Question

A line passing through point A (-5, -4) meet other three 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 line is

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

Answer 1

The correct option is A 2x + 3y + 22 = 0

$(ABcos\theta -5,ABsin\theta -4)$ lie on the line x + 3y + 2 = 0
$\Rightarrow ABcostheta-5+3(ABsin\theta -4)+2=0\Rightarrow cos\theta +3sin\theta =\frac{15}{AB}\underset{–}{}\left(1\right)$
$(ACcos\theta -5,ACsin\theta -4)$ lie on the line 2x + y + 4 = 0 $\Rightarrow 2cos\theta +sin\theta =\frac{10}{AC}\underset{–}{}\left(2\right)$
$(ADcos\theta -5,ADsin\theta -4)$ lie on the line x - y - 5 =0 $\Rightarrow cos\theta -sin\theta =\frac{6}{AD}\underset{–}{}\left(3\right)$
${\left(\frac{15}{AB}\right)}^2+{\left(\frac{10}{AC}\right)}^2={\left(\frac{6}{AD}\right)}^2\Rightarrow Tan\theta =\frac{-2}{3}=m\Rightarrow y+4=-\frac{2}{3}(x+5)\Rightarrow 2x+3y+22=0$