Question

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

• A. $2x+3y+22=0$
• B. $5x-4y+7=0$
• C. $3x-2y+3=0$
• D. $2x+3y+7=0$
  

Answer 1

Answer: (A)

$2x+3y+22=0$

The parametric form of equation of line passing through A(-5,-4)
$\frac{x+5}{\cos \theta }=\frac{y+4}{\sin \theta }=r$
Let $AB=r_1$
Then the coordinates of B are $(r_1\cos \theta -5,r_1\sin \theta -4)$
Since, point B lies on the line $x+3y+2=0$
$\Rightarrow r_1(3\sin \theta +\cos \theta )=15$
$\Rightarrow \frac{15}{r_1}=3\sin \theta +\cos \theta$
Let $AC=r_2$
Coordinates of C are $(r_2\cos \theta -5,r_2\sin \theta -4)$
Since, point C lies on the line $2x+y+4=0$
$\Rightarrow r_2(\sin \theta +2\cos \theta )=10$
$\Rightarrow \frac{10}{r_2}=\sin \theta +2\cos \theta$
Let $AD=r_3$
Coordinates of D are $(r_3\cos \theta -5,r_3\sin \theta -4)$
Since, point D lies on the line $x-y-5=0$
$\Rightarrow r_3(-\sin \theta +\cos \theta )=6$
$\Rightarrow \frac{6}{r_3}=-\sin \theta +\cos \theta$
$(\frac{15}{AB}{)}^2+(\frac{10}{AC}{)}^2=(\frac{6}{AD}{)}^2$
$(\frac{15}{r_1}{)}^2+(\frac{10}{r_2}{)}^2=(\frac{6}{r_3}{)}^2$
$\Rightarrow (3\sin \theta +\cos \theta {)}^2+(\sin \theta +2\cos \theta {)}^2=(-\sin \theta +\cos \theta {)}^2$
$\Rightarrow (3\sin \theta +2\cos \theta {)}^2=0$
$\Rightarrow \tan \theta =-\frac{2}{3}$
So, the equation of required line is
$y+4=-\frac{2}{3}(x+5)$
$\Rightarrow 2x+3y+22=0$