The locus of a point P which moves such that | vedclass

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(A) Let the two perpendicular lines be the $x$-axis and $y$-axis. Let the point be $P(x, y)$. The distance of $P$ from the $x$-axis is $|y|$ and from

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Square

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(A) Let the two perpendicular lines be the $x$-axis and $y$-axis. Let the point be $P(x, y)$. The distance of $P$ from the $x$-axis is $|y|$ and from the $y$-axis is $|x|$.According to the problem,the sum of these distances is $1$,so $|x| + |y| = 1$.This equation represents four line segments in the four quadrants:$1$. In the first quadrant $(x \geq 0, y \geq 0)$,$x + y = 1$.$2$. In the second quadrant $(x \leq 0, y \geq 0)$,$-x + y = 1$.$3$. In the third quadrant $(x \leq 0, y \leq 0)$,$-x - y = 1$.$4$. In the fourth quadrant $(x \geq 0, y \leq 0)$,$x - y = 1$.These four lines form a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$.

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

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