The area of the quadrilateral formed by the l | vedclass

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(B) The given lines are $L_1: x+2y+3=0$,$L_2: 2x+4y+9=0$,$L_3: x-2y+3=0$,and $L_4: 3x-6y+11=0$. Notice that $L_1$ and $L_2$ are parallel,and $L_3$ and

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$\frac{1}{4}$

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(B) The given lines are $L_1: x+2y+3=0$,$L_2: 2x+4y+9=0$,$L_3: x-2y+3=0$,and $L_4: 3x-6y+11=0$. Notice that $L_1$ and $L_2$ are parallel,and $L_3$ and $L_4$ are parallel. Let $L_1: x+2y+3=0$ and $L_2: x+2y+4.5=0$. The distance between these parallel lines is $d_1 = \frac{|4.5-3|}{\sqrt{1^2+2^2}} = \frac{1.5}{\sqrt{5}} = \frac{3}{2\sqrt{5}}$. Let $L_3: x-2y+3=0$ and $L_4: x-2y+\frac{11}{3}=0$. The distance between these parallel lines is $d_2 = \frac{|11/3-3|}{\sqrt{1^2+(-2)^2}} = \frac{2/3}{\sqrt{5}} = \frac{2}{3\sqrt{5}}$. The area of a parallelogram formed by two pairs of parallel lines $a_1x+b_1y+c_1=0, a_1x+b_1y+c_2=0$ and $a_2x+b_2y+d_1=0, a_2x+b_2y+d_2=0$ is given by $\frac{|c_1-c_2||d_1-d_2|}{|a_1b_2-a_2b_1|}$. Here,$a_1=1, b_1=2, c_1=3, c_2=4.5$ and $a_2=1, b_2=-2, d_1=3, d_2=11/3$. The denominator is $|(1)(-2) - (1)(2)| = |-2-2| = 4$. The area is $\frac{|3-4.5| 	imes |3-11/3|}{4} = \frac{1.5 	imes 2/3}{4} = \frac{1}{4}$.

The area of the quadrilateral formed by the lines $x+2y+3=0$,$2x+4y+9=0$,$x-2y+3=0$,and $3x-6y+11=0$ is

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