Is it possible to write an absolute value equation that produces a graph of a straight line?

Question

Title question was asked in a secondary school maths class, and I don't know how to go about answering it. Goal is to write some equation using the absolute value function that produces a straight line that could also be graphed using a linear equation.

Answer 1

Speaking roughly, the absolute value function of a line spits out the input if the input is above zero and spits out the negative of the input if the input is below zero.

For an absolute value equation of a linear function to resemble a straight line itself, the slope of the line has to stay the same for all segments. Since the absolute value makes the slope negative, the only slope that retains its 'straightness' is one where the slope is zero.

$y = mx + b$ when $(mx + b) \geq 0$.

$y = -mx - b$ when $(mx + b) < 0$.

As seen, $m = -m$ only when $m = 0$.

A general equation would be $y = b$ for any constant $b$. $b$ can be anything given the nature of the absolute function, but if you want an equation that gives exactly itself after an absolute value function applied is then $y = b$ for any $b \geq 0$.

Edit: After revising further, The only equations that would work would be $y = |b|$ for any constant $b$.

Answer 2

It looks impossible to succeed under the form of a cartesian equation that would be of the form:

$$y=a_0x+b_0+|a_1x-b_1|+|a_2x-b_2|+...$$

But it is possible to obtain any straight line under the form of an implicit equation containing (mainly) absolute values.

We give below an example which is the red line with implicit equation:

$$|10y-5x|=|x-7y|-3(4x-3y)\tag{1}$$

$$\textit{GeoGebra plot of lines defined by (1) (red line) and (2) (green line).}$$

This line passes through the origin ; from it, one can deduce any other straight line with the same slope ; for example, an up-shift by $a=2$ is obtained by replacing in each of its occurences, $y$ by $y-a$ in equation (1):

$$|10(y-2)-5x|=|x-7(y-2)|-3(4x-3(y-2))\tag{2}$$

which is the implicit equation of the green straight line.

Remark: the vertical blue line with implicit equation $|y-2x|=|-y+x|-3x$ has been instrumental in the obtention of red curve with equation (1) : indeed red line is obtained by "coordinates' rotation" from vertical blue line.

Didactic comment: (answer to the valuable comment by Zoey Pope) : implicit equations like (1) are difficult to understand for ordinary secondary school students. But it can be a profitable exercise for students having a good level : they can be guided into the analysis of regions splitted the different cases. For example : if $(10y-5x)>0$ and $(x-7y)>0$, (defining a plane sector S), eq. (1) becomes $10y-5x=x-7y-3(4x-3)$ giving a straight line that may or not intersect sector S. Once this putative intersection is drawn, treat in the same way the 3 other cases ($(10y-5x)>0$ and $(x-7y)<0$, etc.).

Besides, plotting the locus using Geogebra or Desmos can trigger the interest of curious students.