Let the angles made with the positive x -axis by two straight lines drawn from the point P(2,3) and meeting the line x+y=6 at a distance (2)/(3) from the point P be θ_ 1 and θ_ 2 . Then the value of (θ_ 1 +θ_ 2 ) is:

Correct answer: C. (π)/(2) Using parametric form through P(2,3) : point at distance r = (2)/(3) is (2 + r θ,\ 3 + r θ) . Substituting in x + y = 6 : r( θ + θ) = 1 , so θ + θ = (3)/(2) . 2 (θ + (π)/(4)) = (3)/(2) (θ + (π)/(4)) = 3 2 . θ + (π)/(4) = (π)/(3) or (2π)/(3) . θ_1 = (π)/(12),\ θ_2 = (5π)/(12) . θ_1 + θ_2 = (π)/(12) + (5π)/(12) = (π)/(2) .

JEE Main 202624 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Straight Lines

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Straight Lines

Let the angles made with the positive x -axis by two straight lines...

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:

Options

A $\frac{\pi}{3}$
B $\frac{\pi}{6}$
C $\frac{\pi}{2}$
D $\frac{\pi}{12}$

Correct answer

C. $\frac{\pi}{2}$

Step-by-step solution

Using parametric form through

P(2,3)

: point at distance

r = \sqrt{\frac{2}{3}}

(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}

\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}

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