Example
Claim: \(\mathbb P^2 \setminus \{\text{pt}\}\) is \(\Spec _{\mathbb P^1} \Sym \mathcal O_{\mathbb P^1}(1)\) Consider a point \(x \in \mathbb P^2\) and \(X = \mathbb P^2 \setminus \{x\}\). Choose a line \(\mathbb P^1= \ell \subset X\) and the projection map \(\pi: X\to \ell\) given by projecting \(\mathbb P^2\) to \(\ell\) away from \(x\). The fibers of \(\pi\) are all copies of \(\mathbb A^1\), hence \(X\) is a line bundle over \(\ell\). Since \(\ell \cong \mathbb P^1\), all of whose line bundles take the form of \(\mathcal O(d)\) for some \(d\), we are done if we can show \(d = 1\). Any other line \(\ell'\subset \mathbb P^2\) which avoids \(x\) is a section of \(\pi\), or if you rather, is the vanishing of a section of \(\mathcal O_{\mathbb P^1}(1)\). Since \(\ell \cap \ell' \) is a single point, we conclude that
\begin{align*} \deg \mathcal O_{\mathbb P^1}(d) = 1 \implies d = 1 \end{align*}giving us the result.
We can see this in coordinates in a straightforward way too. Let \(p = [0:0:1] \in \mathbb P^2_{xyz}\) and \(\ell = \{[z_0:z_1:0] \}\) so that the projection \(\pi:X\to \ell = \mathbb P^1\) is given \([x:y:z] \mapsto [x:y]\). Then the fibers of \(\pi\) are given by
\begin{align*} \pi^{-1}([x:y]) = \{[x:y:\lambda] | \lambda \in \mathbb A^1\}. \end{align*}To see that the line bundle \(X \to \ell\) is \(\mathcal O(1)\), take the global section
\begin{align*} [x:y] \mapsto [x:y:x]. \end{align*}It has exactly one intersection with the zero section of \(\pi\) which occurs at \([x:y] = [0:1]\), so the degree of the bundle is indeed \(1\).