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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.

1 Answers 1

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

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    I think you mean $\left|3y+x\right|=\left|3x-y\right|$, can you please clarify the procedure you took to find it – 2017-01-01
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    Write your equations as $4y=2x$ and $2y=-4x$. Rewrite as $3y\pm y=-x\pm3x$. Regroup as $3y+x=\pm(3x-y)$. Done. – 2017-01-01
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    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). – 2017-01-01
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    More generally, the two equations $y=ax+b$ and $y=cx+d$ is related to the single equation $2y=(a+c)x+b+d+|(a-c)x+b-d|$. – 2017-01-01
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    You may be right about the details. – 2017-01-01
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    Thanks for the explanation ,but just to make sure, a single modulus sign gives rise to a "v-shaped" graph while two signs give rise to a perpendicular graph that extends in the four quadrants, so the rule you have described above would always graph a "v-shaped" graph but never an intersection of two perpendicular lines (i.e. the "x-shaped" graph), please correct me if I am wrong – 2017-01-01
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    Sounds good, though there's no restriction to perpendicular lines. That's just the special case where $ac=-1$, but the formulas should work whether that relation holds or not. – 2017-01-01
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    thanks very much, I would also appreciate any source to better understand the formula and how it works – 2017-01-02
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    After experimenting with the formula I've found that it works in the form $|2y-(a+c)x-b-d|=|(a-c)x+b-d|$, I have no mathematical basis for this conclusion but it seems to always work, It seems very interesting and would be happy if you could explain it or help me learn more about the math behind it – 2017-01-02
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    All that's going on here is that $w+z+|w-z|$ is either $2w$ or $2z$, depending on whether $w-z$ is positive or negative. If you let $w$ be $ax+b$, and let $z$ be $cx+d$, you get to where I got to. – 2017-01-03