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Here is my question:

Let $f(x,y)\in {\Bbb C}[x,y]$ be a quadratic polynomial. If $f$ vanishes on three different points (say, $p,q,r$) of a complex line $$ L:=\{(x,y)\in{\Bbb C}^2\mid ax+by-c=0\},\quad a,b,c\in{\Bbb C}, $$ then do we have $f(x,y)=0$ for all $(x,y)\in L$?

The only idea I have come up so far is that on the real plane ${\Bbb R}^2$, three different points define a quadratic polynomial $f(x)\in {\Bbb R}[x]$. This might be a useful to be generalized to the complex case. But I really don't see how to go on.

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up vote 3 down vote accepted

Yes. The restriction $f\left|_L\right.$ is a one-dimensional quadratic polynomial, which vanishes on three different points, i.e. has three distinct zeros. Therefore, $f\left|_L\right.=0$.

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