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A point moves in the $X Y$-plane such that the sum of its distances from two mutually perpendicular lines is always equal to 3 . The area enclosed by the locus of that point is (in sq. units)
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The correct answer is:
18
In figure, $X X^{\prime}$ and $Y Y^{\prime}$ are two perpendicular axes.
$$
\therefore|X|+|Y|=3
$$
So, $A B C D$ is a square coordinate of $B$ and $C$ are $(3,0)$ and $(0,3)$
$\therefore$ Length of $B C=\sqrt{(3-0)^2+(0-3)^2}=\sqrt{9+9}=\sqrt{18}$
Area of square $A B C D=(\text { side })^2=(\sqrt{18})^2=18$ unit $^2$
$$
\therefore|X|+|Y|=3
$$
So, $A B C D$ is a square coordinate of $B$ and $C$ are $(3,0)$ and $(0,3)$
$\therefore$ Length of $B C=\sqrt{(3-0)^2+(0-3)^2}=\sqrt{9+9}=\sqrt{18}$
Area of square $A B C D=(\text { side })^2=(\sqrt{18})^2=18$ unit $^2$
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