Why 5 points determine a conic? How to prove that any five points, of which no 3 are colinear, that 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.)
 A: Here's the outline of one method that uses some linear algebra.
$\def\Ker{\mathop{\mathrm{Ker}}}$


*

*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.

*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.

*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}

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

*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.

*Conclude.
