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Suppose we have $X\subset\mathbb{P}^2$ an irreducible variety, and consider the 2-secant variety $\sigma_2(X)$. In a lecture I read that if X is a curve, and not a line, then $\sigma_2(X)=\mathbb{P}^2$, but I really can't understand why (I just started studying secant varieties and this is considered an easy example). Any hint or suggestion would be helpful!

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If $P\in \mathbb P^2$ is an arbitrary point, any line $L$ through $P$ that is not tangent to $X$ and doesn't go through a singularity of $X$ will cut $X$ in $d$ distinct points: $X\cap L=\{Q_1,\cdots,Q_d\}$.
If $X$ is not a line, $d\geq 2$ so that the the secant line $\overline {Q_1Q_2}$ is equal to $L$.
Thus every point $P$ is indeed on a secant of $X$, namely $\overline {Q_1Q_2}=L$, and this means that the secant variety of $X$ is indeed $\mathbb P^2$.

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