I am having trouble understanding a step in my Professor's Lecture notes
She shows that
Lemma 2.2.4 Let $P_1,\ldots,P_5$ be distinct points in $\mathbb{P}_k^2$. There exists a conic in $\mathbb{P}_k^2$ containing these
points. If no four of these points are on a line, the conic is unique.
What I am having trouble with: In the case where 3 of the points $P_1, P_2, P_3$ are on a line, she asserts that the line itself is in the conic.
Is this just a general fact?: Suppose you have a plane curve given by $\sum_{|\alpha|=n} a_\alpha x^{\alpha_1}y^{\alpha_2}z^{\alpha_3}$ with 3 collinear points say(without loss of generality) $P_1=[0:0:1],P_2=[x_0:y_0:z_0],P_3$ on it. The plane curve will be of the form $\sum_{|\alpha|=n, \alpha_3 <n} a_\alpha x^{\alpha_1}y^{\alpha_2}z^{\alpha_3}$ and the line going through these points is given by the morphism $[t:s] \to [tx_0,ty_0,s]$. The question of whether the $\overline{P_1P_2P_3}$ is automatically contained in the plane surface is posed as, "suppose there an $[s:t]\neq [1:z_0]$ or $[0:0:1]$ such that $\sum_{|\alpha|=n} t^{\alpha_1+\alpha_2}s^{\alpha_3} a_\alpha {x_0}^{\alpha_1}{y_0}^{\alpha_2}{z_0}^{\alpha_3}=0$. Then this is zero for all $[t:s]$.