Question

Inclination of the lines drawn through the point $(1,2)$ and cutting the line $x+y=4$ at a distance of $\frac{\sqrt{6}}{3}$ units from the point $(1,2)$ are

• A. $\frac{\pi }{6}$ and $\frac{\pi }{3}$
• B. $\frac{\pi }{8}$ and $\frac{3\pi }{8}$
• C. $\frac{\pi }{12}$ and $\frac{5\pi }{12}$
• D. none of these

Answer 1

The correct option is C $\frac{\pi }{12}$ and $\frac{5\pi }{12}$

Suppose a line passing through $(1,2)$ makes an angle $\theta$ with $x-$axis. Then, its equation is
$\frac{x-1}{\cos \theta }=\frac{y-2}{\sin \theta }=r$
The coordinates of a point on this straight line at a distance $\frac{\sqrt{6}}{3}=\sqrt{\frac{2}{3}}$ from $(1,2)$ are given by
$x=1+\sqrt{\frac{2}{3}}\cos \theta ,y=2+\sqrt{\frac{2}{3}}\sin \theta$
$\because$ point lies on $x+y=4$
$\therefore 1+\sqrt{\frac{2}{3}}\cos \theta +2+\sqrt{\frac{2}{3}}\sin \theta =4$
$\Rightarrow \cos \theta +\sin \theta =\sqrt{\frac{3}{2}}$
$\Rightarrow \sin (\theta +\frac{\pi }{4})=\frac{\sqrt{3}}{2}$
$\Rightarrow \theta +\frac{\pi }{4}=\frac{\pi }{3},\frac{2\pi }{3}$
$\Rightarrow \theta =\frac{\pi }{12},\frac{5\pi }{12}$