Area of Region Defined by Distances to Two Lines in First Quadrant

Let $P(x,y)$ be any point in the first quadrant of the Cartesian plane, so $x \ge 0$ and $y \ge 0$.

The perpendicular distances $d_1(P)$ and $d_2(P)$ from $P(x,y)$ to the lines $x - y = 0$ and $x + y = 0$ respectively are given by: $d_1(P) = \frac{|x - y|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y|}{\sqrt{2}}$ $d_2(P) = \frac{|x + y|}{\sqrt{1^2 + 1^2}} = \frac{|x + y|}{\sqrt{2}}$

Since $P$ lies in the first quadrant ($x \ge 0, y \ge 0$), we have $x + y \ge 0$, which simplifies $d_2(P)$ to: $d_2(P) = \frac{x + y}{\sqrt{2}}$

The sum of the two distances is: $d_1(P) + d_2(P) = \frac{|x - y| + x + y}{\sqrt{2}}$

We are given that $2 \le d_1(P) + d_2(P) \le 4$. To analyze this region $R$, we divide the first quadrant into two sub-regions separated by the line $y = x$:

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Case 1: $x \ge y \ge 0$

In this sub-region, $|x - y| = x - y$. Thus: $d_1(P) + d_2(P) = \frac{(x - y) + x + y}{\sqrt{2}} = \frac{2x}{\sqrt{2}} = \sqrt{2}x$

The given condition $2 \le d_1(P) + d_2(P) \le 4$ simplifies to: $2 \le \sqrt{2}x \le 4 \implies \sqrt{2} \le x \le 2\sqrt{2}$

For any fixed $x \in [\sqrt{2}, 2\sqrt{2}]$, the $y$-coordinate satisfies $0 \le y \le x$. The area of this sub-region $R_1$ is: $\text{Area}(R_1) = \int_{\sqrt{2}}^{2\sqrt{2}} \int_{0}^{x} dy \, dx = \int_{\sqrt{2}}^{2\sqrt{2}} x \, dx = \left[ \frac{x^2}{2} \right]_{\sqrt{2}}^{2\sqrt{2}} = \frac{(2\sqrt{2})^2 - (\sqrt{2})^2}{2} = \frac{8 - 2}{2} = 3$

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Case 2: $y > x \ge 0$

In this sub-region, $|x - y| = y - x$. Thus: $d_1(P) + d_2(P) = \frac{(y - x) + x + y}{\sqrt{2}} = \frac{2y}{\sqrt{2}} = \sqrt{2}y$

The given condition $2 \le d_1(P) + d_2(P) \le 4$ simplifies to: $2 \le \sqrt{2}y \le 4 \implies \sqrt{2} \le y \le 2\sqrt{2}$

For any fixed $y \in [\sqrt{2}, 2\sqrt{2}]$, the $x$-coordinate satisfies $0 \le x < y$. The area of this sub-region $R_2$ is: $\text{Area}(R_2) = \int_{\sqrt{2}}^{2\sqrt{2}} \int_{0}^{y} dx \, dy = \int_{\sqrt{2}}^{2\sqrt{2}} y \, dy = \left[ \frac{y^2}{2} \right]_{\sqrt{2}}^{2\sqrt{2}} = \frac{(2\sqrt{2})^2 - (\sqrt{2})^2}{2} = \frac{8 - 2}{2} = 3$

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Total Area of Region $R$

Since $R_1$ and $R_2$ intersect only along the line segment $y = x$ (which has zero area), the total area of the region $R$ is: $\text{Area}(R) = \text{Area}(R_1) + \text{Area}(R_2) = 3 + 3 = 6$