Show that the railway metric is a metric

Question

I'm having trouble proving this. I am able to prove other metrics, I think it is possibly the format of the railway metric that is confusing me...

Consider the function $d : \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow [0, \infty)$ given by

$d(x,y)=\begin{cases} d_2(x,y)~\text{if $x,y,0$ are collinear}\\d_2(x,0) + d_2(0,y) ~\text{otherwise}\end{cases}$

I have also tried to attach a photo of the question)

Answer 1

Let d be the Euclidean metric for the real plane.
Let O be the point (0,0).
The railroad metric p for the real plane is
p(x,y) = d(x,y) when x,y,0 are collinear,
. . . . = d(x,0) + d(0,y) otherwise.

This is also called the Frence railroad metric.
All trains go through Pairs.
To get from a to b, take the train from a to Paris and then the train from Paris to b unless b is on the way from a to Paris or visa versa.

Clearly p(x,y) = p(y,x) and p(x,y) = 0 iff x = y.
For the triangle inequality, consider the case when x,y,0, y,z,0 and x,z,O are not collinear.
Then p(x,z) = p(x,0) + p(0,z)
<= p(x,0) + p(0,y) + p(y,0) + p(0,z)
= p(x,y) + p(y,z).
The other cases are left for the diligent reader.