Question

If the slope of one of the pair of lines represented by $2x^2 + 3xy + Ky^2 = 0$ is 2, then the angle between the pair of lines is:

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{6}$
• D. $\frac{\pi}{4}$

Hint:

For a second-degree equation representing a pair of lines, the angle between the lines is given by $\tan \theta = \left| \frac{2\sqrt{H^2 - AB}}{A + B} \right|$.

Answer 1

The Correct Option is A

Step 1: Finding the Slopes of the Pair of Lines The general equation for a pair of straight lines is: $$ax^2 + 2hxy + by^2 = 0$$ For the given equation: $$2x^2 + 3xy + Ky^2 = 0$$ Comparing, we have: $$a = 2, \quad 2h = 3 \Rightarrow h = \frac{3}{2}, \quad b = K$$ The slopes of the lines are given by: $$m = \frac{- h \pm \sqrt{h^2 - ab}}{b}$$
Step 2: Using the Given Slope One slope is given as 2: $$2 = \frac{-\frac{3}{2} + \sqrt{\left(\frac{3}{2}\right)^2 - (2)(K)}}{K}$$
Step 3: Finding the Angle Between the Lines The angle between the two lines is given by: $$\tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b}$$ Using the given values and solving, we find: $$\theta = \frac{\pi}{2}$$
Final Answer: $$\frac{\pi}{2}$$