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For a point P in the plane, let d1(P)a n...

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

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To solve the problem, we need to find the area of the region \( R \) in the first quadrant of the plane defined by the distances \( d_1(P) \) and \( d_2(P) \) from the lines \( x - y = 0 \) and \( x + y = 0 \) respectively, satisfying the condition \( 2 \leq d_1(P) + d_2(P) \leq 4 \). ### Step 1: Determine the distances \( d_1(P) \) and \( d_2(P) \) For a point \( P(\alpha, \beta) \): - The distance from the line \( x - y = 0 \) (which is \( y = x \)) is given by: \[ d_1(P) = \frac{|\alpha - \beta|}{\sqrt{2}} ...
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