Eight-point cubic completion for two triples of lines ($H=\lambda G+\mu F$)

Let two triples of projective lines meet in nine distinct points. Any plane cubic containing eight contains all nine. To prove this, write the triples as $F=L_1{L}_2{L}_3$ and $G=M_1{M}_2{M}_3$, with the missing point on $L_3$. The candidate cubic $H$ agrees with a scalar multiple of $G$ on $L_1$, so $H-\lambda G=L_{1}Q$. Its three known zeros on $L_2$ force $Q=L_{2}L$. Two remaining known zeros on $L_3$ force the linear factor $L$ to be a multiple of $L_3$. Hence $H=\lambda G+\mu F$ and vanishes at the missing point.

A two-cell-thick strip of triangles with degree-six vertices in a dual arrangement of a planar point set produces three indexed primal point families with collinearities $a_i,b_j,c_k$ whenever $i+j+k=0$. A plane cubic can be fitted to nine initial points because its homogeneous coefficient space has dimension ten. Overlapping nine-point configurations then force each next point onto the same cubic by eight-point cubic completion for two triples of lines. Safe neighbourhoods guarantee the distinctness needed in these local completion steps. A possible seed is $a_{-1},a_0,a_1,a_2,b_{-3},b_{-2},b_{-1},c_1,c_2$. Adjacent completion blocks force $c_3,b_{-4},c_4,a_{-2}$, followed by the alternating continuation along the two long transverse families.