Let $E$ be a vector space, $H\subset E$ a vectorial hyperplane, and $\mathcal{H} \neq H$ an affine hyperplane parallel to $H$. Then I know that every vectorial line (one-dimensional subspace) $D \subset E$ not contained in $H$ meets $\mathcal{H}$ in a unique point $P_D$; and inversely, for every $P\in \mathcal{H}$ there is a single vectorial line $D_P$ not contained in $H$ that passes through $P$.
We then have a bijection $\mathcal{H} \rightarrow \mathbb{P}(E) \setminus \mathbb{P}(H)$ such that $P \mapsto D_P$, which allows us to think of $\mathbb{P}(E) \setminus \mathbb{P}(H)$ as an affine space.
I also know that for a fixed $\mathbb{P}(H)$, we can replace $\mathcal{H}$ by a different affine hyperplane $\tilde{\mathcal{H}} \neq H$ parallel to $H$ without modifying in a significant manner the affine structure of $\mathbb{P}(E) \setminus \mathbb{P}(H) \cong \tilde{\mathcal{H}}$ (this amounts to subjecting every element in $\mathcal{H}$ to a vectorial homothety, but doesn’t alter the notions of affine subspaces, parallelism, etc.).
However, if I understood correctly from my lecture notes, the choice of the vectorial hyperplane $H$ as a starting point isn’t canonical and will result in a totally different affine structure for $\mathcal{H}$. That is, if $\mathbb{P}(H) \neq \mathbb{P}(H’)$, then the corresponding affine spaces $\mathcal{H}$ and $\mathcal{H’}$ will be different. My question is: is there any way to see why that is the case, or to relate the choice of $H$ to the affine structure of $\mathcal{H}$? Visual/intuitive explanations would help a lot: for instance, I’ve read that “a statement concerning parallel lines (in $\mathcal{H}$) would be translated to a new statement concerning concurrent lines (in $\mathcal{H’}$)”, but I’m not sure if I understand what this means.