A) \[\frac{{{c}^{2}}}{ab}\]
B) \[\frac{2{{c}^{2}}}{ab}\]
C) \[\frac{{{c}^{2}}}{2ab}\]
D) None of these
Correct Answer: B
Solution :
\[ax\pm by\pm c=0\Rightarrow \frac{x}{\pm c/a}+\frac{y}{\pm c/b}=1\] which meets on axes at \[A\text{ }\left( \frac{c}{a},0 \right)\text{ },\text{ }\]\[C\text{ }\left( -\frac{c}{a},0 \right)\text{ },\text{ }\]\[\text{ }B\text{ }\left( 0,\frac{c}{b} \right)\], \[D\text{ }\left( 0,-\frac{c}{b} \right)\]. Therefore, the diagonals AC and BD of quadrilateral ABCD are perpendicular, hence it is a rhombus whose area is given by \[=\frac{1}{2}AC\times BD=\frac{1}{2}\times \frac{2c}{a}\times \frac{2c}{b}=\frac{2{{c}^{2}}}{ab}\].
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