The locus of a point $P$ which moves such that the sum of its distances from two perpendicular lines is equal to $1$ is a

Point $P$ moves such that the sum of its distances from two perpendicular lines is equal to one.

Let the perpendicular lines are $x-\mathrm{axis}$ and $y-\mathrm{axis}$.

$L_1:x=0$,  $L_2:y=0$ and $P\left(x,y\right)$ then the sum of distances of point $P$ from $L_1 \& L_2$ $=\left|x\right|+\left|y\right|=1$

So, four equations of lines are possible which are

$\left\{\begin{array}{ll}x+y=1& x\ge 0,y\ge 0\\ -x+y=1& x\le 0,y\ge 0\\ -x-y=1& x\le 0,y\le 0\\ x-y=1& x\ge 0,y\le 0\end{array}\right.$

Hence, these lines will form a square.