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I am currently studying graphs of modular equations and was wondering if there exists a modular equation for any 2 perpendicular lines and if this is the case what method would you use to find this modulus equation if you are given the equation of both perpendicular lines.

For example:

the modulus equation $|y|=|x-1|$ could be graphed as the intersection of the two lines $y=x-1$ ad $y=-x+1$.

now if we work backward, would that always work?

For example:

For the two perpendicular lines $y=\frac{x}{2}$ and $y=-2x$ does there exist a modulus equation and how would you go on to find what it is.

Also feel free to edit or add tags.

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$\left|3y+x\right|=\left|3x-y\right|$ gives your equations.

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    $\begingroup$ I think you mean $\left|3y+x\right|=\left|3x-y\right|$, can you please clarify the procedure you took to find it $\endgroup$ – Basem Fouda Jan 1 '17 at 16:09
  • $\begingroup$ Write your equations as $4y=2x$ and $2y=-4x$. Rewrite as $3y\pm y=-x\pm3x$. Regroup as $3y+x=\pm(3x-y)$. Done. $\endgroup$ – Gerry Myerson Jan 1 '17 at 16:13
  • $\begingroup$ but actually $3y+x=\pm(3x-y)$ does not give the required graph, instead $\left|3y+x\right|=\left|3x-y\right| $seems to be te right answer. (I have checked using a graphing tool). $\endgroup$ – Basem Fouda Jan 1 '17 at 16:22
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    $\begingroup$ More generally, the two equations $y=ax+b$ and $y=cx+d$ is related to the single equation $2y=(a+c)x+b+d+|(a-c)x+b-d|$. $\endgroup$ – Gerry Myerson Jan 1 '17 at 16:23
  • $\begingroup$ You may be right about the details. $\endgroup$ – Gerry Myerson Jan 1 '17 at 16:24

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