A line L passing through the point $\mathrm{P}(-5,-4)$ cuts the line $x-y-5=0$ and $x+3 y+2=0$ respectively…
A line L passing through the point $\mathrm{P}(-5,-4)$ cuts the line $x-y-5=0$ and $x+3 y+2=0$ respectively at Q and R such that $\frac{18}{P Q}+\frac{15}{P R}=2$, then slope of the line $L$ is
$\pm 1$
$\pm \frac{1}{\sqrt{3}}$
$\pm \sqrt{3}$
$\pm \frac{2}{\sqrt{3}}$
Solution
Equation of line passing through $\mathrm{P}(-5,-4)$ and making angle $\theta$ with $x$-axis is $\frac{x+5}{\cos \theta}=\frac{y+4}{\sin \theta}=r$ $\Rightarrow x=-5+r \cos \theta, y=-4+r \sin \theta$
Equation of line PQ where Q lies on $x-y-5=0$ is $-5+P Q \cos \theta+4-P Q \sin \theta-5=0$
$\Rightarrow \frac{6}{\mathrm{PQ}}=\cos \theta-\sin \theta$ ...(i)
Equation of line PR where R lies on $x+3 y+2=0$ is $-5+P R \cos \theta-12+3 P R \sin \theta+2=0$
$\Rightarrow \frac{15}{P R}=\cos \theta+3 \sin \theta$ Given condition is $\frac{18}{P Q}+\frac{15}{P R}=2$
Using (i) and (ii)
$\begin{aligned}
& 3 \cos \theta-3 \sin \theta+\cos \theta+3 \sin \theta=2 \\
& \Rightarrow 4 \cos \theta=2 \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=2 n \pi \pm \frac{\pi}{3}
\end{aligned}$ So, slope of the line is $\tan \theta=\tan \left(2 n \pi \pm \frac{\pi}{3}\right)= \pm \sqrt{3}$