Let θ be the angle between the given pair of straight lines x2−3xy+λy2+3x−5y+2=0 ∴θ=tan−1
1
3
(given) or tanθ=
1
3
We know that, tanθ=
2√h2−ab
a+b
where, h=−
3
2
,a=1,b=λ put these value in Eq. (ii), we get
1
3
=
2√
9
4
−λ.1
1+λ
⇒(1+λ)=3√9−4λ On squaring both sides, we get (1+λ)2=(3√9−4λ)2 ⇒1+λ2+2λ=81−36λ ⇒λ2+38λ−80=0 ⇒λ2+38λ−80=0 ⇒λ2+40λ−2λ−80=0 ⇒λ(λ+40)−2(λ+40)=0 ⇒(λ−2)(λ+40)=0 ⇒λ=2,−40⇒λ⇒=2 (∵ λ is non-negative real number))