How do I plot the graph of $|x|+|y|=1?$

Question

When I plot the graph of $|x|+|y|=1$ in Desmos Graphing Calculator it comes like this -

How do I graph this plot? Can someone guide me?

Break the equation down into 4 cases, one for each quadrant. — Ragib Zaman, Apr 19 '22 at 14:42
You can also write it as $|y|=1-|x|$. For $y>0$ (above $x$ axis) it is $y=1-|x|$ and for $y<0$ it is $y=|x|-1$. — Etemon, Apr 19 '22 at 20:19

score 1 · Accepted Answer · answered Apr 19 '22 at 14:48

1

Since $|x|+|y|=1$, you have $|x|\leq 1$. Now separate in 4 cases :

$x,y\geq 0$ : then $|x|+|y|=1$ can be written as $$y=1-x$$ for $0\leq x\leq 1$.
$x\geq 0,y\leq 0$ : then $|x|+|y|=1$ can be written as $$y=x-1$$ for $0\leq x\leq 1$.
$x\leq 0,y\geq 0$ : then $|x|+|y|=1$ can be written as $$y=1+x$$ for $-1\leq x\leq 0$.
$x,y\leq 0$ : then $|x|+|y|=1$ can be written as $$y=-1-x$$ for $-1\leq x\leq 0$.

answered Apr 19 '22 at 14:48

SacAndSac

1,677

3

Or just do the case $x,y\geq0$ and then note the symmetry of the equation. Basically, $(x,y)$ is a point of this graph if and only if $(-x,y)$ is, if and only if $(x,-y)$ is, if and only if $(-x,-y)$ is. – Thomas Andrews Apr 19 '22 at 15:05

score 1 · Answer 2 · answered Apr 19 '22 at 15:01

Break the equation $|x|+|y|=1$ as follows. Case 1: When $x\geq 0$, $y\geq 0$, then $|x|=x$ and $|y|=y$. Above equation turns into $x+y=1$ to give line segment AB.(By intercept form) Case 2: When $x\leq 0$, $y\geq 0$, then $|x|=-x$ and $|y|=y$. Above equation turns into $-x+y=1$ to give line segment AD. Case 3: When $x\leq 0$, $y\leq 0$, then $|x|=-x$ and $|y|=-y$. Above equation turns into $-x-y=1$ to give line segment DC. Case 4: When $x\geq 0$, $y\leq 0$, then $|x|=x$ and $|y|=-y$. Above equation turns into $x-y=1$ to give line segment CB. See the figure below

How do I plot the graph of $|x|+|y|=1?$

2 Answers2