A point moves in the $x y$-plane such that the sum of its distance from two mutually perpendicular lines is…
A point moves in the $x y$-plane such that the sum of its distance from two mutually perpendicular lines is always equal to 5 units. The area (in sq units) enclosed by the locus of the point, is
$\frac{25}{4}$
$25$
$50$
$100$
Solution
In a figure, $x x^{\prime}$ and $y y^{\prime}$ are two perpendicular axes.
$\therefore \quad|x|+|y|=5$
$\therefore$ Coordinates of $B$ and $C$ are $(5,0)$ and $(0,5)$.
$\therefore$ Length of $B C=\sqrt{5^2+5^2}=5 \sqrt{2}$
$\therefore$ Area of square $A B C D=(5 \sqrt{2})^2=50$
