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Math Help - Find the eq of st line

  1. #1
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    Find the eq of st line

    Find the equation of st. line passing through the point (-2,-3) and inclined at 60 degrees to the line x+ \sqrt{2} y = 2
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    Dear TomJerry,

    1)Draw the given line and the point as a rough sketch.

    2)As the given line makes an ange of 45 degrees with the x axis I think by drawing the line and the point you will be able to think of the tangent of the line that you have to find.

    But if you need futher assistance please don't hesitate to reply back.

    Hope this helps.
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    I have already done plotting the graph and only after that when i wasnt able to device the eq had posted this question on the forum to u guys......

    My work ..............what i have done till now

    Eq of a line (y-y_0) = m (x-x_0)

    m is the slope

    m = tan^{-1} \left(\frac{\pi}{3}\right)

    m = \sqrt{3}

    Eq of line is x+\sqrt{3} = 2 \Rightarrow y = \frac{1}{\sqrt{3}}(2-x) here m = \frac{1}{\sqrt{3}}

    tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|

    \sqrt{3} = \left| \frac{m_1 - \frac{1}{\sqrt{3}}}{1 + m_1 \frac{1}{\sqrt{3}}} \right|

    squaring both the terms
    m^2 = \sqrt{\frac{5}{3}}

    equation of line is
    y+3 = \sqrt{\frac{5}{3}} (x+2) ......................Is this correct ???????????
    Last edited by mr fantastic; January 5th 2010 at 06:24 AM.
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    Dear TomJerry,

    Please reconsider the value you have taken to the slope (m) of the line. According to the question relative to what the slope is given?
    Yes you are correct that the question asks only to find the equation not to draw the line. But on the contart since straight lines could be easily drawn and since the questionis about straight lines if you know how to sketch a straight line it would be helpful.
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    Quote Originally Posted by Sudharaka View Post
    Dear TomJerry,

    Please reconsider the value you have taken to the slope (m) of the line. According to the question relative to what the slope is given?
    Yes you are correct that the question asks only to find the equation not to draw the line. But on the contart since straight lines could be easily drawn and since the questionis about straight lines if you know how to sketch a straight line it would be helpful.
    thanks but could u please tell me what is wrng with my result .so that i can rectify it ......
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  6. #6
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    Angle between two lines in coordinate geometry

    Hello TomJerry
    Quote Originally Posted by TomJerry View Post
    Find the equation of st. line passing through the point (-2,-3) and inclined at 60 degrees to the line x+ \sqrt{2} y = 2
    There seems to be a bit of confusion here, so let me see if I can clear it up. Here's a very useful result:
    If line 1 makes an angle \theta with the positive direction of Ox and line 2 makes an angle \phi with the positive direction of Ox, then the angle between line 1 and line 2 is |\phi-\theta|.
    Suppose that the line we've been given makes an angle \theta with the positive direction of Ox, and the line we want makes an angle \phi with Ox. Then, since the angle between these two line is 60^o:
    |\phi-\theta| = 60^o
    In other words:
    \phi-\theta = \pm 60^o
    Note that we can choose either the + sign or the - sign. Which one we take will determine whether the 60^o angle is measured clockwise or anticlockwise from the first line. There will therefore be two possible values of \phi, and two possible lines.

    Next, another fundamental result:
    If a line makes an angle \theta with the positive direction of Ox, its gradient is \tan \theta.
    The line x+\sqrt2 y =2 has a gradient of -\frac{1}{\sqrt 2} (do you agree?), so:
    \tan\theta = -\frac{1}{\sqrt 2}
    (By the way, this means that the line makes an angle of about 35^o below the x-axis, not 45^o.)

    So:
    \phi-\theta = \pm60^o

    \Rightarrow \tan(\phi-\theta) = \pm \sqrt 3

    \Rightarrow \frac{\tan\phi - \tan\theta}{1 + \tan\phi\tan\theta}=\pm\sqrt3

    \Rightarrow \frac{\tan\phi +\dfrac{1}{\sqrt 2}}{1-\dfrac{1}{\sqrt 2}\tan\phi}=\pm\sqrt3

    \Rightarrow \frac{\sqrt2\tan\phi+1}{\sqrt2-\tan\phi}=\pm\sqrt3

    \Rightarrow \sqrt2\tan\phi+1=\pm\sqrt3(\sqrt2-\tan\phi)

    \Rightarrow \tan\phi(\sqrt2\pm\sqrt3) = \pm\sqrt6-1

    \Rightarrow \tan\phi = \frac{\sqrt6-1}{\sqrt2+\sqrt3} or \tan\phi = \frac{\sqrt6+1}{\sqrt3-\sqrt2}
    So there are the two gradients that you need. You can now use the equation y - y_1=m(x-x_1) to get the equations of the two possible lines.

    (Note: you can check the working here if you like. \theta = - 35.26^o. These two values of \tan\phi give \phi = 24.74^o and 84.74^o. The first value gives \phi-\theta = 60^o; the second gives \phi-\theta = 120^o, which is, of course, an angle of 60^o measured in the other direction.)

    Grandad
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