Question

A straight line L drawn through the point A (1, 2) intersects the line $x + y = 4$ at a distance of $\frac{\sqrt{6}}{3}$ units from A. The angle made by L with positive direction of x - axis can be:

• A. $\frac{\pi}{24}$
• B. $\frac{\pi}{6}$
• C. $\frac{\pi}{3}$
• D. $\frac{5\pi}{12}$

Answer 1

Answer: (D)

$\frac{5\pi}{12}$

Answer 2

To find the angle made by line L with the positive direction of the x-axis, we can use the parametric form of a line.

Let the line L pass through the point A(1, 2) and make an angle $\theta$ with the positive x-axis. Any point P on line L at a distance $r$ from A can be represented as:

$P(x, y) = (1 + r \cos \theta, 2 + r \sin \theta)$

We are given that the line L intersects the line $x + y = 4$ at a distance $r = \frac{\sqrt{6}}{3}$ units from A. Substitute the coordinates of P and the distance $r$ into the equation of the line $x + y = 4$:

$(1 + \frac{\sqrt{6}}{3} \cos \theta) + (2 + \frac{\sqrt{6}}{3} \sin \theta) = 4$

$3 + \frac{\sqrt{6}}{3} (\cos \theta + \sin \theta) = 4$

$\frac{\sqrt{6}}{3} (\cos \theta + \sin \theta) = 1$

$\cos \theta + \sin \theta = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}$

Now, we solve the trigonometric equation $\cos \theta + \sin \theta = \frac{\sqrt{6}}{2}$. Rewrite the left side:

$\sqrt{2} (\frac{1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt{2}} \sin \theta) = \sqrt{2} \cos(\theta - \frac{\pi}{4})$

So, $\sqrt{2} \cos(\theta - \frac{\pi}{4}) = \frac{\sqrt{6}}{2}$, which simplifies to:

$\cos(\theta - \frac{\pi}{4}) = \frac{\sqrt{3}}{2}$

The general solution for $\cos x = \frac{\sqrt{3}}{2}$ is $x = 2n\pi \pm \frac{\pi}{6}$. Therefore:

$\theta - \frac{\pi}{4} = \pm \frac{\pi}{6}$

Case 1: $\theta - \frac{\pi}{4} = \frac{\pi}{6} \implies \theta = \frac{\pi}{4} + \frac{\pi}{6} = \frac{5\pi}{12}$

Case 2: $\theta - \frac{\pi}{4} = -\frac{\pi}{6} \implies \theta = \frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12}$

From the given options, $\frac{5\pi}{12}$ is a valid answer.