Distance between the lines represented by the equation `x^2+2sqrt(3)xy+3y^2-3x-3sqrt(3)y-4=0` is
(A) `5/2`
(B) `5/4`
(C) `5`
(D) `0`

Options

A. perpendicular to each other
B. parallel
C. inclined at `45^@` to each other
D. None of the above

Answer

B

Explanation

We have , `x^2+2sqrt(3)xy+3y^2-3x-3sqrt(3)y-4=0`.......(i)
Here , `a=1,h=sqrt(3),b=3`
Now, `h^2=(sqrt(3))^2=3`
and `ab=1xx3=3`
`therefore h^2=ab`
Hence , lines represented by Eq.(i) are parallel.