The pairing $V \times V^* \to K$ gives a -host- of dualities,
(especially for $K=\bR$, as then you get an ordering)!

$$x \mapsto \set{v \st P}$$
...where $P$ is:
$vx=0$: vector duality: closure is vector space closure, dual is perp
$vx=1$: affine duality: closure is affine closure, dual is
  ...it is what it is: it's ``affine `perp' ''.
$vx\leq 1$: convex (vector) duality: closure is convex (vector) closure,
  meaning you have to include $0$, dual is convex dual.

You can get convex affine duality, but then you have to pair
$V$ with $\Aff(V^*)$.
This makes sense: the -affine- dual of a subset $V$ can't be any
-particular- subset of $V^*$: you could move it by any translation,
which corresponds to living in $\Aff$ instead of the original space.
(er, and by $\Aff$ I mean: it's what you get if you include
translations of $V$, together with linear duals)

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Note that affine duality is ultimately coming from projective duality:
points in $\bP V$ correspond to hyperplanes in $\bP V^*$.

(and $n$ generic points determine a hyperplane, while $n$
 generic hyperplanes determine a point:
 nicest for $n=2$, where ``generic'' means ``not equal'',
 hence the classical projective duality of the projective plane)
