The area of the figure formed by the lines $a x+b y+c=0$, ax-by $+c=0, a x+b y-c=0$ and $a x-b y-c=0$ is

• A

  $\frac{\mathrm{c}^2}{\mathrm{ab}}$
• Correct answer
  B

  $\frac{2 \mathrm{c}^2}{\mathrm{ab}}$
• C

  $\frac{\mathrm{c}^2}{2 \mathrm{ab}}$
• D

  $\frac{\mathrm{c}^2}{4 \mathrm{ab}}$

Solution

Area of triangle $\Delta \mathrm{AOB}=\frac{1}{2} \times \frac{\mathrm{c}}{\mathrm{~b}} \times \frac{\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{c}^2}{2 \mathrm{ab}}$
Total area $=4 \times \text { area } \triangle A O B=4 \times \frac{c^2}{2 \mathrm{ab}}=\frac{2 \mathrm{c}^2}{\mathrm{ab}}$

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