Equation of the line equidistant from the parallel lines 9x+6y−7=0 and 3x+2y+6=0 is
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a
6x + 4y + 5 = 0
b
18x + 12y + 11 = 0
c
18x + 12y – 11 = 0
d
12x + 8y + 7 = 0
answer is B.
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Detailed Solution
The given lines are 3x + 2y + 6 = 0 and 3x+2y−73=0Let the required line be 3x + 2y + k = 0. Since it is equidistant from the given lines k+739+4=6−k9+4 ⇒ k=116