Question

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. $\frac{1-\sqrt{7}}{1+\sqrt{7}}$
• D. $\frac{\sqrt{7}-1}{\sqrt{7}+1}$

Answer 1

Answer: (C)

$\frac{1-\sqrt{7}}{1+\sqrt{7}}$

$x=2+r\cos \theta$
$y=3+r\sin \theta$
$\Rightarrow 2+r\cos \theta +3+r\sin \theta =7$
$\Rightarrow r(\cos \theta +\sin \theta )=2$
$\Rightarrow \sin \theta +\cos \theta =\frac{2}{r}=\frac{2}{\pm 4}=\pm \frac{1}{2}$
$\Rightarrow 1+\sin 2\theta =\frac{1}{4}$
$\Rightarrow \sin 2\theta =-\frac{3}{4}$
$\Rightarrow \frac{2m}{1+m^2}=-\frac{3}{4}$
$\Rightarrow 3m^2+8m+3=0$
$\Rightarrow m=\frac{+4\pm \sqrt{7}}{1-7}$
$\frac{1-\sqrt{7}}{1+\sqrt{7}}=\frac{(1-\sqrt{7}{)}^2}{1-7}=\frac{8-2\sqrt{7}}{-6}=\frac{-4+\sqrt{7}}{3}$.