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(B) Let the point $P$ be $(x, y)$. The perpendicular distances from $P$ to the lines $L_1: 2x + y - 3 = 0$ and $L_2: x - 2y + 1 = 0$ are given by $d_1

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$8$

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(B) Let the point $P$ be $(x, y)$. The perpendicular distances from $P$ to the lines $L_1: 2x + y - 3 = 0$ and $L_2: x - 2y + 1 = 0$ are given by $d_1 = \frac{|2x + y - 3|}{\sqrt{2^2 + 1^2}} = \frac{|2x + y - 3|}{\sqrt{5}}$ and $d_2 = \frac{|x - 2y + 1|}{\sqrt{1^2 + (-2)^2}} = \frac{|x - 2y + 1|}{\sqrt{5}}$.Given $d_1 + d_2 = 2$,we have $\frac{|2x + y - 3| + |x - 2y + 1|}{\sqrt{5}} = 2$,which simplifies to $|2x + y - 3| + |x - 2y + 1| = 2\sqrt{5}$.The locus of $P$ forms a square. The distance between the parallel lines formed by the absolute values is $2\sqrt{5}$.The side length $s$ of the square is given by $s = \frac{2\sqrt{5}}{\sqrt{2^2+1^2}} 	imes \sqrt{2} = 2\sqrt{2}$.Alternatively,the area of the square formed by $|ax+by+c_1| + |bx-ay+c_2| = k$ is $\frac{2k^2}{a^2+b^2}$.Here,$a=2, b=1, k=2\sqrt{5}$.Area $= \frac{2(2\sqrt{5})^2}{2^2+1^2} = \frac{2(20)}{5} = \frac{40}{5} = 8 \, sq \, units$.

Let $P$ be a moving point such that the sum of its perpendicular distances from the lines $2x + y - 3 = 0$ and $x - 2y + 1 = 0$ is always $2 \, units$. Then,the area bounded by the locus of point $P$ is:

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