Q.
The distance between the parallel lines given by (x+7y)2+42(x+7y)−42=0 is
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a
4/5
b
42
c
2
d
102
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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