Question

A line through $A(-5,-4)$ meets the line $x+3y+2=0,2x+y+4=0$ and $x-y-5=0$ at the points $B,C$ & $D$ respectively, if ${\left(\frac{15}{AB}\right)}^2+{\left(\frac{10}{AC}\right)}^2={\left(\frac{6}{AD}\right)}^2$

Let the equation of the line be $lx+my+n=0$.

Then find $n-m-l-8$.

Answer 1

$AB=r_1,AC=r_2,AD=r_3$

Using parametric form of line, we get

$B\equiv (5+r_1\cos \theta ,4+r_1\sin \theta )$

It lie on line $x+3y+2=0$

$AB=r_1=\left|\frac{-5-12+2}{\cos \theta +3\sin \theta }\right|=\frac{15}{\cos \theta +3\sin \theta }$

Similarly,

$r_2=\frac{10}{2\cos \theta +\sin \theta }$ & $r_3=\frac{6}{\cos \theta -\sin \theta }$

Now,

${\left(\frac{15}{AB}\right)}^2+{\left(\frac{10}{AC}\right)}^2={\left(\frac{6}{AD}\right)}^2$

$(\cos \theta +3\sin \theta {)}^2+(2\cos \theta +\sin \theta {)}^2=(\cos \theta -\sin \theta {)}^2$

$\Rightarrow 4{\cos }^2\theta +9{\sin }^2\theta +12\sin \theta \cos \theta =0$

$\Rightarrow (2\cos \theta +3\sin \theta {)}^3=0$

$\Rightarrow \tan \theta =-\frac{2}{3}$

Equation of line through $A$ is $(y+4)=-\frac{2}{3}(x+5)$

$\Rightarrow 2x+3y+22=0$

$\Rightarrow l=2,m=3,n=22$

$n-m-l=22-2-3=17$