The area of the parallelogram formed by the l | vedclass

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(D) Given lines are $L_1: \lambda x+4 y+2=0$,$L_2: 3 x+4 y-3=0$,$L_3: 2 x+\mu y+6=0$,$L_4: 2 x+y+3=0$. Since $L_1 \parallel L_2$,their slopes must be

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$3$

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(D) Given lines are $L_1: \lambda x+4 y+2=0$,$L_2: 3 x+4 y-3=0$,$L_3: 2 x+\mu y+6=0$,$L_4: 2 x+y+3=0$. Since $L_1 \parallel L_2$,their slopes must be equal: $-\frac{\lambda}{4} = -\frac{3}{4} \Rightarrow \lambda = 3$. Since $L_3 \parallel L_4$,their slopes must be equal: $-\frac{2}{\mu} = -\frac{2}{1} \Rightarrow \mu = 1$. The area of a parallelogram formed by lines $a_1 x+b_1 y+c_1=0$,$a_1 x+b_1 y+c_2=0$,$a_2 x+b_2 y+d_1=0$,and $a_2 x+b_2 y+d_2=0$ is given by $\frac{|(c_1-c_2)(d_1-d_2)|}{|a_1 b_2 - a_2 b_1|}$. Here,$L_1: 3x+4y+2=0$,$L_2: 3x+4y-3=0$,$L_3: 2x+y+6=0$,$L_4: 2x+y+3=0$. $c_1=2, c_2=-3, d_1=6, d_2=3$. $a_1=3, b_1=4, a_2=2, b_2=1$. Area $= \frac{|(2 - (-3))(6 - 3)|}{|(3)(1) - (2)(4)|} = \frac{|5 	imes 3|}{|3 - 8|} = \frac{15}{|-5|} = \frac{15}{5} = 3$. Thus,the area is $3$ square units.

The area of the parallelogram formed by the lines $L_1 \equiv \lambda x+4 y+2=0$,$L_2 \equiv 3 x+4 y-3=0$,$L_3 \equiv 2 x+\mu y+6=0$,and $L_4 \equiv 2 x+y+3=0$,where $L_1$ is parallel to $L_2$ and $L_3$ is parallel to $L_4$,is

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