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Thread: perpendicular lines

  1. June 25th 2012, 01:52 AM #1
    muddywaters
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    perpendicular lines

    the question with the answers:
    perpendicular lines-untitled0.jpg
    perpendicular lines-untitled.jpg
    perpendicular lines-untitledb.jpg

    what i don't understand is this part:
    perpendicular lines-untitled1.jpg
    how did that happen?
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  2. June 25th 2012, 01:54 AM #2
    muddywaters
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    Re: perpendicular lines

    click to enlarge :O
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  3. June 26th 2012, 11:01 PM #3
    earboth
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    Re: perpendicular lines

    Quote Originally Posted by muddywaters View Post
    the question with the answers:
    Click image for larger version.  Name: untitled0.JPG  Views: 13  Size: 36.4 KB  ID: 24153
    Click image for larger version.  Name: untitled.JPG  Views: 8  Size: 19.2 KB  ID: 24154
    Click image for larger version.  Name: untitledb.JPG  Views: 8  Size: 34.6 KB  ID: 24155

    what i don't understand is this part:
    Click image for larger version.  Name: untitled1.JPG  Views: 11  Size: 9.9 KB  ID: 24156
    how did that happen?
    1. You know from equation 2:

    \displaystyle{y-y_0=-\frac1m(x-x_0)~\implies~(y-y_0)^2=\frac1{m^2}(x-x_0)^2}

    2. \displaystyle{|\overline{PQ}|=\sqrt{(x-x_0)^2+(y-y_0)^2}}

    Replace \displaystyle{(y-y_0)^2} by the term of #1:

    \displaystyle{|\overline{PQ}|=\sqrt{(x-x_0)^2+\frac1{m^2}(x-x_0)^2}}

    \displaystyle{|\overline{PQ}|=\sqrt{\frac{m^2}{m^2 } (x-x_0)^2+\frac1{m^2}(x-x_0)^2}}

    Factor out \displaystyle{(x-x_0)^2}

    \displaystyle{|\overline{PQ}|=\sqrt{\frac{m^2-1}{m^2} (x-x_0)^2}}
    Thanks from muddywaters
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  4. June 26th 2012, 11:15 PM #4
    muddywaters
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    Re: perpendicular lines

    yes. thanks
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