Question

For a point P in the plane, let d₁(P) and d₂(P) be the distances of the point P from the lines x - y = 0 and x + y = 0 respectively. The area of the region R consisting of all points P lying in the first quadrant of the plane and satisfying 2 ≤ d₁(P) + d₂(P) ≤ 4, is _____.

Answer 1

Therefore, according to the question (Fig.)

Since x, y ≥ 0 in the first quadrant.

When x > y (or y - x < 0),

|x - y| = x - y and |x + y| = x + y

Therefore, Eq. (1) is true given that,

2√2 ≤ x - y + x + y ≤ 4√2 ⇒ √2 ≤ x ≤ 2√2

checking with (2, 1) in region x > y, i.e. 2 > 1

Therefore, we shade area below y = x from [2, 2√2].

Area of this region = 1/2(2√2 x 2√2) - (1/2)√2 x √2 = 4 - 1 = 3 sq. units

By symmetry about y = x, total area required = 6 sq. units