$F$ and $G$ each intersect line $L$ at the same points with the same multiplicities. Prove $\exists \alpha\in R$ s.t. $L$ divides $F + \alpha G$.

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Let $F = 0$ and $G = 0$ be distinct projective plane curves of degree $n$. Suppose $L = 0$ is a line such that $F$ and $G$ each intersect $L$ a total of $n$ times with multiplicity, and that $F$ and $G$ each intersect $L$ at the same points with the same multiplicities. Prove that there is some $\alpha\in R$ such that $L$ divides $F + \alpha G$.

Please help me check my attempt.

My attempt:

Suppose $P$ is one of the intersection points in the finite plane. Then consider $I_p(L,F)$ and $I_p(L,G)$. We can find a transformation that transform $P$ to the origin $O$. Then $I_p(L,F)=I_O(L',F')$ and $I_p(L,G)=I_O(L',G')$. Then $I_p(L,F+\alpha G)=I_O(L',F'+\alpha G')$. Since $L'$ passes through the origin, then the equation of $L'$ is either $x=0$ or $y=mx$.

$I_O(y-mx,F'+\alpha G')=I_O(y-mx,F'(x,mx)+\alpha G'(x,mx))= the\ smallest\ degree \ of\ any\ nonzero\ term\ of\ F'(x,mx)+\alpha G'(x,mx)=a'$.

Since $I_O(L',F')=I_O(L',G')=a$, then we can find an $\alpha$ such that $a<a'$. We do this for all the intersection points. Then we will get that the curve $F'+\alpha G'$ intersect L' more that n times with multiplicties. This is impossible unless $L$ divides $F'+\alpha G'$.

If the interstection points are at infinity, we can have a transformation that transform them to the finite plane.

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Pick a point on $L$ that is not one of the points of intersection and then choose $\alpha$ such that $F+\alpha G$ is zero at that point.