Question

If the angle between the pair of straight lines represented by the equation $x^{2} - 3xy + \lambda y^{2} + 3x - 5y + 2 = 0$is $\tan^{- 1}\left( \frac{1}{3} \right)$, where $\lambda$is a non- negative real number, then $\lambda$is

• A. 2
• B. 0
• C. 3
• D. 1

Answer 1

Answer: (A)

2

Answer 2

Given that $\theta = \tan^{- 1}\left( \frac{1}{3} \right) \Rightarrow \tan\theta = \frac{1}{3}$Now, since $\tan\theta = \left| \frac{2\sqrt{h^{2} - ab}}{a + b} \right|$⇒ $\frac{1}{3}$= $\left| \frac{2\sqrt{\left( \frac{- 3}{2} \right)^{2} - \lambda}}{\lambda + 1} \right|$

⇒ $(\lambda + 1)^{2} = 9(9 - 4\lambda)$ ⇒ $\lambda^{2} + 38\lambda - 80 = 0$

⇒ $\lambda^{2} + 40\lambda - 2\lambda - 80 = 0$⇒ $\lambda(\lambda + 40) - 2(\lambda + 40) = 0$

⇒ $(\lambda - 2)(\lambda + 40) = 0$⇒ $\lambda = 2$ or – 40,

but $\lambda$ is a non-negative real number. Hence $\lambda = 2$