Rational sextic plane curves In Coolidge's venerable Treatise on algebraic plane curves, Theorem 28 p. 392 states:

*

*Let $S=\{P_1,\ldots ,P_{10}\} \subset \mathbb{P}^2$  such that for any $i$,   there is a sextic curve (edit: integral) singular along $S\smallsetminus P_i$; then there is a sextic curve singular along $S$.

I am probably missing something, but it seems to me that the text before and after this statement has nothing to do with the Theorem. Does someone know a proof?
 A: Let me try to reconstruct Coolidge's argument at pages 390-392.
Note: there is a misprint at line 4 of p. 392, where "with double points" should be "with triple points".

Proposition 1 [Coolodge, p. 390-391]. Assume that $P_1, \ldots, P_8$ are points on the plane, not three on a line and not six on a conic. Let $\phi_1$, $\phi_2$ be a basis for the pencil of cubics through the $P_i$ and let $f$ be a non-degenerate sextic with the $P_i$ as double points. Then a point $P_9$ is a further double point of a non-degenerate sextic with double points at $P_1, \ldots,  P_8$ if and only if it is a smooth point of the curve $$\frac{\partial(\phi_1, \, \phi_2, \, f)}{\partial(x_1, \, x_2, \, x_3)}=0.$$
This curve has triple points at each of the points $P_1, \ldots, P_8$ and passes simply through the $12$ singular points of the nodal curves of the pencil of cubics spanned by $\phi_1$ and $\phi_2$.


Proposition 2 [Coolidge, Theorem 27, p. 391]. Assume that nine points of the plane are the base locus of a pencil of nodal sextics. Then there are precisely $12$ sextics in the pencil having a further double point.

Using these results, we can now prove the

Theorem. Assume that $P_1, \ldots, P_{10}$ is a set of ten points such that for any nine of them there is a non-degenerate sextic (and hence a pencil of non-degenerate sextics) having nodes there . Then there is a non-degenerate sextic having nodes at all of the ten points.

Proof. Consider the nodal sextics through $P_1, \ldots P_8$. Since by assumption there is a nodal sextic through $P_1, \ldots, P_9$, we deduce that $P_9$ is a smooth point of the curve $C$ of degree $9$ and with triple points at $P_1, \ldots, P_8$ described in Proposition $1$. Analogously, consider the nodal sextics through $P_1, \ldots, P_7, \, P_9$. Then $P_8$ is a smooth point of the curve $D$ of degree $9$ and with triple points at $P_1, \ldots, P_7, \, P_9$.
The $12$ nodal points $Q_1, \ldots, Q_{12}$ of the $12$ singular curves of the pencil of nodal sextics through $P_1, \ldots, P_9$ (see Proposition 2) must belong to both $C$ and $D$. Moreover, since $$9\cdot 9 - 7 \cdot 3 \cdot 3 - 3 \cdot 1 - 1 \cdot 3 = 12,$$ the only intersection points of $C$ and $D$ are $P_1, \ldots, P_9$ and $Q_1, \ldots, Q_{12}$.
Finally, by assumption we also have pencils of nodal sextics through $P_1, \ldots, P_8, \, P_{10}$ and through $P_1, \ldots, P_7, \, P_9, \, P_{10}$. Then $P_{10}$ belongs to the intersection of $C$ and $D$, too. By the remark above, this means that $P_{10}=Q_i$ for some $i$.
Summing up, there is a curve in the pencil of nodal sextics through $P_1, \ldots, P_9$ which has a further node at $P_{10}$. This concludes the proof.
