Affine combination (${\sum }_i{\lambda }_i{P}_i,{\sum }_i{\lambda }_i=1$)

An affine combination is a weighted point $P={\sum }_{i=0}^k{\lambda }_i{P}_i$ with ${\sum }_{i=0}^k{\lambda }_i=1$. The coefficients can be negative. The sum is meaningful in an affine space without selecting a preferred origin: relative to any point $O$, it means $P=O+{\sum }_i{\lambda }_i(P_i-O)$. Affine maps preserve these weighted points. ^[ref]

For an original example in the Euclidean plane, take $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$. Then $P=\frac{1}{4}A+\frac{1}{2}B+\frac{1}{4}C=(1,\frac{1}{2})$ lies inside their triangle. Its coefficients are nonnegative, making it a convex combination. In contrast, $R=-\frac{1}{2}A+\frac{3}{2}B=(3,0)$ is beyond $B$: it is an affine combination but not a convex combination of these three points. Negative weights allow extrapolation, rather than only interpolation inside the convex hull.

For the affine map $T(x,y)=(2x+y+1,y-2)$, the same weights produce $T(P)=(\frac{7}{2},-\frac{3}{2})$ from $T(A),T(B),T(C)$, and produce $T(R)=(7,-2)$ from $T(A),T(B)$. Thus one can compute a weighted point before transforming the vertices or afterwards, obtaining exactly the same point.

This is useful whenever an object is built from weighted control points: its construction commutes with an affine map, so transforming the control points transforms the constructed object as well. The sum of the weights is the essential condition; no assumption about lengths or angles is needed.

Keep the basis fixed and replace the origin by a point with old coordinates $o$. Each point's coordinates change from $p_i$ to $p_i-o$. For coefficients with sum one, the weighted coordinates becomeAdding the new origin $o$ gives the original point. This proves origin independence rather than asserting that the coordinate tuples themselves stay unchanged.

The requirement cannot be omitted. In the example $A=(0,0)$, $B=(2,0)$, blindly treating $A+B$ as a point gives $(2,0)$ in the old coordinates. If the new origin is $o=(1,1)$, the two coordinates become $(-1,-1)$ and $(1,-1)$; adding them and restoring $o$ instead gives $(1,-1)$. The coefficients sum to two, so this procedure has no origin-independent interpretation as a weighted point.