Slope of a line passing through
Question:

Slope of a line passing through $\mathrm{P}(2,3)$ and intersecting the line, $x+y=7$ at a distance of 4 units from $\mathrm{P}$, is

1. $\frac{\sqrt{5}-1}{\sqrt{5}+1}$

2. $\frac{1-\sqrt{5}}{1+\sqrt{5}}$

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

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

Correct Option: , 3

Solution:

$x=2+r \cos \theta$

$y=3+r \sin \theta$

$\Rightarrow 2+r \cos \theta+3+r \sin \theta=7$

$\Rightarrow r(\cos \theta+\sin \theta)=2$

$\Rightarrow \sin \theta+\cos \theta=\frac{2}{r}=\frac{2}{\pm 4}=\pm \frac{1}{2}$

$\Rightarrow 1+\sin 2 \theta=\frac{1}{4}$

$\Rightarrow \sin 2 \theta=-\frac{3}{4}$

$\Rightarrow \frac{2 m}{1+m^{2}}=-\frac{3}{4}$

$\Rightarrow 3 m^{2}+8 m+3=0$

$\Rightarrow \mathrm{m}=\frac{-4 \pm \sqrt{7}}{1-7}$

$\frac{1-\sqrt{7}}{1+\sqrt{7}}=\frac{(1-\sqrt{7})^{2}}{1-7}=\frac{8-2 \sqrt{7}}{-6}=\frac{-4+\sqrt{7}}{3}$