Question

Through the point $P(4,1)$ a line is drawn to meet the line $3x-y=0$ at $Q$ where $PQ=11/\left(2\sqrt{2}\right)$. Determine the equation of the line.

Answer 1

$\frac{x-4}{\cos \theta }=\frac{y-1}{\sin \theta }=r=\frac{11}{2\sqrt{2}}$
It meets the line $3x-y=0$
$\therefore \frac{1}{r}=\frac{3\cos \theta \sin \theta }{3.4-1}=\frac{2\sqrt{2}}{11}$
$\sin \theta =(2\sqrt{2}+3\cos \theta )$
$1-{\cos }^2\theta =8+12\sqrt{2}\cos \theta +9{\cos }^2\theta$
or $10-{\cos }^2\theta +12\sqrt{2}\cos \theta +7=0$
$\therefore \cos \theta =\frac{-12\sqrt{2}\pm \surd (288-280)}{20}$
$\cos \theta =-\frac{1}{\sqrt{2}}$ or $-\frac{7\sqrt{2}}{10}$ then from $\left(1\right)$
$\sin \theta =\frac{1}{\sqrt{2}}$ or $-\frac{\sqrt{2}}{10}$
$\therefore \tan \theta =-1,\frac{1}{7}$
$\therefore$ Lines are $x+y=5$ and $x-7y+3=0$