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Given four points

$(x_1, y_1) (x_2, y_2) (x_3, y_3) (x_4, y_4)$

How does one construct a system of two equations:

$a_1x + a_2x^2 + a_3y + a_4y^2 + a_5xy = c_1$

$b_1x + b_2x^2 + b_3y + b_4y^2 + b_5xy = c_2$

such that the set of solutions of this system is the four original points?

Solving the general systems is growing massively complicated.

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1 Answer 1

up vote 1 down vote accepted

Here's one way to do it, at least when the points are in a general position.

Call the four given points $P_1, P_2, P_3, P_4$. Let $\ell_{ij}$ denote the line connecting $P_i$ and $P_j$, and $L_{ij}$ a linear equation in $x$ and $y$ satisfied by the points on $\ell_{ij}$. Then you can use the system of equations: \begin{equation} L_{12} L_{34} = 0, \\ L_{13} L_{24} = 0. \end{equation}

Indeed, it is easy to see that each of the four points $P_i$ satisfy both equations, so they lie on the intersection. Provided none of the lines are identical (the condition that the points are in general position), then there are exactly four solutions to this system of equations, which you've already found.

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This is very smart –  frogeyedpeas May 15 '13 at 2:45
Can this be generalized to cubis with 9 points etc??? –  frogeyedpeas May 15 '13 at 2:47
In general, you cannot find three lines that pass through 9 points. Also, once two cubics pass through 8 common points, there is a ninth point that they automatically share as well. See –  Michael Joyce May 15 '13 at 2:50
Does that mean there exist certain combinations of 9 points for which no cubic exists? Or we just don't have the mathematical machinery to develop a cubic system for any group of 9 points –  frogeyedpeas May 15 '13 at 2:57
@frogeyedpeas Not necessary. The conclusion is that there are sets of 9 points for which there does not exist 2 cubics that pass through those 9 points. –  Calvin Lin May 15 '13 at 3:03

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