That is not a real answer, as I do not understand your question still, but a small illustration could be useful.
Take the two equations $x+y=0 \ \Leftrightarrow \ a_1=1, b_1=1, c_1=0$ and $x+2y+3=0 \ \Leftrightarrow \ a_2=1, b_2=2, c_2=3$. Or, we can rewrite them as functions $y=-x$ and $y=-\frac{1}{2}x - \frac{3}{2}$.
They intersect at the point $(3,-3)$. Your theorem says, we can find all the lines going through that point using the formula $(x+y) + \lambda (x+2y+3)=0$.
So, let's try this out.
$ x+y + \lambda x + 2\lambda y + 3\lambda=0 \ \Leftrightarrow \ y=\frac{1+\lambda}{-1-2\lambda} x + \frac{3\lambda}{-1-2\lambda}$, if we rewrite this as a function.
Put $x=3$, then $y(3)=\frac{3+3\lambda+3\lambda}{-1-2\lambda}=\frac{3(1+2\lambda)}{-1(1+2\lambda)}=-3$. That means we have found a family of lines, that all go through the point $(3,-3)$, as we see that $y(3)$ is independent from $\lambda$.
Writing this idea in common (not for $x+y$ and $x+2y+3$ in particular) will bring you a proof. You can find the intersection point $\left((\frac{-c_1-c_2}{a_1-a_2}),(\frac{-a_1c_2-c_1a_2}{b_1a_1-b_1a_2}), \ a_1-a_2\ne0, \ b_1a_1-b_1a_2 \ne 0\right)$ and put in the equation $y=\frac{-a_1-\lambda a_2}{b_1+\lambda b_2} \cdot x + \frac{-c_1-\lambda c_2}{b_1 + \lambda b_2}$, which you get, if you transform your given equation from the theorem.