Question

A line through A (-5, -4) meets the lines x + 3y + 2 = 0, 2x + y + 4 = 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

• 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

$\frac{x+5}{\cos \theta }=\frac{y+4}{\sin \theta }=\frac{r_1}{AB}=\frac{r_2}{AC}=\frac{r_3}{AD}$
$(r_1\cos \theta -5,r_1\sin \theta -4)$ lies on x + 3y + 2 = 0
$r_1=\frac{15}{\cos \theta +3\sin \theta }$
and similarly $\frac{10}{AC}=2\cos \theta +\sin \theta$ and
$\frac{6}{AD}=\cos \theta -\sin \theta$
On putting in the given relation, we get
$(2\cos \theta +3\sin \theta {)}^2=0$
$\therefore \tan \theta =-\frac{2}{3}\Rightarrow y+4=-\frac{2}{3}(x+5)$
$\Rightarrow 2x+3y+22=0$