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Prove that all lines $ax+y=b$ such that coefficients $a,$$1,$$b$ constitute arithmetic sequence have one common point.

We know that $1-a=b-1$ and solving for b we get $b=$$2-a$ replacing b in the equation $y = 2-a-ax$ but I am not sure where to go from here that will help me prove my claim.

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Hint: If the claim is true, then you can find the point of intersection by choosing any two values of $a$ and seeing where the two lines meet.

Then, once you have the point, show that it lies on the line for any given $a$.

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  • $\begingroup$ Your hint was very helpful, do you have any suggestions to find if there exists a point on the coordinate plane such that no single line (ax+y=b with coefficients a, 1, b constituting an arithmetic sequence) passes through it. I was trying to prove this by contradiction but I am not sure if it is necessarily the most efficient way. $\endgroup$ – Michelle Stasso Dec 7 '16 at 17:11
  • $\begingroup$ @MichelleStasso Glad to help! For your second question, first consider the family of lines $y=mx$. These are all lines passing through the origin, except for the vertical line. Notice, for example, that there is no $m$ for which the point $(0,1)$ will lie on a line in this family. Now find a similar point for your own question, and you will get your contradiction. $\endgroup$ – Théophile Dec 7 '16 at 17:55
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We have the equation,

$ax+y=b$ $\tag{1}$

also we have

$a,1,b$ in A.P.

therefore $1-a=b-1$

$-a+2=b $ $\tag{2}$

On comparing $(1)$ and $(2)$ we have

The common point $=(-1,2)$

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  • $\begingroup$ @TheLoneWolf The formatting is still crazy here; the labels look like part of the equation, especially with that minus sign. Note that \tag{label me!} is a much better typographical solution (it does need to be displayed math, however, although not necessarily in an align environment). I'm not sure italicized Roman numerals can be used, but... $\endgroup$ – pjs36 Dec 18 '16 at 5:40
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    $\begingroup$ Thanks @pjs36, I didn't knew that. Thank you so much. $\endgroup$ – I am Back Dec 18 '16 at 5:44
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If $y=-ax+2-a=-a(x+1)+2$.

If we plug $x=-1$ the value of $a$ doesn't matter and then all lines will go through the point $(-1,2)$.

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Let $r$ be the reason of the arithmetic sequence. then

$a=1-r$ and $\;b=1+r$.

the equation of the line becomes

$$y=(r-1)x+(1+r)$$ $$=r(x+1)-x+1.$$

the common point is $(-1,2)$.

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