Do any two points in Minkowski spacetime determine a unique line? Any two points in a Euclidean space determine a unique line, but I wasn't sure if this result generalized to Minkowski spacetime given that the latter is not a Euclidean 4-space, but is, instead, a Euclidean 3-space plus a fourth temporal dimension. 
 A: Let us start from the notion of affine space next focussing on the Euclidean $3$-dimensional physical space and finally coming to Minkowski spacetime.
An affine (real) $n$-dimensional space is a triple $(\mathbb A,\vec{\cdot}, V)$, where $\mathbb A$ is a set whose elements are called points, $V$ is a real $n$-dimensional vector space and  $\vec{\cdot} : \mathbb A \times \mathbb A \to V$ is a map associating a pair of points $P,Q \in \mathbb A$ with a vector $\vec{PQ}\in V$. The following requirements must hold.


*

*$\vec{PQ}+ \vec{QR} = \vec{PR}\:$ if $P,Q,R \in \mathbb A$.

*For every $P\in \mathbb A$ and $v\in V$, there exists exactly one $Q \in \mathbb A$ such that $\vec{PQ}=v$. 
These requirements permits one to define the notion of Cartesian coordinate system on $\mathbb A$. 
This is nothing but a particular bijective map  $\psi : \mathbb A \to \mathbb R^n $. 
To this end, fix an origin, i.e. a preferred point  $O\in \mathbb A$ and a system of axes, i.e., a vector basis $e_1,\ldots, e_n \in V$. 
The bijective map $\psi : \mathbb A^n \to \mathbb R^n $ is the one associating $P\in \mathbb A$ with the components $(x_1(P), \ldots, x_n(P)) \in \mathbb R^n$ of the vector $\vec{OP}$ with respect to the basis $e_1,\ldots, e_n \in V$.
An example of $3$-dimensional affine space is $\mathbb R^3$ itself. However this space has  much more structure than a generic affine $3$-dimensional space. For instance a preferred point $(0,0,0)$ and a preferred basis, the canonical one, of the space of translations $V= \mathbb R^3$. For these reasons $\mathbb R^3$ is not a good mathematical representation of the physical Euclidean space, in a sense it includes too much mathematical structure with no corresponding physical objects. On the other hand, even an affine $3$-dimensional space is not adequate to describe the physical space, since we also need further  metrical structures to properly mathematically describe the physical space.
The appropriate structure for describing the physical space is the notion  of Euclidean space.
An Euclidean  $n$-dimensional space $\mathbb E^n$ is a quadruple  $(\mathbb E,\vec{\cdot}, V, < , >)$, where $(\mathbb E,\vec{\cdot}, V)$ is an affine space and $<,> : V \times V \to \mathbb R$ is a symmetric 
positively-defined scalar product.
This structure selects a preferred distance function $d: \mathbb E \times \mathbb E \to [0,+\infty)$ defined as 
$$d(P,Q) = \sqrt{<\vec{PQ}, \vec{PQ}>}\:.$$
This distance is, by construction, translationally invariant: $\vec{P'Q'}= \vec{PQ}$ implies $d(P,Q)= d(P',Q')$.  
The definition of orthonormal Cartesian coordinate system is now obtained by specializing  the notion of  Cartesian coordinate system to the case of an orthonormal basis $e_1,\ldots, e_n \in V$ with respect to $<,>$.
For $n=3$, $\mathbb E^n$ is just the model of the physical Euclidean space.
The notion of (straight) line is given in an affine space and, in fact, two distinct points determine a unique line. If $P,Q \in \mathbb A$ with $P\neq Q$, the associated line is $r = \{R \in \mathbb A \:|\: \vec{RQ} = \lambda \vec{PQ}\:\: \forall \lambda \in \mathbb A \}$.
As you see no metrical notion is necessary.
Let us eventually come to Minkowski spacetime $\mathbb M^4$ which differs form an Euclidean space only in the metrical structures. 
First of all $\mathbb M^4$ is a $4$-dimensional affine space, whose points are called events.
(Already with this part of the definition,   the notion of straight line does make sense.)
Secondly, $\mathbb M^4$ is equipped with a Lorentzian scalar product in the space of translations $V$.    This is a bilinear map $<,> : V \times V \to \mathbb R$, which is symmetric ($<u,v>= <v,u>$), non-degenerate ($<u,v>=0$ for every $v\in V$ implies $u=0$) and with Lorentzian signature (there are bases $e_1,e_2,e_3,e_4$ where $<,>$ is represented by the matrix 
$diag(-1,1,1,1)$).
A: The concept of a line is totally independent of the inner product and makes sense for a general vector space; mathematically a line is an affine one-dimensional subspace. So yes, two points do determine a unique line, as long as you're in a flat vector space. If you introduce curvature as in General Relativity, things change.
The only difference with regular 3D space is that your line in Minkowski space might correspond to faster than light movement, which would be unphysical if you want some physical object to follow the line. But the mathematical concept of line itself is perfectly well defined.
