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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.

An affine map combines a linear map with a translation. After choosing coordinate origins, it has the form $T:{\mathbb{R}}^p\to {\mathbb{R}}^q$, $T(x)=Mx+b$, where the $q\times p$ matrix $M$ is its linear part of an affine map and $b$ is a fixed vector. Equivalently, it preserves every affine combination. ^[ref]

An original planar example isIt stretches horizontally, applies a shear mapping, and then translates. The vertices $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$ become $A^{\prime }=(1,-2)$, $B^{\prime }=(5,-2)$ and $C^{\prime }=(3,0)$. A square grid becomes a grid of parallelograms; straight edges and their weighted points still fit together.

This map is not a linear map with respect to the displayed origins, because $T(0,0)=(1,-2)$ rather than zero. It also is not a Euclidean similarity: $AB$ has length $2$ and $A^{\prime }{B}^{\prime }$ has length $4$, while the right angle at $A$ becomes $45^{\circ }$. Its determinant is $2$, so it doubles planar area. These metric changes explain why affine geometry studies straightness, incidence and weighted points rather than treating every transformation as an isometry.

The linear part is an invertible matrix, giving the concise inverse $T^{-1}(u,v)=((u-v-3)/2,v+2)$. More generally an affine map is bijective exactly when its linear part of an affine map is bijective. Between equal finite-dimensional spaces, this means an invertible matrix; the translation cannot repair a loss of rank of a linear map.

A control point is an editable geometric handle used to specify a curve or surface through a basis or refinement rule. The handle is part of the representation; it need not be a point on the resulting curve. In a Bézier curve, the control points are the geometric coefficients of the Bernstein basis. Joining them in order gives a control polygon, which makes the handles easy to manipulate without drawing the curve as a chain of line segments.^[ref]

For example, use the cubic Bézier curve with control pointsIts Bernstein basis representation givesThe greatest height is $1.5$, so neither interior control point, both at height $2$, lies on this curve. The endpoint control points do lie on it: $C(0)=P_0$ and $C(1)=P_3$. This separates interpolation, which requires passage through prescribed data, from an approximating representation, which uses handles to shape a curve without requiring passage through every handle. If measured points must be interpolated, choose control points to satisfy those constraints; simply using the measured points as the handles of a Bézier curve does not enforce interpolation.

Now move only $P_1$ by the vector $h=(0,1)$, giving ${\tilde{P}}_1=(1,3)$. With the Bernstein basis and parameter $t$ unchanged, the new curve isAt $t=1/2$ this gives $\tilde{C}(1/2)=(2,15/8)$, a displacement $(0,3/8)$ despite moving the handle by a full unit. Both endpoints stay fixed because the coefficient $3t(1-t{)}^2$ vanishes there. Thus dragging a control point shapes a whole range of the curve; it does not generally drag an arbitrary selected point on the curve by the same vector. The control-point influence on a curve tells us exactly how much each parameter value moves.

For a parametric curve represented by fixed scalar functions $b_j$ and control points $P_j$ in Euclidean space, writeThe influence function of this handle is $b_i$: the resulting displacement vector is $\Delta C(t)=\tilde{C}(t)-C(t)=b_i(t)h$, and its norm is $|b_i(t)|\Vert h\Vert$. These are exact finite changes, not only a first-order estimate. Where $b_i$ is negative the displacement points opposite to $h$. Where $b_i(t)=0$, the curve is unchanged at that parameter.

For the cubic Bézier curve above, the Bernstein basis influence function is $b_{1,3}(t)=3t(1-t{)}^2$. Its derivative isIt increases up to $t=1/3$ and decreases thereafter on $[0,1]$. Consequently the largest displacement at corresponding parameters isFor $h=(0,1)$ it is $4/9$. This is a comparison at the same parameter. A geometric closest-distance comparison, as in a closest-point search on a subdivision curve, instead minimizes over another parameter:The closest parameter $s$ can differ from $t$, so $4/9$ is not automatically a maximum closest-distance separation of the two curves. Neither parameter is assumed to be arc length.

The support of a function $b_i$ is the closure of the parameter values where $b_i\ne 0$. It bounds where a nonzero handle move can act, but may include boundary points where the actual displacement is zero. Here the support of a function $b_{1,3}$ is all of $[0,1]$ relative to the parameter domain, while the endpoints are unaffected. In contrast, a B-spline of order $k$ has support within $[t_i,t_{i+k}]$ in its spline knot sequence, so its handle affects only that parameter range; the support and Gram bandwidth of B-splines describes the same locality in another application. Changing the spline knots or the basis along with the handle requires a different comparison.

When the basis functions form a partition of unity, ${\sum }_j{b}_j(t)=1$, each evaluation is an affine combination of the control points. Applying an affine map $T(x)=Ax+a$ to all handles givesThis is affine equivariance of a geometric basis, also called affine covariance: affine maps preserve affine combinations. In particular, translating every handle by $h$ translates the whole curve by $h$, whereas moving just one handle produces $b_i(t)h$. The Bernstein basis has this partition of unity property.^[ref] Nonnegative weights further make evaluations convex combinations, giving the convex hull bound illustrated by the control polygon; nonnegativity is not needed for affine covariance.

For a proof outline of the influence formula, subtract the two basis representations: every unchanged coefficient cancels, leaving $b_i(t)(P_i+h)-b_i(t)P_i=b_i(t)h$. The same subtraction with several moved handles gives $\Delta C(t)={\sum }_i{b}_i(t)h_i$. No claim about interpolation, closest-point search on a subdivision curve or endpoint motion follows until the actual basis values are examined.