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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

16

4

10

6

Solution:

2 ≤ d1(p) + d2(p) ≤ 4
For P(α, β), α > β ⇒ 2√2 ≤ 2α ≤ 4√2
√2 ≤ α ≤ 2√2 ⇒ Area of region = ((2√2)² - (√2)²) = 8 - 2 = 6 sq. units