Let the angles made with the positive x-axis by two straight lines drawn from the point � and meeting the line � at a distance � from the point � be � and �. Then the value of � is:

Let the angles made with the positive x-axis by two straight lines drawn from the point P(2,3)

Q. 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 $$ \sqrt{\frac{2}{3}} $$ from the point $ P $ be $ \theta_1 $ and $ \theta_2 $. Then the value of $ (\theta_1 + \theta_2) $ is:

(A) $ \dfrac{\pi}{2} $

(B) $ \dfrac{\pi}{3} $

(D) $ \dfrac{\pi}{6} $

Correct Answer: $ \dfrac{\pi}{2} $

Explanation

Let the equation of a line passing through $ P(2,3) $ making an angle $ \theta $ with the x-axis be:

$$ y-3 = \tan\theta (x-2) $$

Rewrite in general form:

$$ \tan\theta \, x - y + (3 - 2\tan\theta) = 0 $$

The distance between this line and the given line $ x + y = 6 $ is:

$$ \frac{|a_1c_2 - a_2c_1|}{\sqrt{a_1^2 + b_1^2}\sqrt{a_2^2 + b_2^2}} $$

Using distance between two parallel lines formula and equating it to $ \sqrt{\frac{2}{3}} $, we get a quadratic in $ \tan\theta $.

On solving, the sum of roots $ \tan\theta_1 + \tan\theta_2 $ comes out to be:

$$ \tan\theta_1 + \tan\theta_2 = 0 $$

Hence,

$$ \theta_1 + \theta_2 = \frac{\pi}{2} $$

Therefore, the correct answer is $ \dfrac{\pi}{2} $.

Explanation

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