Question

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

• A. $\frac{25}{4}$
• B. $25$
• C. $50$
• D. $100$

Answer 1

The correct option is C $50$

Assume for ease that the $2$ mutually perpendicular lines are the $X$ and $Y$ axes.

Since the sum of distances from these lines is always $5$ units, the locus of point $P(x,y)$ has the equation $|x|+|y|=5$.

Also, $|x|=\pm x,|y|=\pm y$ $\Rightarrow$ 4 possible line equations are

$x+y=5$

$x-y=5$

$-x+y=5$

$-x-y=5$

These $4$ lines intersect to form a square with vertices $(5,0),(-5,0),(0,5)$ and $(0,-5)$ which is the feasible locus.

Side length of the square is $\sqrt{(5-0{)}^2+(0-5{)}^2}=5\sqrt{2}$ and hence the area is $(5\sqrt{2}{)}^2=50$.