$a x \pm b y \pm c=0 \Rightarrow \frac{x}{\pm c / a}+\frac{y}{\pm c / b}=1$
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 $A C$ and $B D$ of quadrilateral $A B C D$ are perpendicular, hence it is a rhombus whose area is given by
$=\frac{1}{2} A C \times B D=\frac{1}{2} \times \frac{2 c}{a} \times 2 \frac{c}{b}=\frac{2 c^{2}}{a b}$