Question

For a point P in the plane, let $d_1$(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
$2\le d_1\left(P\right)+d_2\left(P\right)\le 4,$ is ___

Answer 1

Let the point $P$ be $(x,y)$
Then $d_1\left(P\right)=\left|\frac{x-y}{\sqrt{2}}\right|\text{and}{d}_2\left(P\right)=\left|\frac{x+y}{\sqrt{2}}\right|$
For $P$ lying in first quadrant $x>0,y>0.$
Also $2\le d_1\left(P\right)+d_2\left(p\right)\le 4\Rightarrow 2\le \left|\frac{x-y}{\sqrt{2}}\right|+\left|\frac{x+y}{\sqrt{2}}\right|\le 4\text{If}x>y,\text{then}2\le \frac{x-y+x+y}{\sqrt{2}}\le 4\text{or}\sqrt{2}\le x\le 2\sqrt{2}\text{If}x<y,\text{then}2\le \frac{y-x+x+y}{\sqrt{2}}\le 4\text{or}\sqrt{2}\le y\le 2\sqrt{2}$
The required region is the shaded region in the figure given below.


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