Question

If \(θ\) is the acute angle between the lines \(2x^2+7xy+3y^2 = 0\), then the value of \(\frac{2cosθ-3sinθ}{4sinθ+5cosθ} = \ldots\)

• A. \(1\)
• B. \(\frac{5}{9}\)
• C. \(-\frac{1}{9}\)
• D. \(\frac{1}{9}\)

Hint:

Factor the pair and find tan theta between the lines.

Answer 1

The Correct Option is C

Step 1: Factor the equation
\(2x^2+7xy+3y^2 = (2x+y)(x+3y)\).

Step 2: Slopes
The lines are \(2x+y=0\) with slope \(m_1=-2\) and \(x+3y=0\) with slope \(m_2=-\frac13\).

Step 3: Angle between them
\[ \tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right| = \left|\frac{-2+\frac13}{1+\frac23}\right| = \frac{5/3}{5/3} = 1 \]
So \(\theta = 45^{\circ}\) and \(\sin\theta = \cos\theta = \frac{1}{\sqrt2}\).

Step 4: Evaluate
\[ \frac{2\cos\theta-3\sin\theta}{4\sin\theta+5\cos\theta} = \frac{2-3}{4+5} = -\frac19 \]

Final Answer:
The value is -1/9. \[ \boxed{\text{(C)}\ -\frac19} \]