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: Use the formula:
Here $a=1$, $2h=-3$, $b=2$. So $h=-\tfrac32$ and $h^2-ab=\tfrac94-2=\tfrac14$.

Step 2: Compute:
$\tan\theta=\dfrac{2\cdot\frac12}{1+2}=\dfrac13$.

Step 3: Triangle method:
Take $\sin\theta=\tfrac1{\sqrt{10}}$ and $\cos\theta=\tfrac3{\sqrt{10}}$. Then $3\sin\theta+2\cos\theta=\tfrac{9}{\sqrt{10}}$ and $3\sin\theta-2\cos\theta=-\tfrac{3}{\sqrt{10}}$.

Step 4: Ratio:
$\tfrac{9}{-3}=-3$, option (C).

Final Answer:
Using sin and cos from tan theta = 1/3 gives -3. \[ \boxed{C} \]