Given the definition of a convex set:
A set $X\in \mathbb{R}^n$ is called convex if for all $x^1 \in X$ and $x^2 \in X$ it contains all points $$\alpha x^1 + (1-\alpha)x^2, \quad 0 < \alpha < 1.$$
Let $n =2$.
I notice that if i choose two points $x^1 = [p,q]^{\intercal}$ and $x^2 = [r,s]^{\intercal}$,
if $p + q = r + s$,
then for $\alpha \cdot [p,q]^{\intercal} + (1-\alpha)\cdot[r,s]^{\intercal} = [t,u]^{\intercal}, \quad 0 < \alpha < 1$,
$t + u = p + q = r + s$.
This also appears to hold for any $n$.
Can anyone give me some insight as to why this is the case? (It might be very trivial!) Or if there are extensions beyond the case where $p + q = r + s$?
I'll designate a point as $(u,v)$. For a fixed value of $k$, the curve defined by $u+v = k$ is a line. What you've shown is that if two points both lie on such a line, then all the points on the line segment connecting them also lie on the same line.
In 3 dimensions, where a point is $(u,v,w)$, $u+v+w = k$ defines a plane, and you've shown that the line segment connecting two points lies in any plane containing the two end points.
And yes, there are extensions! The set of lines that you can describe with an equation like $u+v=k$ is a small subset of all lines. To describe an arbitrary line in $R^2$, you can find constants $a,b,$ and $k$, so that the set of $(u,v)$ satisfying the equation $au+bv = k$ is your line. If you calculate $au+bv$ (just like you calculated $u+v$) you'll get the same result for an arbitrary line: if it contains the end points, then it contains the entire segment between them.