SATHEE: Straight Line Question 300

Straight Line Question 300

Question: The equation of the lines which passes through the point (3, - 2) and are inclined at $ 60^{o} $ to the line $ \sqrt{3}x+y=1 $ [IIT 1974; MP PET 1996]

Options:

A) $ y+2=0,\ \ \sqrt{3}x-y-2-3\sqrt{3}=0 $

B) $ x-2=0,\ \ \sqrt{3}x-y+2+3\sqrt{3}=0 $

C) $ \sqrt{3}x-y-2-3\sqrt{3}=0 $

D)None of these

Show Answer

Answer:

Correct Answer: A

Solution:

The equation of any straight line passing through(3, ?2) is $ y+2=m(x-3) $?..(i) The slope of the given line is $ -\sqrt{3} $ . So, $ \tan 60^{o}=\pm \frac{m-(-\sqrt{3})}{1+m(-\sqrt{3})} $ On solving, we get $ m=0 $ or $ \sqrt{3} $ Putting the values of m in (i), the required equation of lines are $ y+2=0 $ and $ \sqrt{3}x-y=2+3\sqrt{3} $ .

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Straight Line Question 300

Question: The equation of the lines which passes through the point (3, - 2) and are inclined at $ 60^{o} $ to the line $ \sqrt{3}x+y=1 $ [IIT 1974; MP PET 1996]

Options:

Answer:

Solution:

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