Let $\mathcal{E}$ be a rank two vector bundle on $\mathbb{P}^2$ fitting in the following exact sequence


$$0\rightarrow \mathcal{O}\_{\mathbb{P}^2}\rightarrow \mathcal{E}\rightarrow \mathcal{I}\_p(-1)\rightarrow 0$$


where $\mathcal{I}\_p$ is the ideal sheaf of a point $p\in\mathbb{P}^2$. Then $c\_1(\mathcal{E})=-1$ and $c\_2(\mathcal{E})=1$.


Let $\pi:X = \mathbb{P}(\mathcal{E})\rightarrow\mathbb{P}^2$, and $-K\_X$ the anti-canonical divisor of $X$. Finally, let $C\subset X$ be an irreducible curve such that $\pi(C) = L\_p$ is a line through $p\in\mathbb{P}^2$. 


How can we compute the intersection number $-K\_X\cdot C$ ?



Intersection numbers in $\mathbb{P}^1$-bundles
Here are the details.


Let $L\_p\cong\mathbb{P}^1$ be a line in $\mathbb{P}^2$ through $p$. Your exact sequence restricted to $L\_p$ yields the following exact sequence
$$0\rightarrow\mathcal{O}\_{\mathbb{P}^1}(1)\rightarrow\mathcal{E}\_{|L\_p}\rightarrow\mathcal{O}\_{\mathbb{P}^1}(-2)\rightarrow 0
$$
and hence $\mathcal{E}\_{|L\_p}\cong\mathcal{O}\_{\mathbb{P}^1}(1)\oplus\mathcal{O}\_{\mathbb{P}^1}(-2)$. Therefore, $\pi^{-1}(L\_p) = \mathbb{P}(\mathcal{O}\_{\mathbb{P}^1}(1)\oplus\mathcal{O}\_{\mathbb{P}^1}(-2))$ is the Hirzebruch surface $\mathbb{F}\_3$. Let $\overline{\xi}$ be the class of the only curve with self-intersection $-3$ on $\mathbb{F}\_3$, and let $f$ be the class of a fiber of $\pi\_{|\mathbb{F}\_3}:\mathbb{F}\_3\rightarrow\mathbb{P}^1$. Then if $\widetilde{H} = \pi^{\*}\mathcal{O}\_{\mathbb{P}^2}(1)$ and $\xi = c\_1(\mathcal{O}\_X(1))$ we have $\widetilde{H}\_{|\mathbb{F}\_3} = f$ and $\xi\_{|\mathbb{F}\_3} = f+\overline{\xi}$. 


Since $c\_1(\mathcal{E})=-1$ and $c\_2(\mathcal{E}) = 1$ the canonical divisor of $X$ is given by $-K\_X = 4\widetilde{H}+2\xi$ and we may compute the intersection $-K\_W\cdot\overline{\xi}$ inside $\mathbb{F}\_3$, that is
$$-K\_X\cdot\overline{\xi} =
(4\widetilde{H}+2\xi)\_{|\mathbb{F}\_3}\cdot\overline{\xi} = (4f+2(f+\overline{\xi}))\cdot\overline{\xi} = (6f+2\xi) = 6+2(-3) = 0.$$


