If the lines represented by the equation 4xy+6x-8y+c=0 form a rectangle with the coordinate axis then the area of the rectangle (in sq. units) is

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$| c |$

$| \frac{c}{2} |$

$| \frac{c}{4} |$

$| \frac{c}{8} |$

answer is C.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 4 x y + 6 x - 8 y + c = 0 - - (1) \\ C o m p a r i n g w i t h a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0 \\ a = 0, h = 2, b = 0, g = 3, f = - 4, c = c \\ P . I . = (\frac{h f - b g}{a b - h^{2}}, \frac{g h - a f}{a b - h^{2}}) \\ = (\frac{- f}{h}, \frac{- g}{h}) \\ (1) Re p r e s e n t p a i r o f l i n e t h e n Δ = 0 \\ a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0 \\ 2 f g h - c h^{2} = 0 \\ \Rightarrow c h = 2 f g \Rightarrow f g = \frac{c h}{2} \\ A r e a o f O A P B = | \frac{f}{h} | \times | \frac{g}{h} | = | \frac{f g}{h^{2}} | \\ = | \frac{c h}{2 h^{2}} | = | \frac{c}{2 h} | = | \frac{c}{4} | \end{array}$