A point moves in the xy-plane such that the s | vedclass

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(B) Let the two mutually perpendicular lines be the coordinate axes $x=0$ and $y=0$.Let the point be $(x, y)$. The distances from these lines are $|x|

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$25$

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(B) Let the two mutually perpendicular lines be the coordinate axes $x=0$ and $y=0$.Let the point be $(x, y)$. The distances from these lines are $|x|$ and $|y|$.The given condition is $|x| + |y| = 5$.This equation represents a square with vertices at $(5, 0), (0, 5), (-5, 0),$ and $(0, -5)$.The length of the diagonal of this square is the distance between $(5, 0)$ and $(0, 5)$,which is $\sqrt{(5-0)^2 + (0-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$.The area of a square with diagonal $d$ is given by $\frac{1}{2} d^2$.Area $= \frac{1}{2} 	imes (5\sqrt{2})^2 = \frac{1}{2} 	imes 50 = 25$ sq units.

$A$ point moves in the $xy$-plane such that the sum of its distances from two mutually perpendicular lines is always equal to $5$ units. The area (in sq units) enclosed by the locus of the point is

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