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Math Help - Linear Relations

  1. #1
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    Linear Relations

    This is a simple question, but for some reason I still can't get it.
    Find the equation of a straight line which asses through the point (2,3) and is inclined at 30 degrees to the positive direction of the x-axis.

    How do I solve it? Also I think you need to find the gradient from the 30 degrees but how do you do that? (with and without a scientific calc)
    When they say "Positive direction of the x-axis, what do they mean by that).

    Thanks in advance
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  2. #2
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    Use the rule \tan (\theta) = m where m is the gradient of the line
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  3. #3
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    Thanks, i've done that and plugged it into the formula "y-y1=a(x-x1), but the answer is root 3y - x = 3 root 3 - 2. I couldn't get that answer, how on earth?
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  4. #4
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    What did you get? Chances are your expression is equivalent to the one they give...
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  5. #5
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    Don't mean to ask but I just typed in tan(30) into my graphics calculator and I got root 3 divided by 3? How did they get this, I think thats why my question is wrong because I got 0.57.

    EDIT - Nevermind, just drew up a triangle. I will post my answer in a second

    EDIT 2 - I got m to equal 1 divided by root 3, do I need to rationalize it? I think thats where I stuffed up, coz I wasn't supposed to rationalize it?
    Last edited by theleontan; June 4th 2011 at 04:41 AM.
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  6. #6
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    Hi theleontan,
    The point slope formula gives y-3/x-2 = 1/rad3 Just put this in the form y=mx+b to solve for the y intercept
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  7. #7
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    Can you "see" that the line crosses x-axis at (2+3sqrt(3), 0) ?
    Then, using similar triangles, crosses y-axis at (0, 3+2/sqrt(3)) ?
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