Question

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

• A. $\frac{c^2}{ab}$
• B. $\frac{2c^2}{ab}$
• C. $\frac{c^2}{2ab}$
• D. $\frac{c^2}{4ab}$

Answer 1

The correct option is B $\frac{2c^2}{ab}$

The figure shown by the lines

$ax+by+c=0,ax-by+c=0,ax+by-c=0$ and $ax-by-c=0$ is shown below.

Clearly, $ABCD$ is a parallelogram and its area

$=\frac{1}{2}\times$ Product of diagonals

$=\frac{1}{2}\times \frac{2c}{b}\times \frac{2c}{a}$

$=\frac{2c^2}{ab}$

Hence, B is the correct option.