Question

A line through $A(-5,-4)$ meets the lines $x+3y+2=0,2x+y+4=0$ and $x-y-5=0$ at the points $B,C$ and $D$ respectively. If $(15/AB{)}^2+(10/AC{)}^2=(6/AD{)}^2$, find the equation of the line.

Answer 1

Any line through $A(-5,-4)$ is
$y+4=\tan \theta (x+5)$
or $\frac{x+5}{\cos \theta }=\frac{y+4}{\sin \theta }=r_1=r_2=r_3$
for $B$ $C$ $D$
$B,C,D$ lie on the given lines respectively, where
$r_1={AB,r}_2=AC=r_3=AD$.
$r_1=-\frac{a{x}_1+b{y}_1+c}{a\cos \theta +b\sin \theta }$
for $B$ which lies on $x+3y+2=0$
$AB=\frac{1(-5)+3(-4)+2}{\cos \theta +3\sin \theta }=\frac{15}{\cos \theta +3\sin \theta }$
or $\frac{15}{AB}=\cos \theta +3\sin \theta$.
Similarly $\frac{10}{AC}=2\cos \theta +\sin \theta$
and $\frac{6}{AD}=\cos \theta -\sin \theta$.
Hence from the given relation
${(\cos \theta +3\sin \theta )}^2+{(2\cos \theta +\sin \theta )}^2={(\cos \theta -\sin \theta )}^2$
or $4{\cos }^2\theta +12\sin \theta \cos \theta +9{\sin }^2\theta =0$
or ${(2\cos \theta +3\sin \theta )}^2=0\therefore \tan \theta =-2/3$
Hence from $\left(1\right)$, the equation of the line through $A$ is
$y+4=-\frac{2}{3}(x+5)$
or $2x+3y+22=0$.