Question

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

• A. $x+y=5,x-7y+3=0$
• B. $x-y=5,x-7y+3=0$
• C. $x+y=5,x+7y+3=0$
• D. $x-y=5,x+7y+3=0$

Answer 1

Answer: (A)

By parametric form $Q\left(4+\frac{11}{2\sqrt{2}}\cos \theta ,1+\frac{11}{2\sqrt{2}}\sin \theta \right)$
it lies on $3x-y=0\Rightarrow 12+\frac{33}{2\sqrt{2}}\cos \theta -1-\frac{11}{2\sqrt{2}}\sin \theta =0$
$\Rightarrow 1+\frac{3}{2\sqrt{2}}\cos \theta -\frac{\sin \theta }{2\sqrt{2}}=0\Rightarrow 3\cos \theta -\sin \theta =-2\sqrt{2}$
squaring both sides
$9{\cos }^2\theta +{\sin }^2\theta -6\sin \theta \cos \theta =8\left({\sin }^2\theta +{\cos }^2\theta \right)\Rightarrow {\cos }^2\theta -6\sin \theta \cos \theta -7{\sin }^2\theta =0$
$7{\tan }^2\theta +6\tan \theta -1=0\Rightarrow \tan \theta =-1,\frac{1}{7}.$
Hence required line are $x+y=5,x-7y+3=0$