The distance between the parallel lines given by (x+7y)2+42(x+7y)−42=0 is

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answer is C.

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Detailed Solution

The lines given by the equation are (x+7y−32)(x+7y+72)=0⇒ x+7y−32=0 and x+7y+72=0distance between these lines =72−(−32)12+72=10252=2.

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