Show that the following two lines intersect at a point with $z \neq 0$

Question

Consider the following two lines in complex projective space $\mathbf{P^2}$:

$a_{1}x + b_{1}y + c_{1}z = 0$ and $a_{2}x + b_{2}y + c_{2}z = 0$.

We suppose further that $a_{1}b_{2} - b_{1}a_{2} \neq 0$. We must show these two lines intersect at a point with $z \neq 0$.

I understand that if there is a point $(x:y:z)$ that lies on both lines, it must follow that $z \neq 0$, since if $z = 0$ we have "linearly independent" equations and so the only solution would be $(0:0:0)$ which is not a point in $\mathbf{P^2}$.

So to show that there is a solution with $z \neq 0$ i'm not sure what to do. I think by doing gaussian elimination on the two equations above, we can find conditions on $x$ and $y$ in terms of $z$ to get that a point of the form $(\alpha z, \beta z, z)$ is a solution to both equations and so the two lines intersect at $(\alpha z: \beta z: z) \in \mathbf{P^2}$ where $z \neq 0$.

Is this the correct way to approach this problem?

Answer 1

Hint: When working in the subspace $z\neq0$ of $\Bbb P^2$, you can assume $z=1$, and thus project that subspace isomorphically onto the $xy$-plane.

Answer 2

Hint:

put it into matrix terms $$\bf {Ax} = z\, \bf{c}$$

You are supposing that the determinant of $\bf A$ is not null, so it is invertible:
so for whichever $z$ and $\bf c$ you get a unique solution for $\bf x$.

And if $z \ne 0$ ...