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

• A

  $\frac{25}{4}$
• B

  $25$
• Correct answer
  C

  $50$
• D

  $100$

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 $A B C D=(5 \sqrt{2})^2=50$

Hence, the correct answer is option: 'C'.