Straight Line Question 61

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 $ {{( \frac{15}{AB} )}^{2}}+{{( \frac{10}{AC} )}^{2}}={{( \frac{6}{AD} )}^{2}}, $ then the equation of the line is [IIT 1993]

Options:

A) $ 2x+3y+22=0 $

B) $ 5x-4y+7=0 $

C) $ 3x-2y+3=0 $

D)None of these

Show Answer

Answer:

Correct Answer: A

Solution:

  • $ \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 $ .\ $ r_1=\frac{15}{\cos \theta +3\sin \theta } $ Similarly $ \frac{10}{AC}=2\cos \theta +\sin \theta $ and $ \frac{6}{AD}=\cos \theta -\sin \theta $ Putting in the given relation, we get $ {{(2\cos \theta +3\sin \theta )}^{2}}=0 $ \ $ \tan \theta =-\frac{2}{3}\Rightarrow y+4=-\frac{2}{3}(x+5) $ Þ $ 2x+3y+22=0. $