Graph of $|x| + |y| = 1$

Question

Can anybody explain as how to plot a graph of $|x| + |y| = 1$. Here $|x|$ and $|y|$ are absolute values.

Answer 1

HINT:

As $|y|\ge0,|x|=1-|y|\le1\implies-1\le x\le1$

Now use $$|x|=\begin{cases} x &\mbox{if } x\ge 0 \\-x & \mbox{if } x<0\end{cases}$$

Answer 2

Two solutions here (at least).

1. Notice that $|x|+|y|=1$ is the boundary of the ball of radius $1$ for the 1-norm and knowing what it does look like.
2. Or just split the cases for every quadrant ${\lbrace x \geq 0,y \geq 0 \rbrace},{\lbrace x \geq 0,y \leq0 \rbrace},\dots$

For example in ${\lbrace x \geq 0,y \geq 0 \rbrace}$, your equation is $x+y=1$.

Answer 3

Following the method given by lab bhattacharjee, this is the graph you get :