Question

If \(θ\) is the acute angle between the lines represented by the equation \(x^2-3xy+2y^2 = 0\), then \(\frac{3sinθ+2cosθ}{3sinθ-2cosθ} =\)

• A. \(-\frac{1}{2}\)
• B. \(\frac{1}{2}\)
• C. \(-3\)
• D. \(3\)

Hint:

Factor the pair of lines, find tan(theta), then divide by cos(theta).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A homogeneous equation \(ax^2+2hxy+by^2=0\) represents two lines through the origin. The acute angle between them satisfies \(\tan\theta=\left|\dfrac{2\sqrt{h^2-ab}}{a+b}\right|\).

Step 2: Find the lines:
\(x^2-3xy+2y^2=(x-y)(x-2y)=0\), so the lines are \(y=x\) and \(y=\tfrac{x}{2}\). Their slopes are \(1\) and \(\tfrac12\).

Step 3: Find tan theta:
\[ \tan\theta=\left|\frac{1-\frac12}{1+1\cdot\frac12}\right|=\frac{1/2}{3/2}=\frac13 \]

Step 4: Evaluate the expression:
Divide numerator and denominator by \(\cos\theta\):
\[ \frac{3\sin\theta+2\cos\theta}{3\sin\theta-2\cos\theta}=\frac{3\tan\theta+2}{3\tan\theta-2}=\frac{1+2}{1-2}=-3 \]

Step 5: Choose:
Option (C).

Final Answer:
tan theta = 1/3, so the expression equals -3. \[ \boxed{-3} \]