Question

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

• A. 23x + 14y - 40 = 0
• B. 5x - 4y + 7 = 0
• C. 3x - 2y + 3 = 0
• D. None of these

Answer 1

The correct option is A 23x + 14y - 40 = 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
$\therefore r_1=\frac{15}{cos\theta +3sin\theta }$
Similarly $\frac{10}{AC}=2cos\theta +sin\theta and\frac{6}{AD}=cos\theta -sin\theta$
Putting in the given relation, we get $(2cos\theta +3sin\theta {)}^2=0$
$\therefore tan\theta =-\frac{2}{3}\Rightarrow y+4=-\frac{2}{3}(x+5)\Rightarrow 2x+3y+22=0$.