Can we predict if two line segments on the plane will intersect each other?

Question

Let $X$ be a non-empty set of 4 points in the plane, and $(X,\mathrm{d})$ the euclidean metric on $X$. Let $(x_0,y_0)\in X^2$ be any pair of points in $X$ and $\overline{x_0y_0}$ a straight line joining them. Can we tell if the line $\overline{x_1y_1}$ between the remaining points will intersect $\overline{x_0y_0}$ using only points in $X$ and distances in the metric space? (Angles too, but no line extending is allowed).

I've been doing a lot of thinking, but the only logical conclussion i came up with is that this is impossible: to know if two line segments do not intersect each other i would need to prove they are either parallel or the intersection point lies outside the convex hull of $X$, but these are all properties of lines in the plane, not points in $X$.

Answer 1

You can indeed determine this just from the distances.

To see this, note that after applying an isometry of the plane, we can assume with no loss of generality that $x_0=(0,0)$, $y_0=(0,d)$ for some $d>0$, and $x_1$ lies in the (closed) upper half plane. Then $x_1$ is uniquely determined by its distance to $x_0$ and $y_0$, and similarly $y_1$ is uniquely determined by its distance to $x_0$ and $y_0$, up to a reflection across the horizontal axis.

But then from the two possible choices, $y_1$ is uniquely determined by its distance from $x_1$, except in the special case that both possible locations are equidistant from $x_1$, in which case $x_1$ must lie on the horizontal axis. In this latter case, reflection across the horizontal axis gives an isometry preserving $x_0$, $y_0$, and $x_1$, so we may assume that $y_1$ is in the upper half plane as well, and is still uniquely determined.

Since all four points have uniquely determined locations in the plane up to a planar isometry, one can determine the intersection (or lack thereof) using planar geometry.

Remark

The number $4$ isn't really important in this problem. That is, if we have any subset $X\subseteq \mathbb R^2$, then either:

1. All triples of points in $X$ satisfy $d(x,y)+d(y,z)=d(x,z)$ (for some permutation of the points). In this case all of the points are colinear, and we may map some $x$ to $0$, some $y$ to $d(x,y)$, and the distances to $x$ and $y$ uniquely determine a location on the real line (or the horizontal axis) for every other point in $X$.

or

2. There are three points with $d(x,y)+d(y,z)>d(x,z)$ for all permutations, in which case after a planar isometry we may assume $x=(0,0)$, $y=(0,d)$, $z$ is in the open upper half plane (uniquely determined by its distance from $x$ and $y$), and every other point in $X$ is uniquely determined by its distances to $x$, $y$, and $z$.

The upshot is that any isometry between two planar subsets $X_1$ and $X_2$ extends to a linear isometry of the plane itself. As a corollary, any property of a planar subset $X$ that is invariant under planar isometries will be determined solely by the inherited metric on $X$. (A similar statement is also true in higher dimensions as well.)