Two lines are given by `(x-2y)^2 + k (x-2y) = 0` . The value of k, so that the distance between them is 3, is :

Options

A. `(1)/(sqrt(5))`
B. `+-(2)/(sqrt(5))`
C. `+-3sqrt(5)`
D. None of these

Answer

C

Explanation

We have , `(x-2y)^2+k(x-2y)=0` ........(i)
and `d=3`
From Eq.(i) , we get
`x^2+4y^2-4xy+k(x-2y)=0`
`rArr x^2-4xy+4y^2kx-2ky=0`
Here, `a=1,b=4,c=0, g=(k)/(2),h=-2`
Now , `h^2-ab=(-2)-1xx4=0`
Thus , given lines are parallel , so the distance between two parallel lines is `d=2sqrt((g^2-ac)/(a(a+b)))rArr 3=2sqrt(((k^2)/(4)-0)/(1(1+4)))`
`rArr 3=2sqrt((k^2)/(4xx5))rArr 3=+-(2k)/(sqrt(20))`
`rArr 3=+-(k)/sqrt(5)`
`therefore k=3sqrt(5) or -3sqrt(5)` .