If the sum of the distances of a point P(x, y | vedclass

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(A) Let the two perpendicular lines be the coordinate axes $x = 0$ and $y = 0$. The distance of point $P(x, y)$ from the line $x = 0$ is $|x|$ and fro

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rhombus

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(A) Let the two perpendicular lines be the coordinate axes $x = 0$ and $y = 0$. The distance of point $P(x, y)$ from the line $x = 0$ is $|x|$ and from the line $y = 0$ is $|y|$. According to the problem,$|x| + |y| = 1$. This equation represents four line segments in the four quadrants: $1$) $x + y = 1$ for $x > 0, y > 0$ $2$) $-x + y = 1$ for $x  0$ $3$) $-x - y = 1$ for $x $4$) $x - y = 1$ for $x > 0, y These four segments form a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$. $A$ square is a special type of rhombus. Thus,the locus of $P$ is a rhombus.

If the sum of the distances of a point $P(x, y)$ from two perpendicular lines in a plane is $1$,then the locus of $P$ is a

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