Straight Line Question 101

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{5-2}{\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_2:d_1 $ i.e., $ 7:3 $ .



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