If the angle between the pair of straight lines represented by the equation `x^2-3xy+lambday^2+3x-5y+2=0` is `tan^(-1)(1//3)`, where `lambda` is non - negative real number , then `lambda` is equal to

Options

A. 2
B. 0
C. 3
D. 1

Answer

A

Explanation

We have , `x^2-3xy+lambday^2+3x-5y+2=0` ....(i)
Let `theta` be the angle between the lines represented by Eq.(i).
Then , `theta=tan^(-1)((1)/(3))rArr tan theta=(1)/(3)`
Now, `tan theta = |(2sqrt(h^2-ab))/(a+b)|`
`rArr tan theta = |(2sqrt((9)/(4)-lambda))/(1+lambda)|rArr (1)/(3)=|(2sqrt(9-4lambda)/(2))/(1+lambda)|`
`rArr(1)/(3)=|sqrt(9-4lambda)/(1+lambda)|`
`rArr sqrt(9-4lambda)/(1+lambda)=+-(1)/(3)`
Taking positive sign , we get `(sqrt(9-4lamda))/(1+lambda)=(1)/(3)rArr 9-4lambda=((1+lambda)^2)/(9)`
`rArr 81-36lambda =1+lambda^2+2lambda`
`rArr lambda^2+38lambda-80 =0`
`rArr lambda^2+40lambda-2lambda-80=0`
`rArr lambda(lambda+40)-2(lambda+40)=0`
`rArr (lambda+40)(lambda-2)=0`
`rArr lambda= 2` .