Question

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 $\frac{\sqrt{6}}{3}$ from the given point ?

Answer 1

Equation is,

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

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

$\left(\frac{\sqrt{6}\cos \theta }{3}+1,\frac{\sqrt{6}\sin \theta }{3}+2\right)$

$x+y=4$

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

$\sqrt{6}\cos \theta +\sqrt{6}\sin \theta =3$

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

${\left(\cos \theta +\sin \theta \right)}^2=3/2$

${\cos }^2\theta +{\sin }^2\theta +2\sin \theta \cos \theta =3/2$

$2\sin \theta \cos \theta =1/2$

${\sin }^2\theta =\frac{1}{2}$

$2\theta =\pi /6\text{or}{30}^{\circ }$

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

Required line is the positive direction of either 15^o or 75^o to the x-axis.