Question

Find the direction in which a straight line must be drawn through the point $(1,2)$, so that its point of intersection with the line $x+y=4$ is at a distance $\sqrt{\frac{2}{3}}$ from the given point.

Answer 1

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

Hence its equation is

$\frac{x-1}{\cos \theta }=\frac{y-2}{\sin \theta }=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}\therefore x=\frac{\sqrt{6}\cos \theta }{3}+1$ and $y=\frac{\sqrt{6}\sin \theta }{3}+2or,(\frac{\sqrt{6}\cos \theta }{3}+1,\frac{\sqrt{6}\sin \theta }{3}+2)$

This point lies on the line $x+y=4\therefore (\frac{\sqrt{6}\cos \theta }{3}+1\frac{\sqrt{6}\sin \theta }{3}+2)=4\Rightarrow \frac{\sqrt{6}\cos \theta }{3}+\frac{\sqrt{6}\sin \theta }{3}=1\Rightarrow \sqrt{6}\cos \theta +\sqrt{6}\sin \theta =3\Rightarrow \cos \theta +\sin \theta =\frac{3}{\sqrt{6}}=\sqrt{\frac{3}{2}}$

Squaring on both sides:

$(\cos \theta +\sin \theta {)}^2=\frac{3}{2}\Rightarrow {\cos }^2\theta +\sin \theta +2\sin \theta \cos \theta =\frac{3}{2}\Rightarrow 2\sin \theta \cos \theta =\frac{3}{2}-1\Rightarrow \sin 2\theta =\frac{1}{2}\Rightarrow 2\theta =\frac{\pi }{6}or,30°\therefore \theta =15°or,90°-15°=75°$