Q43 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics
Let the angles made with the positive $x$-axis by two straight lines drawn from the point $\mathrm{P}(2,3)$ and meeting the line $x+y=6$ at a distance $\sqrt{\frac{2}{3}}$ from the point P be $\theta_{1}$ and $\theta_{2}$. Then the value of $\left(\theta_{1}+\theta_{2}\right)$ is:
- A$\frac{\pi}{3}$
- B$\frac{\pi}{6}$
- Correct answerC$\frac{\pi}{2}$
- D$\frac{\pi}{12}$
Solution
Using parametric form through $P(2,3)$: point at distance $r = \sqrt{\frac{2}{3}}$ is $(2 + r\cos\theta,\ 3 + r\sin\theta)$.
Substituting in $x + y = 6$: $r(\cos\theta + \sin\theta) = 1$, so $\cos\theta + \sin\theta = \sqrt{\frac{3}{2}}$.
$\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right) = \sqrt{\frac{3}{2}} \Rightarrow \sin\left(\theta + \frac{\pi}{4}\right) = \frac{\sqrt{3}}{2}$.
$\theta + \frac{\pi}{4} = \frac{\pi}{3}$ or $\frac{2\pi}{3}$.
$\theta_1 = \frac{\pi}{12},\ \theta_2 = \frac{5\pi}{12}$.
$\theta_1 + \theta_2 = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{\pi}{2}$.
Substituting in $x + y = 6$: $r(\cos\theta + \sin\theta) = 1$, so $\cos\theta + \sin\theta = \sqrt{\frac{3}{2}}$.
$\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right) = \sqrt{\frac{3}{2}} \Rightarrow \sin\left(\theta + \frac{\pi}{4}\right) = \frac{\sqrt{3}}{2}$.
$\theta + \frac{\pi}{4} = \frac{\pi}{3}$ or $\frac{2\pi}{3}$.
$\theta_1 = \frac{\pi}{12},\ \theta_2 = \frac{5\pi}{12}$.
$\theta_1 + \theta_2 = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{\pi}{2}$.
Hence, the correct answer is option: 'C'.