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(D) The centre of the square is the intersection of $x+2y-3=0$ and $2x-y-1=0$. Solving these,we get $x=1, y=1$. The centre is $(1,1)$.Since the area o

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$(2x-y-1-2\sqrt{5})(x+2y-3+2\sqrt{5})=0$

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(D) The centre of the square is the intersection of $x+2y-3=0$ and $2x-y-1=0$. Solving these,we get $x=1, y=1$. The centre is $(1,1)$.Since the area of the square is $16$,the side length is $s = \sqrt{16} = 4$.The distance from the centre $(1,1)$ to each side is $d = s/2 = 2$.The sides are parallel to the given lines $x+2y-3=0$ and $2x-y-1=0$.Let the equations of the sides be $x+2y+C_1=0$ and $2x-y+C_2=0$.The distance from $(1,1)$ to $x+2y+C_1=0$ is $\frac{|1+2(1)+C_1|}{\sqrt{1^2+2^2}} = 2$ $\Rightarrow |3+C_1| = 2\sqrt{5}$ $\Rightarrow C_1 = -3 \pm 2\sqrt{5}$.The distance from $(1,1)$ to $2x-y+C_2=0$ is $\frac{|2(1)-1+C_2|}{\sqrt{2^2+(-1)^2}} = 2$ $\Rightarrow |1+C_2| = 2\sqrt{5}$ $\Rightarrow C_2 = -1 \pm 2\sqrt{5}$.Choosing one pair of sides,we have $x+2y-3+2\sqrt{5}=0$ and $2x-y-1-2\sqrt{5}=0$.The combined equation is $(2x-y-1-2\sqrt{5})(x+2y-3+2\sqrt{5})=0$.

The combined equation of a possible pair of adjacent sides of a square with area $16 \text{ square units}$ whose centre is the point of intersection of the lines $x+2y-3=0$ and $2x-y-1=0$ is

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