The area of the parallelogram formed by the lines $\mathrm{L}_1 \equiv \lambda x+4 y+2=0, \mathrm{~L}_2 \equiv 3 x+4 y-3=0$, $\mathrm{L}_3 \equiv 2 x+\mu y+6=0, \quad \mathrm{~L}_4 \equiv 2 x+y+3=0$, where $\mathrm{L}_1$ is parallel to $L_2$ and $L_3$ is parallel to $L_4$ is

• A

  $9$
• B

  $7$
• C

  $5$
• Correct answer
  D

  $3$

Solution

$\begin{aligned} & \lambda x+4 y+2=0 \text { is parallel to } 3 x+4 y-3=0 \\ & \Rightarrow \frac{-\lambda}{4}=\frac{-3}{4} \Rightarrow \lambda=3 \\ & 2 x+\mu y+6=0 \text { is parallel to } 2 x+y+3=0 \\ & \Rightarrow \frac{-2}{\mu}=-2 \Rightarrow \mu=1 \\ & A \equiv\left(\frac{-22}{5}, \frac{14}{5}\right), C \equiv(-3,3), \quad D \equiv\left(\frac{-27}{5}, \frac{24}{5}\right) \\ & \text { Area of } \triangle A D C\end{aligned}$
$\Rightarrow$ Area of parallelogram $=2 \mathrm{Ar}(\triangle A D C)=3$.

Hence, the correct answer is option: 'D'.