The only operations defined on points in an affine space are

• point-vector addition. this yields a new point.
• point-point subtraction. this yields a vector.

This can be extended to an affine sum, $$\sum_i \alpha_i P_i$$, where $$\sum_i\alpha_i = 1$$.

However, it is possible to derive an expression using these operations that should otherwise be undefined.

Imagine points, $$P$$, $$Q$$, $$R$$, and $$S$$ such that

$$P - Q = R - S$$

$$P + S = Q + R$$

Each side of the equation is now a point-point addition which is not defined. It is also not an affine sum.

How should each side of this equation be interpreted?

• On the other hand, $\frac{1}{2} P + \frac{1}{2} S = \frac{1}{2} Q + \frac{1}{2} R$ is a valid equation of affine combinations, and one would hope it could be proved using the axioms of affine space that it's equivalent to $P-Q = R-S$. – Daniel Schepler Feb 28 at 1:38
• I also don't have any references, but I think I saw a presentation once that given an affine space $A$ with vector space $V$, you can put a natural vector space structure on $V \sqcup ((\mathbb{R} \setminus \{ 0 \}) \times A)$ where you treat $V$ as being "weight 0" elements and $(\lambda, a)$ where $\lambda \ne 0$ as being "weight $\lambda$" elements. You also embed $A$ into this vector space via $a \mapsto (1, a)$, i.e. as "weight 1" elements. So then, $P+S$ and $Q+R$ become weight 2 elements of that vector space. – Daniel Schepler Feb 28 at 1:49

This leads to weighted points in affine space. The weight of a point must be nonzero and usual affine points have weight one by definition. Given weighted points $$\,aP\,$$ and $$\,b\,Q\,$$ their sum is $$\,aP+b\,Q\,$$ which has weight $$\,c:=a+b\,.$$ If $$\,c\,$$ is nonzero then this is the weighted point $$\,c\frac{aP+b\,Q}c.\,$$ If $$\,c=0\,$$ then the sum is the vector $$\,a(P-Q).\,$$ In general a "point" with weight $$0$$ is just a vector of the underlying vector space. That is, all weighted points have nonzero weights, but if a sum of weighted points has weight zero, then it becomes a vector instead.