Question

A line passing through $A(-5,-4)$ meets the line $x+3y+2=0,2x+y+4=0$ and $x-5-y=0$ at $B,C$ and $D$ respectively. If ${\left(\frac{15}{AB}\right)}^2+{\left(\frac{10}{AC}\right)}^2={\left(\frac{6}{AD}\right)}^2$, then the equation of line is

• A. $2x+3y+22=0$
• B. $5x+4y+7=0$
• C. $3x-2y+3=0$
• D. None of these

Answer 1

The correct option is A $2x+3y+22=0$

Equation of line passing through $A(-5,-4)$ and making an angle $\theta$ with $x-$ axis is
$\frac{x+5}{\cos \theta }=\frac{y+4}{\sin \theta }=r\Rightarrow x=-5+r\cos \theta ,y=-4+r\sin \theta$
So, parametric coordinates of B is $x=-5+AB\cos \theta ,y=-4+AB\sin \theta$
Line $AB$,
$x+3y+2=0\Rightarrow -5+AB\cos \theta +3(-4+AB\sin \theta )+2=0\Rightarrow \frac{15}{AB}=\cos \theta +3\sin \theta \cdots \left(1\right)$

similarly for line $AC$,
$2x+y+4=0\Rightarrow 2(-5+AC\cos \theta )+(-4+AC\sin \theta )+4=0\Rightarrow \frac{10}{AC}=2\cos \theta +\sin \theta \cdots \left(2\right)$

for line $AD$,
$x-5-y=0\Rightarrow (-5+AD\cos \theta )-5-(-4+AD\sin \theta )=0\Rightarrow \frac{6}{AD}=\cos \theta -\sin \theta \cdots \left(3\right)$

${\left(\frac{15}{AB}\right)}^2+{\left(\frac{10}{AC}\right)}^2={\left(\frac{6}{AD}\right)}^2$
Using equation $\left(1\right),\left(2\right)$ and $\left(3\right)$,
$\Rightarrow {\cos }^2\theta +9{\sin }^2\theta +6\sin \theta \cos \theta +4{\cos }^2\theta +{\sin }^2\theta +4\sin \theta \cos \theta ={\cos }^2\theta +{\sin }^2\theta -2\sin \theta \cos \theta \Rightarrow (2\cos \theta +3\sin \theta {)}^2=0\Rightarrow 2\cos \theta +3\sin \theta =0\Rightarrow \tan \theta =-\frac{2}{3}$

Therefore the equation of line is
$y+4=-\frac{2}{3}(x+5)\Rightarrow 3y+12=-2x-10\therefore 2x+3y+22=0$