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

Answer

6

Explanation

For P(x,y), we have
`2led_(1)(P)+d_(2)(P)le4`
`rArr" "2le(|x-y|)/(sqrt(2))+(|x+y|)/(sqrt(2))le4`
`rArr" "2sqrt(2)le |x-y|+|x+y|le4sqrt(2)`
IN first quadrant if `xgty,` we have
`2sqrt(2)lex-y+x+yle4sqrt(2)`
`"or "sqrt(2)lexle2sqrt(2)`
The region of points satisfying these inequalities is

In first quadrant if `xlty`, we have
`2sqrt(2)ley-x+x+yle4sqrt(2)`
`"or "sqrt(2)leyle2sqrt(2)`
The region of points satisfying these inequalities is

Combining above two regions we have

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