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Q.

The distance between the pair of parallel lines 2x2+4xy+2y2+3x+3y+1=0  is

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a

122

b

12

c

12

d

2

answer is A.

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Detailed Solution

The distance between the parallel lines ax2+2hxy+by2+2gx+2fy+c=0 is 2g2−acaa+bFor the equation 2x2+4xy+2y2+3x+3y+1=0,the values are a=2,b=2,c=1,g=32Hence the distance required is dd=294−224=29−8244=2142=122Therefore, the required distance is 122
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