Question

The pair of straight lines represented by the equation \(\sqrt{3}x^2-4xy+\sqrt{3}y^2 = 0\) are

• A. coincident.
• B. perpendicular.
• C. inclined at \(30^{\circ}\) to each other.
• D. inclined at \(60^{\circ}\) to each other.

Hint:

Use \(\tan\theta = \frac{2\sqrt{h^2-ab}}{|a+b|}\).

Answer 1

The Correct Option is C

Step 1: Key Formula:
For \(ax^2+2hxy+by^2 = 0\), the acute angle between the lines satisfies \(\tan\theta = \dfrac{2\sqrt{h^2-ab}}{|a+b|}\).

Step 2: Read off values:
\(a = \sqrt3\), \(b = \sqrt3\), \(2h = -4\) so \(h = -2\).

Step 3: Calculate:
\(h^2 - ab = 4 - 3 = 1\). Then \(\tan\theta = \frac{2\cdot1}{2\sqrt3} = \frac{1}{\sqrt3}\), so \(\theta = 30^{\circ}\).
The lines are real and distinct since \(h^2 > ab\). They are not perpendicular since \(a+b \ne 0\).

Final Answer:
The lines are inclined at \(30^{\circ}\), option (C). \[ \boxed{30^{\circ}} \]