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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Mathematics
Chapter
Straight Lines

BITSAT 2021 · Mathematics · Straight Lines

The distance between the parallel lines 3 x-4 y+7=0 and 3 x-4 y+5=0...

The distance between the parallel lines 3x−4y+7=03 x-4 y+7=0 and 3x−4y+5=03 x-4 y+5=0 is ab\frac{a}{b}. Value of a+ba+b is

Options

  1. A2
  2. B5
  3. C7
  4. D3

Correct answer

C. 7

Step-by-step solution

Given parallel lines are 3x−4y+7=03 x-4 y+7=0 and 3x−4y+5=03 x-4 y+5=0 Required distance =∣7−5∣(3)2+(−4)2=25⇒a=2,b=5=\frac{|7-5|}{\sqrt{(3)^{2}+(-4)^{2}}}=\frac{2}{5}\Rightarrow a=2, b=5
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