# Showing that pencil of conics through four points in general position in $\mathbb{P}^2$ forms a line when it's considered in $\mathbb{P}^5$

I know that there is a correspondence between points in $$\mathbb{P}^5$$ and conics in $$\mathbb{P}^2$$.

How do you show that family of conics through four points (pencil of conics) in general position forms a line when it's considered in $$\mathbb{P}^5$$?

If I consider a general conic given by $$ax^2+by^2+cxy+dx+ey+f=0$$ we can divide both sides by $$a$$ if $$a$$ is nonzero $$x^2+b'y^2+c'xy+d'x+e'y+f'=0,$$ and if I'm given the coordinates of four points, I can solve this for four variables, where the other one is a free variable. That's where the linearity comes from.

You don't have to worry so much about whether $$a$$ is zero or nonzero: you can just write down what the conditions "passing through $$p$$" mean. Explicitly, if $$p=[p_0,p_1,p_2]$$, then a conic $$ax^2+by^2+cxy+dxz+eyz+fz^2$$ passes through $$p$$ iff $$ap_0^2+bp_1^2+cp_0p_1+dp_0p_2+ep_1p_2+fp_2^2=0$$, which is a linear equation in $$a,b,c,d,e,f$$. Assuming the points are in general position gives you $$4$$ independent linear equations, so all you need to do is to understand that $$V(l_1,l_2,l_3,l_4)$$ is a $$\Bbb P^1\subset \Bbb P^5$$. (I should note that your equation of a conic is an affine equation and I'm upgrading to a projective equation - you should too!)