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# In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 is at a distance $\large\frac{\sqrt 6}{3}$ from the given point

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

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A)
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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.