So I'm trying to prove that any five points, of which no 3 are colinear, there is a single conic that passes through al of them. I don't want to use projective geometry but rather, only analytic geometry.

Clearly I have to start out with the equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ and substitute, but I don't know what to do afterwards.

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    $\begingroup$ There are five equations and six unknowns, plus an overall scaling factor on the equation that eliminates one degree of freedom. Thus given that the system is not singular (which is the collinearity condition), there is a unique solution. $\endgroup$ – Mario Carneiro Feb 4 '15 at 2:28
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    $\begingroup$ @MarioCarneiro What you write only guarantees the kernel of the associated linear transformation has dimension at least 1, but it doesn't show it can't be bigger. In other words, this shows existence, not uniqueness. $\endgroup$ – hjhjhj57 Nov 25 '16 at 2:35

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