# Lines passing through the same point

What is the easiest proof to prove that all lines of the form $(a+2)x - (a+1)y - 2a - 3 = 0$ pass through some common point, where $a$ is a real number, and how to find that point. I tried taking $a_1$ and $a_2$ and somehow to prove it with determinants, but I got stuck.

• It is quite trivial that if we take $x=1$ and $y=-1$... – Jack D'Aurizio Sep 18 '16 at 19:56
• Solution is obvious, yes, but how can I get to that solution? – Darko Dekan Sep 18 '16 at 19:58
• For which values of $x,y$ the LHS does not depend on $a$? – Jack D'Aurizio Sep 18 '16 at 19:58
• the part containing $a$ is : $a(x-y-2)$. So, $x-y-2=0$ must hold. Furthermore, $2x-y-3=0$ must hold. This gives the unique solution. – Peter Sep 18 '16 at 19:59
• Among such values, which $(x,y)$ couples lead to an LHS equal to zero? – Jack D'Aurizio Sep 18 '16 at 19:59

I would choose values for $a$ that eliminate one variable. If $a=-2$ you get an equation in $y$ only. If $a=-1$, you get an equation in $x$ only. The solutions to these two equations are the coordinates you want.