# Why 5 point determine a conic?

How to prove that any five points, of which no 3 are colinear, there is a single conic that passes through all of them ?

(I have to start out with the equation $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ but i don't know what to do afterward)

Help me pls.

• without loss of generality you can set $A=1$ - then using your 5 points you can generate 5 equations for the 5 unknowns B,C,D,E,F – WW1 Oct 29 '15 at 2:43
• @WW1 Yes, but the question is why those 5 linear equations in 5 unknowns have exactly one solution. – jathd Oct 29 '15 at 2:46
• Maybe cramer's rule can help? – najayaz Oct 30 '15 at 4:45

Here's the outline of one method that uses some linear algebra. $\def\Ker{\mathop{\mathrm{Ker}}}$

1. Consider the set of polynomials \begin{align} V = \{ Ax^2+Bxy+Cy^2+Dx+Ey+F \mid A,\ldots,F\in\mathbf R \} \end{align} and show that this is a $6$-dimensional vector space.

2. Show that for two vectors $f(x,y)$ and $g(x,y)$ in $V\setminus\{0\}$, the conics defined by $f(x,y)=0$ and $g(x,y)=0$ are the same if, and only if, $f$ and $g$ are colinear.

3. For any point $P=(a,b)$ in the plane, consider the map $\varphi_P:V\to\mathbf R$ defined by $\varphi_P(f) = f(a,b)$ for any $f(x,y)\in V$. Show that it is a surjective linear map. From the surjectivity, deduce that \begin{align} \dim(\Ker\varphi_P) = 5. \end{align}

4. Show that a point $P$ is on the conic $f(x,y)=0$ if, and only if, $\varphi_P(f) = 0$.

5. Consider five points $P_1,\ldots,P_5$. Show that \begin{align} \dim\left(\bigcap_{i=1}^5 \Ker\varphi_{P_i}\right) = 1\end{align} if, and only if, no three of the points are colinear.

6. Conclude.