Question

The slope of a line passing through P$(2,3)$ and intersecting the line, $x+y=7$ at a distance of 4 units from P, is:

• A. $\frac{\sqrt{5}-1}{\sqrt{5}+1}$
• B. $\frac{1-\sqrt{5}}{1+\sqrt{5}}$
• C. $\frac{\sqrt{7}-1}{\sqrt{7}+1}$
• D. $\frac{1-\sqrt{7}}{1+\sqrt{7}}$

Answer 1

The correct option is D $\frac{1-\sqrt{7}}{1+\sqrt{7}}$

By distance form of line, we know
$\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }=r\Rightarrow x=2+r\cos \theta ,y=3+r\sin \theta$
This point lies on line $x+y=7$
$\Rightarrow 2+r\cos \theta +3+r\sin \theta =7\Rightarrow \cos \theta +\sin \theta =\frac{2}{r}=\pm \frac{2}{4}=\pm \frac{1}{2}$
On squaring, we get
$1+\sin 2\theta =\frac{1}{4}\Rightarrow \sin 2\theta =-\frac{3}{4}\Rightarrow \frac{2\tan \theta }{1+{\tan }^2\theta }=-\frac{3}{4}\Rightarrow \frac{2m}{1+m^2}=-\frac{3}{4}\Rightarrow 3m^2+8m+3=0\Rightarrow m=\frac{-8\pm 2\sqrt{7}}{6}\Rightarrow m=\frac{-4\pm \sqrt{7}}{3}Where,\frac{1-\sqrt{7}}{1+\sqrt{7}}=\frac{(1-\sqrt{7}{)}^2}{1-7}=\frac{8-2\sqrt{7}}{-6}=\frac{-4+\sqrt{7}}{3}$