Area of a parallelogram formed by lines ax pm by pm c = 0 , is

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9.Straight Line

easy

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

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

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

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

None of these

(IIT-1973)

Solution

(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{\rm{ }}\left( {\frac{c}{a},0} \right){\rm{ }},{\rm{ }}$$C{\rm{ }}\left( { – \frac{c}{a},0} \right){\rm{ }},{\rm{ }}$${\rm{ }}B{\rm{ }}\left( {0,\frac{c}{b}} \right)$, $D{\rm{ }}\left( {0, – \frac{c}{b}} \right)$.

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}}$.

Standard 11

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