A point moves in the x-y plane such that the sum of its distances from two mutually perpendicular line is always equal to 21/2 units. The area enclosed by the locus of the point is?
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Dear Student,
Please find below the solution to the asked query:

Let mutually perpendicular line be X and Y axis and point be x,yHenceDistance of point from X-axis=yDistance of point from Y-axis=xAccording to question:x+y=2Nowx=±x and y=±yHence four lines will be:x+y=2x-y=2-x+y=2-x-y=2

As you can see enclosed area ABCD is a square.Let side of square be a unitsDiagonal=a2=AC=AO+OC=2+2=22a2=22a=2 unitsHence enclosed area=a2=22=4 unit 2 Answer
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I got the answer...no need to solve it. :)
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