The area of a parallelogram formed by the lin | vedclass

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(B) The given lines are $ax + by + c = 0$,$ax + by - c = 0$,$ax - by + c = 0$,and $ax - by - c = 0$.These can be rewritten in intercept form as $\frac

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

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(B) The given lines are $ax + by + c = 0$,$ax + by - c = 0$,$ax - by + c = 0$,and $ax - by - c = 0$.These can be rewritten in intercept form as $\frac{x}{\pm c/a} + \frac{y}{\pm c/b} = 1$.The vertices of the parallelogram are the intersection points of these lines:$A(\frac{c}{a}, 0)$,$B(0, \frac{c}{b})$,$C(-\frac{c}{a}, 0)$,and $D(0, -\frac{c}{b})$.The diagonals of this quadrilateral are $AC$ and $BD$,which lie along the $x$-axis and $y$-axis respectively.Since the axes are perpendicular,the diagonals are perpendicular,making the parallelogram a rhombus.The length of diagonal $AC = |\frac{c}{a} - (-\frac{c}{a})| = \frac{2c}{a}$.The length of diagonal $BD = |\frac{c}{b} - (-\frac{c}{b})| = \frac{2c}{b}$.The area of the rhombus is $\frac{1}{2} 	imes AC 	imes BD = \frac{1}{2} 	imes \frac{2c}{a} 	imes \frac{2c}{b} = \frac{2c^2}{ab}$.

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

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