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I know, that for any four points, $(x_0,y_0,z_{00})$, $(x_1,y_0,z_{10})$, $(x_0,y_1,z_{01})$ and $(x_1,y_1,z_{11})$, if $x_0\ne x_1$ and $y_0\ne y_1$, there is a unique surface of the form $z(x,y)=a xy+b x+c y+d$ passing through these points.

Can someone please guide me to a resource that discusses these type of surfaces at an introductory level? I am trying to guide high school students to do some experiment in this area without me telling them what to do. An elementary introduction that starts with the equation of a plane through three points and moves towards how to modify a plane equation if a fourth point is added would be ideal. I tried to search on the internet, but without much success.

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Existence and uniqueness go with this linear problem you don't seem to be mentioning: $$ \left( \begin{array}{rrrr} x_0 y_0 & x_0 & y_0 & 1 \\ x_0 y_1 & x_0 & y_1 & 1 \\ x_1 y_0 & x_1 & y_0 & 1 \\ x_1 y_1 & x_1 & y_1 & 1 \end{array} \right) \left( \begin{array}{r} a \\ b \\ c \\ d \end{array} \right) = \left( \begin{array}{r} z_{00} \\ z_{01} \\ z_{10} \\ z_{11} \end{array} \right) $$

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It's called Bilinear interpolation. The wiki page has a good introduction, then top google results on that name should give you enough to choose from.

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  • $\begingroup$ Yes, I know how bilinear interpolation works. My problem is that I would like to guide a student to find this out herself without me telling her the form of the function she should look for. She has four points on the plane with values attached to these, and she needs a surface matching these values. She knows how to find the equation of the plane matching three of these values, the question is what to do next. All the resources I found are of the form: "look for this and this type of surface, and you will find it", but does not guide the reader how to come up with the form of the surface. $\endgroup$ – Ferenc Beleznay Dec 4 '16 at 6:32
  • $\begingroup$ It's a 2d equivalent of a 1d linear interpolation, a tensor product linear interpolation: $s(x,y)=(ax+b)(cy+d)$. The 4 unknown coefficients are just enough to be determined by 4 points (uniquely if the points are unisolvent). $\endgroup$ – rych Dec 4 '16 at 8:50

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