Q70 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
If $\theta$ is the acute angle between the lines $2x^2 + 7xy + 3y^2 = 0$, then the value of $\dfrac{2\cos\theta - 3\sin\theta}{4\sin\theta + 5\cos\theta} = \ldots$
- A$1$
- B$\dfrac{5}{9}$
- Correct answerC$-\dfrac{1}{9}$
- D$\dfrac{1}{9}$
Solution
Given equation of the pair of straight lines is $2x^2 + 7xy + 3y^2 = 0$.
Comparing with $ax^2 + 2hxy + by^2 = 0$, we get $a = 2$, $h = \dfrac{7}{2}$, and $b = 3$.
The acute angle $\theta$ between the lines is given by:
$\tan\theta = \left| \dfrac{2\sqrt{h^2 - ab}}{a + b} \right|$
$\tan\theta = \left| \dfrac{2\sqrt{\left(\dfrac{7}{2}\right)^2 - (2)(3)}}{2 + 3} \right|$
$\tan\theta = \left| \dfrac{2\sqrt{\dfrac{49}{4} - 6}}{5} \right| = \left| \dfrac{2\sqrt{\dfrac{25}{4}}}{5} \right| = \dfrac{2 \times \dfrac{5}{2}}{5} = 1$
We need to find the value of $\dfrac{2\cos\theta - 3\sin\theta}{4\sin\theta + 5\cos\theta}$.
Dividing the numerator and the denominator by $\cos\theta$, we get:
$\dfrac{2 - 3\tan\theta}{4\tan\theta + 5}$
Substituting $\tan\theta = 1$:
$\dfrac{2 - 3(1)}{4(1) + 5} = \dfrac{-1}{9} = -\dfrac{1}{9}$
Answer: $-\dfrac{1}{9}$
Comparing with $ax^2 + 2hxy + by^2 = 0$, we get $a = 2$, $h = \dfrac{7}{2}$, and $b = 3$.
The acute angle $\theta$ between the lines is given by:
$\tan\theta = \left| \dfrac{2\sqrt{h^2 - ab}}{a + b} \right|$
$\tan\theta = \left| \dfrac{2\sqrt{\left(\dfrac{7}{2}\right)^2 - (2)(3)}}{2 + 3} \right|$
$\tan\theta = \left| \dfrac{2\sqrt{\dfrac{49}{4} - 6}}{5} \right| = \left| \dfrac{2\sqrt{\dfrac{25}{4}}}{5} \right| = \dfrac{2 \times \dfrac{5}{2}}{5} = 1$
We need to find the value of $\dfrac{2\cos\theta - 3\sin\theta}{4\sin\theta + 5\cos\theta}$.
Dividing the numerator and the denominator by $\cos\theta$, we get:
$\dfrac{2 - 3\tan\theta}{4\tan\theta + 5}$
Substituting $\tan\theta = 1$:
$\dfrac{2 - 3(1)}{4(1) + 5} = \dfrac{-1}{9} = -\dfrac{1}{9}$
Answer: $-\dfrac{1}{9}$
Hence, the correct answer is option: 'C'.