Graph of $\{(x,y)\in \Bbb R: |x|+|y|=1\}$
If I use graph transformation to solve this que. Transformation
$|x|+|y|=1 \to |Y|=1-|X| \to Y = 1-|X|\to Y = 1-X$
Using this transformation I got a graph which is available in the attachment.
I wonder is there any other approachs to do the same.
Hint Substitution shows that the defining equation is unchanged under the transformations $X \mapsto -X$ and $Y \mapsto -Y$, so its graph is symmetric about the $y$- and $x$-axes respectively. Thus, to determine the graph it's enough to determine the part of it in the (closed) first quadrant and then reflect that part across the axes.
In the closed first quadrant we have $x, y \geq 0$ and so there $|x| = x$ and $|y| = y$. Thus, the part of the graph in the closed first quadrant coincides with the part of the graph of $x + y = 1$ in that quadrant, but this just defines a line segment.