Conic bundle through 4 give points I need to find the equation of a bundle of conics that passes through 
$A(0,0)$
$B(1,0)$
$C(0,1)$
$(2,1)$.
I know I need to assume that the bundle is generated by 3 possibles degerate conics which are lines that go through the points listed above:
$\overline{AB}$-$\overline{CD}$,
$\overline{CB}$-$\overline{AD}$,
$\overline{AC}$-$\overline{BD}$

The first degenerate conic I chose is the one that has line $\overline{CB}$ and $\overline{AD}$ and its equations is:
$(x+y-1)(2y-x)=0$
The second one I chose is the one that has line $\overline{AC}$ and $\overline{BD}$ and its equation is:
$(x)(x-y-1)=0$
I don't get what I need to do at this point, what should I do with those equations?  
 A: A conic in $\mathbb{R}^2$ is represented by an equation like
$$ ax^2+bxy+cy^2+dx+ey+f = 0 $$
where $[a,b,c,d,e,f]\in\mathbb{P}^5(\mathbb{R})$. By imposing that $A,B,C,D$ are points of such conic, we get:
$$ f=0,\quad a+d+f=0,\quad c+e+f=0, \quad 4a+2b+c+2d+e+f=0 $$
hence the conic has the following equation:
$$ ax^2-axy+cy^2-ax-cy = 0 $$
that can be written as
$$ a\cdot\underbrace{(x^2-xy-x)}_{AC,BD}+c\cdot\underbrace{(y^2-y)}_{AB,CD} = 0 $$
where the conics $y^2-y=0$ and $x^2-xy-x=0$ are double lines, namely the double line $\{y=0\}\cup\{y=1\}$ and the double line $\{x=0\}\cup \{x=y+1\}$. You may reach the same equation by considering first such double lines, then a linear combination of their equations.
A: It is sufficient to consider two pairs of lines. (And since you edited your question after I wrote this, I made my choice different from you.) For example, one pair would be $AB: y=0$ and $CD: y=1$. Now you reformulate them to have a zero on the right hand side, so the second becomes $y-1=0$. Then you multiply them: $y(y-1)=0$ or $y^2-y=0$. That's your degenerate conic consisting of these two lines. Any point on one or the other of the lines will cause one of the factors to be zero, so the set of points satisfying this equation is the union of both lines.
Then take another pair, e.g. $BC:x+y=1$ and $AD:x-2y=0$. Multiply and you get a second degenerate conic. If you want to, you can expand that, but for many applications you don't have to. Here is the expansion:
$$(x+y-1)(x-2y)=x^2+xy-x-2xy-2y^2+2y=x^2-xy-x-2y^2+2y=0$$
Now take arbitrary linear combinations of these:
$$\lambda y(y-1)+\mu(x+y-1)(x-2y)=0$$
for $\lambda,\mu\in\mathbb R$ and $(\lambda,\mu)\neq(0,0)$ so at least one of them has to be non-zero. Then those form the pencil of conics through these four points. Each of these four points makes both of the summands zero, so the sum will be zero for them as well.
The pair $(\lambda,\mu)$ is a homogeneous coordinate vector: scalar multiples of such a vector represent the same conic, since it's just a scalar multiple of the corresponding homogeneous equation.
You may choose one of them to be one, e.g. $\mu=1$. If you do, you still get the same set of conics, except for the single degenerate conic which corresponds to $\lambda=1,\mu=0$. So by fixing one of the variables to a non-zero constant value, you are taking the pencil of conics, which topologically is a projective line (i.e. a circle) and by removing one point turn it into an affine line (like the real numbers line).
