If some four of given five distinct points in projective plane are collinear , then there are more than one conic passing through the five points

Question

Let $k$ be an algebraically closed field, $p_1,...,p_5$ be distinct points in $P^2(k)$ such that there is a unique conic (i.e. a homogeneous equation of degree 2 in $k[X,Y,Z]$) passing through $p_1,...,p_5$ ; then how to show that no four of the five points $p_1,...,p_5$ are collinear ?

I can show the converse i.e. if no four of the five points $p_1,...,p_5$ are collinear then there is a unique conic passing through the five points. But I don't know how to prove the direction I mentioned.

Please help. Thanks in advance

Answer 1

I wish I answer to the right question.

Suppose you have 4 aligned points, $A_k (k=1 \cdots 4)$, and a fifth one $A_5$ which is not aligned with them.

Let $ux+vy+wt=0$ be the projective equation of the line $(L_0)$ through them and $(L_k)$ with equation $u_kx+v_ky+w_kt=0$ be the line passing through $A_k$ and $A_5$.

Then, at least four (degenerate) conic curves defined by product equations :

$$(u_kx+v_ky+w_kt)(ux+vy+wt)=0$$

are passing through the five points.