Let V(f) be a plane curve of degree d over an algebraically closed field having a point of multiplicity d. Prove that V(f) consists of d distinct lines.
If V(f) is d distinct lines then $V(f)=V(f_1) \cup ... \cup V(f_d)$.
The point of multiplicity $d$ would imply that $f$ vanishes $d$ times at p (which require the curve to be at least degree d).
I am not sure how to connect these.
Let $p$ be the point of multiplicity $d$, and let $q$ be any other point on $V(f)$. Then the line $\overline{pq}$ must be contained in $V(f)$, because otherwise it would have intersection number at least $d+1$ with $V(f)$, which is impossible.
Since every point of $V(f)$ lies on a line through $p$, this shows that in fact $V(f)$ is a union of lines through $p$. Since $V(f)$ has degree $d$, there must be exactly $d$ of these lines. (One might worry that the lines need not be distinct, but in that case $V(f)$ would be cut out by a polynomial of degree less than $d$.)