$\begin {array} {1 1} (A)\;15^{\circ} \: or\: 75^{\circ} & \quad (B)\;30^{\circ} \: or \: 60^{\circ} \\ (C)\;45^{\circ} & \quad (D)\;60^{\circ} \: or \: 120^{\circ} \end {array}$

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- Equation of a line passing through $(x_1, y_1)$ and making an angle $\theta $ is $ \large\frac{x-x_1}{\cos \theta }$$ = \large\frac{y-y_1}{\sin \theta}$$=r$ where $r$ is the distance of the point from the given line.

Step 1:

Let the required line make an angle $ \theta $ with $x$ - axis.

Hence its equation is

$\large\frac{x-1}{\cos \theta}$$=\large\frac{y-2}{\sin \theta}$$=\large\frac{\sqrt 6}{3}$

$ \therefore x = \large\frac{\sqrt 6 \cos \theta}{3}$$+1$ and $ y=\large\frac{\sqrt 6 \sin \theta}{3}$$+2$

or $ \bigg( \large\frac{\sqrt 6 \cos \theta}{3}$$+1, \large\frac{\sqrt 6 \sin \theta }{3}$$+2 \bigg)$

Step 2 :

This point lies on the line $x+y=4$

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

$ \Rightarrow \large\frac{\sqrt 6 \cos \theta}{3}$$+\large\frac{\sqrt 6 \sin \theta}{3}$$=1$

$ \Rightarrow \sqrt 6 \cos \theta + \sqrt 6 \sin \theta=3$

(i.e) $ \cos \theta + \sin \theta = \large\frac{3}{\sqrt 6}$

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

Step 3 :

Squaring on both sides :

$ ( \cos \theta + \sin \theta )^2 = \large\frac{3}{2}$

(i.e) $ \cos^2\theta + \sin^2\theta + 2\sin \theta \cos \theta = \large\frac{3}{2}$

But $ \sin^2\theta + \cos^2\theta=1$

$ \therefore 2 \sin \theta \cos \theta = \large\frac{3}{2}$$-1$

$ 2\sin \theta \cos \theta = \sin2\theta$

$ \therefore \sin2\theta = \large\frac{1}{2}$

$ \Rightarrow 2\theta = \large\frac{\pi}{6}$ or $30^{\circ}$

$ \therefore \theta = 15^{\circ}$ or $ 90^{\circ} - 15^{\circ} = 75^{\circ}$

Hence the required line is in the positive direction of either $15^{\circ}$ or $75^{\circ}$ to the $x$ axis.

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