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

Explanation

For `P(x, y)`, we have
`" "2 le d_(1)(P)+ d_(2)(P) le 4`
`rArr" " 2le (|x-y|)/(sqrt2)+ (|x+y|)/(sqrt2) le 4` ltBrgt `rArr " "2sqrt2 le |x-y|+ |x+y|le 4sqrt2`
In the first quadrant if `x gt y`, we have
`" "2sqrt2 le x-y + x+y le 4sqrt2`
or `" "sqrt2 le x le 2sqrt2`
The region of points satisfying these inequalities is as follows.

In the first quadrant if `x lt y`, we have
`" "2sqrt2 le y -x + y le 4sqrt2`
or `" "sqrt2 le y le 2sqrt2`
The region of points satisfying these inequalities is as follows.

Combining the above two regions, we have the following.

Area of the shaded region `= ((2sqrt2)^(2)- (sqrt2)^(2))`
`" "= 8-2= 6` sq.units