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 $x + 3y + 2 = 0$
$\therefore -5+r_{i}\,cos\,\theta+3\left(-4+r_{1}\,sin\,\theta \right)+2=0$
$\frac{15}{r_{1}}=cos\,\theta +3\,sin\,\theta$
$C$ lies on $2x + y + 4 = 0$
$\therefore 2\left(-5+r_{2}\,cos\,\theta \right)+\left(-4+r_{2}\,sin\,\theta \right)+4=0$
$\Rightarrow \frac{10}{r_{2}}=2\,cos\,\theta +sin\,\theta$
$D$ 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, $\left(cos\,\theta +3\,sin\,\theta\right)^{2}+\left(2\,cos\,\theta+sin\,\theta\right)^{2}=\left(cos\,\theta-sin\,\theta\right)^{2}$
$\Rightarrow \left(2\,cos\,\theta+3\,sin\,\theta\right)^{2}=0 \Rightarrow tan\,\theta=-\frac{2}{3}$
$\therefore $ Equation of the line is
$y+4=-\frac{2}{3}\left(x+5\right) \Rightarrow 2x+3y+22=0$