A point moves in the x-y plane such that the sum of its distances from two mutually perpendicular line is always equal to 2^1/2 units. The area enclosed by the locus of the point is?
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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Let} \mathrm{mutually} \mathrm{perpendicular} \mathrm{line} \mathrm{be} \mathrm{X} \mathrm{and} \mathrm{Y} \mathrm{axis} \mathrm{and} \mathrm{point} \mathrm{be} \left(\mathrm{x},\mathrm{y}\right) \\ \mathrm{Hence} \\ \mathrm{Distance} \mathrm{of} \mathrm{point} \mathrm{from} \mathrm{X}-\mathrm{axis}=\left|\mathrm{y}\right| \\ \mathrm{Distance} \mathrm{of} \mathrm{point} \mathrm{from} \mathrm{Y}-\mathrm{axis}=\left|\mathrm{x}\right| \\ \mathrm{According} \mathrm{to} \mathrm{question}: \\ \left|\mathrm{x}\right|+\left|\mathrm{y}\right|=\sqrt{2} \\ \mathrm{Now} \\ \left|\mathrm{x}\right|=\pm \mathrm{x} \mathrm{and} \mathrm{y}=\pm \mathrm{y} \\ \mathrm{Hence} \mathrm{four} \mathrm{lines} \mathrm{will} \mathrm{be}: \\ \mathrm{x}+\mathrm{y}=\sqrt{2} \\ \mathrm{x}-\mathrm{y}=\sqrt{2} \\ -\mathrm{x}+\mathrm{y}=\sqrt{2} \\ -\mathrm{x}-\mathrm{y}=\sqrt{2} \end{gathered}$$


$$\begin{gathered} \mathrm{As} \mathrm{you} \mathrm{can} \mathrm{see} \mathrm{enclosed} \mathrm{area} \mathrm{ABCD} \mathrm{is} \mathrm{a} \mathrm{square}. \\ \mathrm{Let} \mathrm{side} \mathrm{of} \mathrm{square} \mathrm{be} \mathrm{a} \mathrm{units} \\ \mathrm{Diagonal}=\mathrm{a}\sqrt{2}=\mathrm{AC}=\mathrm{AO}+\mathrm{OC}=\sqrt{2}+\sqrt{2}=2\sqrt{2} \\ \Rightarrow \mathrm{a}\sqrt{2}=2\sqrt{2} \\ \Rightarrow \mathrm{a}=2 \mathrm{units} \\ \mathrm{Hence} \mathrm{enclosed} \mathrm{area}={\mathrm{a}}^2=2^2=\mathbf{4}\mathbf{ }\mathbf{unit}{\mathbf{ }}^{\mathbf{2}}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$
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