Straight Line Question 109

Question: Distance between the parallel lines $ 3x+4y+7=0 $ and $ 3x+4y-5=0 $ is [RPET 2003]

Options:

A) $ \frac{2}{5} $

B) $ \frac{12}{5} $

C) $ \frac{5}{12} $

D) $ \frac{3}{5} $

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ d_1= $ distance of perpendicular from $ (0,0) $ to $ 3x+4y+7=0 $ $ d_1=\frac{3\times 0+4\times 0+7}{\sqrt{3^{2}+4^{2}}}=\frac{7}{5} $ $ d_2=\frac{3\times 0+4\times 0+(-5)}{\sqrt{3^{2}+4^{2}}}=\frac{-5}{5} $ \ Required distance $ =| d_1-d_2 | $ $ =| \frac{7}{5}-( \frac{-5}{5} ) |=\frac{12}{5}. $

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