'\in\mathbb R$ and $q_2'\in \mathbb R$ being unrestricted) and also 
\begin{align*}
|f(P')-f(Q')|=&\,|f(p_1',p_2')-f(q_1',q_2')|=|f(2,p_2')-f(2,q_2')|\\=&\,|(3-p_2')-(3-q_2')|=|q_2'-p_2'|=d((p_1',p_2'),(q_1',q_2'))=d(P',Q'),\end{align*}
since we are using the taxicab metric.


The second one can be solved using a similar reasoning, but finding the equation for the line $\ell$ connecting $P$ and $Q$ requires some more effort.
