• # question_answer A line through $A(-5,-\ 4)$ meets the lines $x+3y+2=0,$ $2x+y+4=0$ and $x-y-5=0$ 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 the line is                                                    [IIT 1993]  A)            $2x+3y+22=0$                          B)            $5x-4y+7=0$ C)            $3x-2y+3=0$                            D)            None of these

$\frac{x+5}{\cos \theta }=\frac{y+4}{\sin \theta }=\frac{{{r}_{1}}}{AB}=\frac{{{r}_{2}}}{AC}=\frac{{{r}_{3}}}{AD}$            $({{r}_{1}}\cos \theta -5,{{r}_{1}}\sin \theta -4)$ lies on $x+3y+2=0$.            \${{r}_{1}}=\frac{15}{\cos \theta +3\sin \theta }$            Similarly $\frac{10}{AC}=2\cos \theta +\sin \theta$ and $\frac{6}{AD}=\cos \theta -\sin \theta$            Putting in the given relation, we get ${{(2\cos \theta +3\sin \theta )}^{2}}=0$            \$\tan \theta =-\frac{2}{3}\Rightarrow y+4=-\frac{2}{3}(x+5)$Þ$2x+3y+22=0.$