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)

$x = 2 + rcos \theta$
$y = 3 + rsin \theta$
$\Rightarrow $ $2 + r cos \theta + 3 + rsin \theta = 7$
$\Rightarrow $ $r(cos\theta + sin \theta$) = 2
$\Rightarrow \, \, sin\theta + cos \theta = \frac{2}{r} = \frac{2}{\pm 4} = \pm \frac{1}{2}$
$\Rightarrow \, \, 1 + sin2 \theta \, = \frac{1}{4}$
$\Rightarrow \, \, \, sin2\theta = -\frac{3}{4}$
$\Rightarrow \, \frac{2m}{1+m^2} = -\frac{3}{4}$
$\Rightarrow \, 3m^2 + 8m + 3 =0$
$\Rightarrow \, \, m = \, \frac{-4 \pm \sqrt{7}}{1-7}$
$\frac{1-\sqrt{7}}{1+\sqrt{7}} = \frac{(1-\sqrt{7})^2}{1-7} = \frac{8-2\sqrt{7}}{-6} = \frac{-4 + \sqrt{7}}{3}$