Question

For a point in the plane, let $d_1\left(p\right)and{d}_2\left(p\right)$ be the distance of the point from the line x - y = 0 and x+ y =0 respectively. The area of the region R consisting of all point 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

$p(h,k)$ $ax+by+c=0$

$d=\left|\frac{ah+bk+c}{\sqrt{a^2+b^2}}\right|$

$2\le d_{1}P+d_2\left(P\right)\le 4$

$2\le \left|\frac{h-k}{\sqrt{2}}\right|+\left|\frac{h+k}{\sqrt{2}}\right|\le 4$

$2\sqrt{2}\le |h-k|+|h+k|\le 4\sqrt{2}$

if $h>k$ if $h<k$

$2\sqrt{2}\le h-k+h+k\le 4\sqrt{2}$ $2\sqrt{2}\le -h+k+h+k$

$2\sqrt{2}\le 2\sqrt{2}\le 4\sqrt{2}$ $\sqrt{2}\le k\le 2\sqrt{2}$

$\sqrt{2}\le h\le 2\sqrt{2}$

area $=(2\sqrt{2}{)}^2-(\sqrt{2}{)}^2=6unit$