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 square units) enclosed by the locus of the point is

Let the two mutually perpendicular lines be coordinate axis

Let (x,y) be the locus point

$\mathrm{Perpendicular} \mathrm{distance} \mathrm{from} (\mathrm{x},\mathrm{y}) \mathrm{to} \mathrm{x} \mathrm{axis} =\left|\mathrm{y}\right|$

$\mathrm{Perpendicular} \mathrm{distance} \mathrm{from} (\mathrm{x},\mathrm{y}) \mathrm{to} y \mathrm{axis} =\left|x\right|$

sum of its distances  from two  mutually perpendicular lines is always equal to 5 units

$\left|x\right|+\left|y\right|=5$

Area of the square formed by the given lines=$\frac{2c^2}{\left|ab\right|}=2{\left(5\right)}^2=50$