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 $L_4: 3x-6y+11=0$.Note that $L_1$ and $L_2$ are parallel, and $L_3$ an

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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 $L_4: 3x-6y+11=0$.Note that $L_1$ and $L_2$ are parallel, and $L_3$ and $L_4$ are parallel.Rewrite the lines as:$L_1: x+2y+3=0$$L_2: x+2y+4.5=0$$L_3: x-2y+3=0$$L_4: x-2y+\frac{11}{3}=0$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 $	ext{Area} = \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 $a_2=1, b_2=-2, d_1=3, d_2=\frac{11}{3}$.The denominator is $|(1)(-2) - (1)(2)| = |-2-2| = 4$.The area is $\frac{|3-4.5| 	imes |3-\frac{11}{3}|}{4} = \frac{|-1.5| 	imes |-\frac{2}{3}|}{4} = \frac{1.5 	imes \frac{2}{3}}{4} = \frac{1}{4}$.

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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