SOLUTION:

Let the coordinates of point P be (a, b)

Since, the sum of the distance from the axes to the point is always 1

\[\begin{array}{*{35}{l}}

\therefore ~\left| x \right|\text{ }+\text{ }\left| y \right|\text{ }=\text{ }1  \\

\Rightarrow ~\pm x\text{ }\pm \text{ }y\text{ }=\text{ }1  \\

\Rightarrow ~\text{ }-x\text{ }\text{ }-y\text{ }=\text{ }1,\text{ }x\text{ }+\text{ }y\text{ }=\text{ }1,\text{ }-x\text{ }+\text{ }y\text{ }=\text{ }1\text{ }and\text{ }x\text{ }-\text{ }y\text{ }=\text{ }1  \\

\end{array}\]

Therefore, these equations gives us the locus of the point P which is a square.