For a point `P` in the plane, let `d_1(P)a n dd_2(P)` be the distances of the point `P` from the lines `x-y=0a n dx+y=0` respectively. The area of the region `R` consisting of all points `P` lying in the first quadrant of the plane and satisfying `2lt=d_1(P)+d_2(P)lt=4,` is

Options

A. 4 sq.units
B. 6 units
C. 8 sq. units
D. 2 sq. units

Answer

B

Explanation

Let P `(alpha , beta)` be a point in the first quadrant . Then two cases arise .
CASE I When P lies in the region satisfying `y le x` :
In this case , `beta le alpha `
So , `d_(1) (P) + d_(2) (P) = (|alpha - beta|)/(sqrt2) + (|alpha + beta|)/(sqrt2)`
= `(alpha - beta)/(sqrt2) + (alpha + beta)/(sqrt2) = (2 alpha )/(sqrt2) = sqrt2 alpha `
`therefore 2 le d_(1) (P) + d_(2) (P) le 4 implies 2 le sqrt2 alpha le 4 implies sqrt2 le alpha le 2sqrt2 ... (i)`

CASE II When P lies in the region satisfying `y ge x` :
In this case , we have ` beta ge alpha `.
So , `d_(1) (P) + d_(2) (P) = (|alpha - beta|)/(sqrt2) + (|alpha + beta|)/(sqrt2) = (beta - alpha)/(sqrt2) + (alpha + beta)/(sqrt2) = sqrt2 beta`
`therefore 2 le d_(1) (P) + d_(2) (P) le 4 implies 2 le sqrt2 beta le 4 implies sqrt2 le beta le 2 sqrt2` ... (ii)
The shaded region shown in fig satisfies (i) and (ii) . So , region R is the shaded region in Fig.
Area of region R = Area of square OBDQ - Area of square OACP
`= (2sqrt2)^(2) - (sqrt2)^(2) = 6`sq. units
Hence , the locus of Q is `(x - 1)^(2) + (y-2)^(2) = (sqrt2)^(2)` which is a circle of radius `sqrt2` .