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 July 3rd, 2013, 12:23 AM #1 Newbie   Joined: Jul 2013 Posts: 1 Thanks: 0 straight lines Find the equation of straight lines which passes through the point (1,2) and makes an angle theta with the positive direction of the x-axis, where cos (theta) = -1/3. Last edited by skipjack; September 4th, 2015 at 02:59 AM.
 July 3rd, 2013, 12:31 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: straight lines What is the relationship between the angle of inclination and the slope of the line?
 September 4th, 2015, 02:58 AM #3 Senior Member   Joined: Sep 2015 From: 4th Dimension Posts: 146 Thanks: 13 Math Focus: Everything (a little bit) A =(theta) Cos A = -1/3 Therefore Tan A = 8^(1/2)=m (slope of line ) (Y-2/x-1)=m=8^(1/2) Hence, 2(2^(1/2))x - y - 2(2^(1/2)) + 2 =0
 September 4th, 2015, 03:03 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,927 Thanks: 2205 As the question doesn't say otherwise, tan(θ) can be positive or negative.
September 4th, 2015, 06:29 AM   #5
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Quote:
 Originally Posted by skipjack As the question doesn't say otherwise, tan(θ) can be positive or negative.

The slope ranges from [0 ,pi )
Now as cos theta is -ve it lies from ( 90° , 180°)
Therefore tan theta is also -ve

September 4th, 2015, 06:35 AM   #6
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Quote:
 Originally Posted by Yash Malik A =(theta) Cos A = -1/3 Therefore Tan A = 8^(1/2)=m (slope of line ) (Y-2/x-1)=m=8^(1/2) Hence, 2(2^(1/2))x - y - 2(2^(1/2)) + 2 =0
Sorry
Tan theta is -ve
So eqn shd be
2(2^(1/2))x + y - 2(2^(1/2)) - 2 =0

Btw thnx fr reminding me

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