Question: The ratio in which the line $ 3x+4y+2=0 $ divides the distance between $ 3x+4y+5=0 $ and $ 3x+4y-5=0 $ , is
Options:
A) $ 7:3 $
B)3 : 7
C) $ 2:3 $
D)None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Lines $ 3x+4y+2=0 $ and $ 3x+4y+5=0 $ are on the same side of the origin. The distance between these lines is $ d_1=| \frac{2-5}{\sqrt{3^{2}+4^{2}}} |=\frac{3}{5} $ . Lines $ 3x+4y+2=0 $ and $ 3x+4y-5=0 $ are on the opposite sides of the origin. The distance between these lines is $ d_2=| \frac{2+5}{\sqrt{3^{2}+4^{2}}} |=\frac{7}{5} $ . Thus $ 3x+4y+2=0 $ divides the distance between $ 3x+4y+5=0 $ and $ 3x+4y-5=0 $ in the ratio $ d_1:d_2 $ i.e., $ 3:7 $ .