Dimension of the linear system of conics passing through four points. Let $P_1$,$P_2$,$P_3$,$P_4$ $\in \mathbb{P}^2$. Let $V$ be the linear system of conics passing through these points. Then show that if  $P_1$,$P_2$,$P_3$,$P_4$ lie on a line, then $\dim(V)=2$ and if not, then $\dim(V)=1$.

[my attempt]
The definition of $V$ ,the linear system of conics passing through given points, is  $$V:= \{ [a:b:c:d:e:f] \in \mathbb{P}^5 \mid ax^2+by^2+cz^2+dxy+eyz+fxz=0 \text{  passing through  } P_1 , P_2 ,P_3, P_4\}$$
Now, let's denote $P_i=:[l_i:m_i:n_i]$ and consider the followings :
$$ \begin{pmatrix} 
l_1^2 & m_1^2 & n_1^2 & l_1m_1 & m_1n_1 & l_1n_1 \\
l_2^2 & m_2^2 & n_2^2 & l_2m_2 & m_2n_2 & l_2n_2 \\
l_3^2 & m_3^2 & n_3^2 & l_3m_3 & m_3n_3 & l_3n_3 \\
l_4^2 & m_4^2 & n_4^2 & l_4m_4 & m_4n_4 & l_4n_4 \\
\end{pmatrix}
\begin{pmatrix} 
a \\ b \\ c \\ d \\ e \\ f
\end{pmatrix}=
\begin{pmatrix} 
0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0
\end{pmatrix}$$
Suppose $P_1$,$P_2$,$P_3$,$P_4$ lie on a line (say, $\alpha x+ \beta y +\gamma z=0$).
Then $$ \begin{pmatrix} 
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2 \\
l_3 & m_3 & n_3 \\
l_4 & m_4 & n_4 \\
\end{pmatrix}
\begin{pmatrix} 
\alpha \\ \beta  \\\gamma 
\end{pmatrix}=
\begin{pmatrix} 
0 \\ 0 \\ 0 
\end{pmatrix}$$
Therefore the rank of this matrix is less than equal $2$. (This is beacause ($\alpha , \beta , \gamma $) is a non-trivial solution of this linear system) so $P_1$,$P_2$,$P_3$,$P_4$ are linearly dependent.
However, I stuck in this point. How to solve this?
Thank you.
 A: This is best understood without equations IMO. The first two points are completely free choices that cut us down to a $\mathbb P^3$ of conics (i.e. the only way $P_2$ can fail to be general is if it is equal to $P_1$). If we add a third point lying on the line between the first two (call this Step 3a), then the line must be part of the conic (otherwise Bezout is violated); in this case, the remaining two points determine a second line, so we get a (generally) unique reducible conic.
If, instead, we take $P_3$ to not lie on the line $L := \overline{P_1 P_2}$ (call this Step 3b), there is no issue with Bezout so we still have a $\mathbb P^2$ of conics whose general member is irreducible. Continuing with one more general point, we get a $\mathbb P^1$ of conics whose general member is irreducible.
On the other hand, if we continue from Step 3a and impose a fourth point on the line, then we don't "lose" any conics this time; since all conics from Step 3a contain the line $L$, they all pass through $P_4$, so we still have a $\mathbb P^2$ of conics by choosing any line to be the second component.
Finally (on the third hand, I guess?), if we continue from Step 3a but let $P_4$ be a general point, then we are cut down to a $\mathbb P^1$ of reducible conics: choose a line passing through $P_4$ (there is a $\mathbb P^1$ of these) and take its union with $L$.
