Question

For a point $M$ in the plane, let $d_1(M)$ and $d_2(M)$ be the distances of the point $M$ from the lines $x-3 y=0$ and $x+3 y=0$ respectively. If area of the region $R$ consisting of all points $M$ lying in the first quadrant of the plane and satisfying $6 \leq d_1(M)+d_2(M) \leq 12$ is $A$, then find the sum of digits of [A].
[Note: [k] denotes greatest integer function less than or equal to k.]

Answer 1

$\Theta 6 \leq\left|\frac{ x -3 y }{\sqrt{10}}\right|+\left|\frac{ x +3 y }{\sqrt{10}}\right| \leq 12$
$\Theta $ In first quadrant $x, y>0$
$\therefore 6 \sqrt{10} \leq| x -3 y |+ x +3 y \leq 12 \sqrt{10}$
(i) If $x \geq 3 y$ $6 \sqrt{10} \leq 2 x \leq 12 \sqrt{10}$
$\Rightarrow 3 \sqrt{10} \leq x \leq 6 \sqrt{10}$
(ii)$ 6 \sqrt{10} \leq 6 y \leq 12 \sqrt{10} $
$\Rightarrow \sqrt{10} \leq y \leq 2 \sqrt{10}$
$\therefore $ Required area $=$ shaded area
$ =6 \sqrt{10} \times 2 \sqrt{10}-3 \sqrt{10} \times \sqrt{10} $
$\therefore A =90 $
$\therefore \text { Sum of digits }=9$