Question

The area of a parallelogram formed by the lines $ax\pm by\pm c=0$, is

• A. $\frac{c^2}{ab}$
• B. $\frac{2c^2}{ab}$
• C. $\frac{c^2}{2ab}$
• D. None of these

Answer 1

Answer: (B)

$ax\pm by\pm c=0\Rightarrow \frac{x}{\pm c/a}+\frac{y}{\pm c/b}=1$
which meets on axes at $A\left(\frac{c}{a},0\right),C\left(\frac{-c}{a},0\right)$,
$B\left(0,\frac{c}{b}\right)$ and $D\left(0,\frac{-c}{b}\right).$
Therefore, the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ are perpendicular, hence it is a rhombus whose area is given by
$=\frac{1}{2}AC\times BD=\frac{1}{2}\times \frac{2c}{a}\times 2\frac{c}{b}=\frac{2c^2}{ab}$