For a point in the plane, let and be the distance of the point from the lines and respectively. The area of the region consisting of all points lying in the first quadrant of the plane and satisfying is [2014]

Question

For a point $P$ in the plane, let $d_{1} (P)$ and $d_{2} (P)$ be the distance 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 $2 \leq d_{1} (P) + d_{2} (P) \leq 4,$ is [2014]

Smart Achievers · Accepted Answer

(6)

Let the point $d_{1} (P)$ be $(x, y)$ .

$Then d_{1} (P) =$ $| \frac{x - y}{\sqrt{2}} |$ $and d_{2} (P) =$ $| \frac{x + y}{\sqrt{2}} |$

For $P$ lying in first quadrant $x > 0, y > 0$

Now $2 \leq d_{1} (P) + d_{2} (P) \leq 4$

$\Rightarrow 2 \leq$ $| \frac{x - y}{\sqrt{2}} |$ $+$ $| \frac{x + y}{\sqrt{2}} |$ $\leq 4$

$If x > y, then 2 \leq \frac{x - y + x + y}{\sqrt{2}} \leq 4 \Rightarrow \sqrt{2} \leq x \leq 2 \sqrt{2}$

$If x < y, then 2 \leq \frac{y - x + x + y}{\sqrt{2}} \leq 4 \Rightarrow \sqrt{2} \leq y \leq 2 \sqrt{2}$

The required region is the shaded region in the figure given below.

$∴$ $Required area = {(2 \sqrt{2})}^{2} - {(\sqrt{2})}^{2} = 8 - 2 = 6 sq units.$

Solution

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Solution

Q. For a point P in the plane, let d1(P) and d2(P) be the distance 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 [2014]

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