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

• A. 2x + 3y + 22 = 0
• B. 5x - 4y + 7 = 0
• C. 3x - 2y + 3 = 0
• D. None of these

Answer 1

Answer: (D)

The parametric equation of a line through A is
$\frac{x+5}{cos\theta }=\frac{y+4}{sin\theta }=r$
Let $AB=r_1,AC=r_2$ and $AD=r_3$
Then the coordinates of B, C, D are
$\left(-5,+r_{1}cos\theta -4+r_{i}sin\theta \right),i=1,2,3$
Now B lies on the line $x+3y+2=0$
$\therefore -5+r_{i}cos\theta +3\left(-4+r_{1}sin\theta \right)+2=0$
$\frac{15}{r_1}=cos\theta +3sin\theta$
$C$ lies on $2x+y+4=0$
$\therefore 2\left(-5+r_{2}cos\theta \right)+\left(-4+r_{2}sin\theta \right)+4=0$
$\Rightarrow \frac{10}{r_2}=2cos\theta +sin\theta$
$D$ lies on $x-y-5=0$
$\therefore -5+r_{3}cos\theta +4-r_{3}sin\theta -5=0\Rightarrow \frac{6}{r_3}=cos\theta -sin\theta .$
From the given condition ${\left(\frac{15}{AB}\right)}^2+{\left(\frac{10}{AC}\right)}^2={\left(\frac{6}{AD}\right)}^2$
we get, ${\left(cos\theta +3sin\theta \right)}^2+{\left(2cos\theta +sin\theta \right)}^2={\left(cos\theta -sin\theta \right)}^2$
$\Rightarrow {\left(2cos\theta +3sin\theta \right)}^2=0\Rightarrow tan\theta =-\frac{2}{3}$
$\therefore$ Equation of the line is
$y+4=-\frac{2}{3}\left(x+5\right)\Rightarrow 2x+3y+22=0$