Question

Angles made with the $x$-axis by two lines drawn through the point $(1,2)$ and cutting the line $x+y=4$ at a distance $\sqrt{\frac{2}{3}}$ 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. $\frac{\pi }{4}$ and $\frac{\pi }{2}$

Answer 1

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

Let $P\equiv (1,2)$
Using the parametric form , the coordinates of the point at a distance of $\sqrt{\frac{2}{3}}$ is given by,
$(1+\sqrt{\frac{2}{3}}\cos \theta ,2+\sqrt{\frac{2}{3}}\sin \theta )$
This point lies on the line $x+y=4$
Hence,
$3+\sqrt{\frac{2}{3}}(\sin \theta +\cos \theta )=4$
On squaring and simplifying, we get
${\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =\frac{3}{2}$
$1+\sin 2\theta =\frac{3}{2}$
$\sin 2\theta =\frac{1}{2}$
The possible values of $\theta$ are $\frac{\pi }{12}$ and $\frac{5\pi }{12}$