I know that:
- straight lines in the $2$-norm are just regular straight lines
- straight lines in the $1$-norm are lines made from horizontal and vertical segments
- straight lines in the $\infty$-norm are lines made from segments at multiples of $45$ degrees
But what do they look like for other values of $p$?
By "straight line", I mean a function $f : \mathbb{R} \to X$ such that $d(f(x), f(y)) = |x - y|$ for all $x, y \in \mathbb{R}$.
I guess, you are asking, given two points $A$, $B$, what is the set of points $C$ satisfying the triangle equality $|AC|+|CB| = |AB|$. In an euclidean space, it is simply the segment $AB$.
In the case of 1-norm, it is an entire rectangle with corners in $A, B$ aligned with the axes, in the case of $\infty$-norm, it is a rectangle with corners in $A, B$ under the angle $45^\circ$.
However, note that the original (euclidean) segment $AB$ is still contained inside these rectangles. This is not a coincidence. By the very fact we are dealing with a vector space with a norm, ve have for every $\alpha\in(0,1)$ and a vector $v$ the following: $|\alpha v| + |(1-\alpha)v| = \alpha |v| + (1-\alpha)|v| = |v|$. So in any norm, the standard (euclidean) segment will always satisfy the triangle equality.
So your question is basically asking whether there is something extra to the standard segment in a general $p$-norm, as it is in the case of a 1-norm, or $\infty$-norm, and the answer is: No. Only the degeneracy of 1-norm and $\infty$-norm allows other points to satisfy the triangle equality.
Therefore, for any $p\in(1;\infty)$, the segments in the $p$-norm look the same as segments in 2-norm.