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 $P$, is:
Correct Option: , 2
Solution:
Since point at 4 units from $\mathrm{P}(2,3)$ will be
$\mathrm{A}(4 \cos \theta+2,4 \sin (\theta+3)$ and this point will satisfy
the equation of line $x+y=7$
$\Rightarrow \cos \theta+\sin \theta=\frac{1}{2}$
On squaring
$\Rightarrow \sin 2 \theta-\frac{3}{4} \Rightarrow \frac{2 \tan \theta}{1+\tan ^{2} \theta}=-\frac{3}{4}$
$\Rightarrow 3 \tan ^{2} \theta+8 \tan \theta+3=0$
$\Rightarrow \tan \theta=\frac{-8 \pm 2 \sqrt{7}}{6} \quad($ ignoring $-$ ve sign $)$
$\Rightarrow \tan \theta=\frac{-8+2 \sqrt{7}}{6}=\frac{1-\sqrt{7}}{1+\sqrt{7}}$