question_answer

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) \[\frac{2{{c}^{2}}}{ab}\]

C) \[\frac{{{c}^{2}}}{2ab}\]

D) \[\frac{{{c}^{2}}}{4ab}\]

Correct Answer: B

Solution :

[b] Area of triangle \[\Delta AOB=\frac{1}{2}\times \frac{c}{b}\times \frac{c}{a}=\frac{{{c}^{2}}}{2ab}\] Total area \[=4\times area\Delta AOB\] \[=4\times \frac{{{c}^{2}}}{2ab}=\frac{2{{c}^{2}}}{ab}\]