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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.

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

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  • $\begingroup$ Another way, perhaps even more elegant. $\endgroup$ – Peter Sep 18 '16 at 20:04

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