The title sounds trivial but bear with me.
We know from elementary geometry that every nondegenerate triangle $ABC$ (ie. points $A, B, C$ are all distinct) has an orthocenter $H$, defined as the intersection between two altitudes from a vertex to the opposite side. Importantly and well known (but not obvious to a beginner to prove) is that all three altitudes concur at the same point.
However, an issue arises when we consider a degenerate triangle, one in which (say) $A$ lies on segment $BC$. Then, the altitudes are all parallel lines, and in the standard Euclidean plane, these lines do not intersect at all, rendering our pedestrian definition of an orthocenter useless and leaving $ABC$ without an orthocenter.
My question is, can we expand / redefine the plane such that every "triangle" $ABC$ (nondegenerate or degenerate) has an orthocenter $H$ in the expanded plane?
My first guess was the real projective plane $\mathbf{P}^2$, which (ignoring the details of the $\mathbb{R}^3$ manipulation) adds one point at infinity for every direction with angle in $[0, \pi)$, and two parallel lines intersect at the point at infinity corresponding to the direction of the lines. (a direction is the same as the opposite direction.) This construction works for now - if $P$ is a "finite" point whereas $Q$ a point at infinity, then the line $PQ$ is defined as the (Euclidean) line through $P$ running in the direction of $Q$.
However, this breaks the following lemma from conventional Euclidean geometry: If $H$ is the orthocenter of $ABC$ (degenerate or otherwise), then (by definition) $A$ is the orthocenter of $HBC$ (using properties of perpendicularity and collinearity). If we had two distinct prospective orthocenters $H_1$ and $H_2$ on segment $BC$, then each of $H_1BC$ and $H_2BC$ would have orthocenter $A$, the point at infinity corresponding to the direction parallel to line $BH_1H_2C$. Then using our lemma, triangles $ABC$ could have orthocenter either $H_1$ or $H_2$, contradicting our desire that every triple of points has a unique orthocenter. For a given segment $BC$, all degenerate triangles $ABC$ with $A$ on line $BC$ would have the same orthocenter! Evidently, we need our points at infinity to carry some information not just about direction but also affine displacement in the direction perpendicular to that direction.
Is there a version of our pedestrian Euclidean plane that can ensure every triple of points has a defined and unique orthocenter?