The area of a parallelogram formed by the lin | vedclass

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(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){

Vedclass · Accepted Answer

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

Vedclass · 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{m{ }}\left( {\frac{c}{a},0} ight){m{ }},{m{ }}$$C{m{ }}\left( { - \frac{c}{a},0} ight){m{ }},{m{ }}$${m{ }}B{m{ }}\left( {0,\frac{c}{b}} ight)$, $D{m{ }}\left( {0, - \frac{c}{b}} ight)$.

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 	imes BD = \frac{1}{2} 	imes \frac{{2c}}{a} 	imes \frac{{2c}}{b} = \frac{{2{c^2}}}{{ab}}$.

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

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