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Math Help - equation of a line - help

  1. #1
    chg
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    equation of a line - help

    Hi,

    I'm stuck on a problem and I need help... I'm supposed to find the general equation of a line (Ax+By+C=0) from two points. I did a couple of these problems and everything went well but not for this one... so here goes:

    points: (-sqrt(2)/2, sqrt(2)/2), (1/2, sqrt(3)/2)

    We are given the answer but I can't figure out how to get there... Here is the answer:

    (sqrt(3) - sqrt(2))x - (1 + sqrt(2))y + (sqrt(2) + sqrt(6)) / 2 = 0

    Anyone can help me with the steps to get to this answer?

    This is my first post so go easy on me if there is a better way to format equations (just tell me how to).

    Thanks!
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  2. #2
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    Re: equation of a line - help

    Hi

    You have these points: \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right); \left(\frac{1}{2},\;\frac{}{\sqrt{3}}{2}\right)

    The slope of the line is: \mathrm{m}=\dfrac{\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}}{\frac{1}{2}+\frac{\sqrt{2}}{2}  }

    Now consider a general point \mathrm{(x,\;y)} of this line, which must have the same slope:

    \mathrm{m=\dfrac{\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}}{\frac{1}{2}+\frac{\sqrt{2}}{2}  }=\dfrac{y-\frac{\sqrt{2}}{2}}{x+\frac{\sqrt{2}}{2}}}

    Now you got it

    Greetings
    Last edited by darthjavier; November 13th 2012 at 05:47 AM.
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  3. #3
    chg
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    Re: equation of a line - help

    Hi and thanks for your answer.

    I'm pretty new to maths so what might seem obvious for you is still a mystery for me

    I did understand the slope formula but I can't see why the y difference (sqrt(3) - sqrt(2)) transfers to the A variable in Ax+By+C=0 (same thing for the x difference that becomes B...) Also, how you get the C variable from these differences is beyond me.

    Sorry if this is too much explanation to have me understand something simple but I just don't get it

    Thanks
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  4. #4
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    Re: equation of a line - help

    \mathrm{m} is the slope

    \mathrm{m=\dfrac{\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}}{\frac{1}{2}+\frac{\sqrt{2}}{2}  }=\dfrac{y-\frac{\sqrt{2}}{2}}{x+\frac{\sqrt{2}}{2}}}

    or

    \mathrm{m=\dfrac{\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}}{\frac{1}{2}+\frac{\sqrt{2}}{2}  }=\dfrac{y-\frac{\sqrt{3}}{2}}{x-\frac{1}{2}}}

    Now multiply in each side

    \\\mathrm{\dfrac{\frac{\sqrt{3}}{\rlap{/}2}-\frac{\sqrt{2}}{\rlap{/}2}}{\frac{1}{\rlap{/}2}+\frac{\sqrt{2}}{\rlap{/}2}}=\dfrac{y-\frac{\sqrt{3}}{2}}{x-\frac{1}{2}}}}\\\mathrm{x(\sqrt{3}-\sqrt{2})-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}=y(1+\sqrt{  2})-\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{6}}{2}}\\\mathrm{x(\sqrt{3}-\sqrt{2})-y(1+\sqrt{2})+\dfrac{\sqrt{2}+\sqrt{6}}{2}}=0

    \mathrm{A=\sqrt{3}-\sqrt{2}}

    \mathrm{B=-(1+\sqrt{2})}

    \mathrm{C=\dfrac{\sqrt{2}+\sqrt{6}}{2}}

    Last edited by darthjavier; November 13th 2012 at 07:22 AM.
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  5. #5
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    Re: equation of a line - help

    You give the equation as Ax+ By= C. Do you realize those numbers are not "unique"? You could multiply the entire equation by any constant. You could, for example, Divide the entire equation by C to get (A/C)x+ (B/C)y= 1. Rewriting that as A'x+ B'y= 1 (A'= A/C and B'= B/C), putting x= -\sqrt{2}/2, y= \sqrt{2}/2 we have \frac{\sqrt{2}}{2}A'+ \frac{\sqrt{2}}{2}B'= 1 and multiplying both sides by \frac{2}{\sqrt{2}}= \sqrt{2}, A'+ B'= \sqrt{2}.

    Now put x= 1/2, y= \sqrt{3}{2} in the equation: A'/2+ \sqrt{3}B'/2= 1 so that A'+ \sqrt{3}B'= 2. Now, subtracting the previous equation from that, the A' is eliminated: (\sqrt{3}- \sqrt{2})B'= 1. Solve that for B', then put that into either of the previous equations.
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  6. #6
    chg
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    Re: equation of a line - help

    Thank you very much guys, that was very helpful (a lot more than my exercise book). Looks like I still have some basics to grasp, but with some help like this, I'll get there eventually

    Cheers!
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