Question

A line passing through the point $A(-6,5)$ meets the lines $x+3y+2=0$, $2x+y+4=0$ and $x-y-5=0$ at $P,Q$ and $R$ respectively. If ${\left(\frac{11}{AP}\right)}^2+{\left(\frac{16}{AR}\right)}^2={\left(\frac{3}{AQ}\right)}^2$,
then slope of the line can be,

• A. $\pm \frac{\sqrt{2}}{3}$
• B. $\pm 1$
• C. $\pm \frac{\sqrt{3}}{2}$
• D. $\pm \frac{2}{5}$

Answer 1

The correct option is A $\pm \frac{\sqrt{2}}{3}$

$\frac{x+6}{\cos \theta }=\frac{y-5}{\sin \theta }=AP...\left(1\right)$
$\frac{x+6}{\cos \theta }=\frac{y-5}{\sin \theta }=AQ...\left(2\right)$
$\frac{x+6}{\cos \theta }=\frac{y-5}{\sin \theta }=AR...\left(3\right)$

$(AP\cos \theta -6,AP\sin \theta +5)$ lies on line $x+3y+2=0$
$\Rightarrow \frac{-11}{AP}=\cos \theta +3\sin \theta$

Similarly,
$\frac{3}{AQ}=2\cos \theta +\sin \theta$
$\frac{16}{AR}=\cos \theta -\sin \theta$

$\because {\left(\frac{11}{AP}\right)}^2+{\left(\frac{16}{AR}\right)}^2={\left(\frac{3}{AQ}\right)}^2$

$\Rightarrow 2{\cos }^2\theta -9{\sin }^2\theta =0\Rightarrow \tan \theta =\pm \sqrt{\frac{2}{9}}=\pm \frac{\sqrt{2}}{3}$