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(A) Let the two perpendicular lines be the coordinate axes,$X$-axis and $Y$-axis. Let the point be $P(x, y)$.The distance of point $P$ from the $X$-ax

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Square

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(A) Let the two perpendicular lines be the coordinate axes,$X$-axis and $Y$-axis. Let the point be $P(x, y)$.The distance of point $P$ 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 $|x| + |y| = 1$.If the point lies in the first quadrant,$x > 0$ and $y > 0$,so $x + y = 1$.If the point lies in the second quadrant,$x  0$,so $-x + y = 1$.If the point lies in the third quadrant,$x If the point lies in the fourth quadrant,$x > 0$ and $y These four equations represent the sides of a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$.Thus,the locus is a square.

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

If the sum of the distances of a point from two perpendicular lines in a plane is $1$,then its locus is

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