Why can a trigonal genus 5 curve be represented as a plane quintic with a node?

Suppose $X$ is a smooth projective curve of genus 5 which a) is not hyperelliptic and b) possesses a $g_3^1$. I want to show that $X$ can be represented as a plane quintic with a node, i.e. there's a birational map $f:X\to \Bbb P^2$ which has image a plane quintic with a node. I think I can show most of this, but the main sticking point for me is why the singular point can be chosen to be a node and not a cusp.

Here's what I have so far: suppose $D$ is an effective divisor in the $g_3^1$. Then $l(D)=2$ (either because one assumes $g_3^1$ to be complete or by Clifford's theorem), and by Riemann-Roch $l(K-D)=3$. Now let $P\in X$ be arbitrary and consider $D+P$: $l(D+P)\geq l(D)=2$, and by Clifford's theorem, $l(D+P)\leq 2$, so $l(D+P)=2$. Therefore by Riemann-Roch, $l(K-D-P)=2$ and so $K-D$ is base-point free. Thus $K-D$ gives a map $f:X\to\Bbb P^2$, and $f(X)$ has degree 5 in $\Bbb P^2$. By the degree-genus formula, a plane quintic has arithmetic genus 6, so $f(X)$ must have one singular point with $\delta=1$ (where $\delta$ is the dimension of the integral closure of $\mathcal{O}_{f(X),x}$ modulo $\mathcal{O}_{f(X),x}$). But I know that both nodes and cusps have $\delta=1$, so how come I can get a node and not a cusp? I suspect I'm supposed to be able to deduce this from some geometry of the canonical embedding, but I don't see it yet.

I'm also potentially a little shaky on why $f$ should be birational on to its image: I'd love to use the criteria that $l(K-D-P-Q)=l(K-D)-2$ for almost all $P,Q$ (really, all $P,Q$ except the two points over the singularity) but I'm honestly unsure how to make this argument successfully. Again, it seems I'm missing an understanding of the geometry of the canonical embedding of $X$ in $\Bbb P^4$.

Solution 1:

If you take a general plane quintic curve with one cusp, its normalization has genus 5, and the projection from the cusp provides it with a $g_3^1$.