Is there always a line that intersects a given Zariski-closed set in finite number of points?

Assume that we're working in an algebraically closed field. Let $X \subsetneq \mathbb{A}^n$ be a Zariski-closed set. Is there a line in $\mathbb{A}^n$ that intersects $X$ in finitely many points? I need a hint.

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Is $X$ supposed to be properly contained in $\mathbb{A}^n$? And should "algebraically closed set" be "algebraically closed field"? – Qiaochu Yuan Dec 29 '12 at 7:31
@QiaochuYuan it was a typo :) About $X$, assume you're talking to someone who doesn't know the first thing about schemes. – Alexei Averchenko Dec 29 '12 at 7:32
@Alexei: I think Qiaochu's point is that if $X=\mathbb{A}^n$ the answer is trivially no (since any algebraically closed field is infinite). – Zev Chonoles Dec 29 '12 at 7:36
@ZevChonoles Oh :) Yeah, I meant a proper subset. – Alexei Averchenko Dec 29 '12 at 7:37

Suppose that a line $\ell$ intersects $X$ in infinitely many points. The intersection is an infinite Zariski closed subset of $X$ and $\ell$. Now $X$ may have lots of different infinite Zariski closed subsets (depending on what it is), but what about $\ell$?

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Once again, AG is so easy once someone explained it :D Thanks! – Alexei Averchenko Dec 29 '12 at 7:38