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 point $B,C$ and $D$ respectively. If ${\left(\frac{15}{AB}\right)}^2+{\left(\frac{10}{AC}\right)}^2={\left(\frac{6}{AD}\right)}^2$. Find the equation of the line.

Answer 1

Given that, Aline through A(-5,4) meets the lines $x+3y+2=0$, $2x+y-5=0$ at the point B,C and D respectively. If ${\left(\frac{15}{AB}\right)}^2+{\left(\frac{10}{AC}\right)}^2={\left(\frac{6}{AD}\right)}^2$.

We know that,

$\frac{x+5}{\cos \theta }=\frac{y+4}{\sin \theta }=\frac{r_1}{AB}=\frac{r_2}{AC}=\frac{r_3}{AD}$

$(r_1\cos \theta -5,\sin \theta -4)$ lies on x+3y+2=0

$\therefore r_1=\frac{15}{\cos \theta +3\sin \theta }$

And similarly, $\frac{10}{AC}=2\cos \theta +\sin \theta$

And $\frac{6}{AD}=\cos \theta -\sin \theta$

Now, put these value in given relation, we get

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

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

$y+4=-\frac{2}{3}(x+5)$

$2x+3y+7=0$

Hence, this is the answer.