If the sum of distances from a point P to two | vedclass

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(C) Let the two mutually perpendicular lines be the coordinate axes, $x = 0$ and $y = 0$.
Let the point $P$ be $(x, y)$.
The distance of $P$ from th

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an ellipse

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(C) Let the two mutually perpendicular lines be the coordinate axes, $x = 0$ and $y = 0$.
Let the point $P$ be $(x, y)$.
The distance of $P$ from the line $x = 0$ is $|x|$ and the distance of $P$ from the line $y = 0$ is $|y|$.
According to the problem, the sum of these distances is $1$, so $|x| + |y| = 1$.
This equation represents a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$.
Since a square is a special type of rhombus and is not listed in the options, we must re-examine the problem statement. If the question implies the sum of the squares of the distances is constant, it would be a circle. However, for the sum of distances $|x| + |y| = 1$, the locus is a square. Given the standard nature of such problems, if this were a multiple-choice question where one must be selected, there is a discrepancy in the provided options as none represent a square.

If the sum of distances from a point $P$ to two mutually perpendicular straight lines is $1$ unit, then the locus of $P$ is

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