SATHEE: Straight Line Question 89

Straight Line Question 89

Question: The distance between the lines $ 3x+4y=9 $ and $ 6x+8y=15 $ is [MNR 1982; RPET 1995; MP PET 2002]

Options:

A)3/2

B)3/10

C)6

D)None of these

Show Answer

Answer:

Correct Answer: B

Solution:

Here the lines are, $ 3x+4y-9=0 $ ……(i) and $ 6x+8y-15=0 $ ……(ii) Now distance from origin of both the lines are $ \frac{-9}{\sqrt{3^{2}+4^{2}}}=-\frac{9}{5} $ and $ \frac{-15}{\sqrt{6^{2}+8^{2}}}=-\frac{15}{10} $ Hence distance between both the lines are $ | -\frac{9}{5}-( -\frac{15}{10} ) |=\frac{3}{10} $ . Ailter: Put $ y=0 $ in the first equation, we get $ x=3 $ therefore, the point (3, 0) lies on it. So the required distance between these two lines is the perpendicular length of the line $ 6x+8y=15 $ from the point (3, 0). i.e., $ \frac{6\times 3-15}{\sqrt{6^{2}+8^{2}}}=\frac{3}{10} $ .

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Straight Line Question 89

Question: The distance between the lines $ 3x+4y=9 $ and $ 6x+8y=15 $ is [MNR 1982; RPET 1995; MP PET 2002]

Options:

Answer:

Solution:

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