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
![\[\begin{array}{*{35}{l}} \therefore ~\left| x \right|\text{ }+\text{ }\left| y \right|\text{ }=\text{ }1 \\ \Rightarrow ~\pm x\text{ }\pm \text{ }y\text{ }=\text{ }1 \\ \Rightarrow ~\text{ }-x\text{ }\text{ }-y\text{ }=\text{ }1,\text{ }x\text{ }+\text{ }y\text{ }=\text{ }1,\text{ }-x\text{ }+\text{ }y\text{ }=\text{ }1\text{ }and\text{ }x\text{ }-\text{ }y\text{ }=\text{ }1 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9731c545c965539d9d0bc4706898b89b_l3.png "Rendered by QuickLaTeX.com")
Therefore, these equations gives us the locus of the point P which is a square.
If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.
If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.