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Math Help - Perpendicular line equation!

  1. #1
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    Question Perpendicular line equation!

    Hi There!

    Was hoping for some help.

    Set up an equation for the line (2,3) which is perpendicular to the line (2, -1) and (-1.-3). Don't understand this at all. Can someone please help me?!

    Thx!
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  2. #2
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    Quote Originally Posted by Walkman View Post
    Hi There!

    Was hoping for some help.

    Set up an equation for the line (2,3) which is perpendicular to the line (2, -1) and (-1.-3). Don't understand this at all. Can someone please help me?!

    Thx!
    Use the equation of a straight line: y=m_1x+c

    where m_1 = \frac{y_2-y_1}{x_2-x_1}

    You have the two points for a line so you can work this out.

    For two lines to be perpendicular their gradients m_1, m_2 must equal -1 and so m_2 = -\frac{1}{m_1}

    Now m_2 is known use the point (2,3) to get an equation of the form y = mx+c
    Last edited by e^(i*pi); December 12th 2009 at 12:48 PM. Reason: lates
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    Use the equation of a straight line: [YOUTUBE]y=m_1x+c[/YOUTUBE]

    where m_1 = \frac{y_2-y_1}{x_2-x_1}

    You have the two points for a line so you can work this out.

    For two lines to be perpendicular their gradients m_1, m_2 must equal -1 and so m_2 = -\frac{1}{m_1}

    Now m_2 is known use the point (2,3) to get an equation of the form y = mx+c

    Hm. Maybe i took the wrong way solving this. But just before you replied i got the equation fot the line between the two given points.

    y = (2 x)/3-7/3.


    dont know if it's correct though...
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  4. #4
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    Quote Originally Posted by Walkman View Post
    Hm. Maybe i took the wrong way solving this. But just before you replied i got the equation fot the line between the two given points.

    y = (2 x)/3-7/3.


    dont know if it's correct though...
    m_1 = \frac{-3-(-1)}{-1-2} = \frac{-2}{-3} = \frac{2}{3}

    Using the point (2,-1): -1 = \frac{2}{3}(2)+c \: \rightarrow \: c = -\frac{7}{3}

    line 1: y = \frac{2x-7}{3}

    Indeed you are right


    As you now know the gradient of the above line is \frac{2}{3} so m_2 = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}

    Are you ok to do the same as you did before but with the point (2,3) and the gradient -1.5
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  5. #5
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    Quote Originally Posted by e^(i*pi) View Post
    m_1 = \frac{-3-(-1)}{-1-2} = \frac{-2}{-3} = \frac{2}{3}

    Using the point (2,-1): -1 = \frac{2}{3}(2)+c \: \rightarrow \: c = -\frac{7}{3}

    line 1: y = \frac{2x-7}{3}

    Indeed you are right


    As you now know the gradient of the above line is \frac{2}{3} so m_2 = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}

    Are you ok to do the same as you did before but with the point (2,3) and the gradient -1.5

    Here's where my problem always come in. I never solves the perpendicular line. I know how to do it, but for some reason i can't do it... I'll try..
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  6. #6
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    Just as i thought. Can't get it right...

    Keep getting wierd answers all the time...

    help please..?
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  7. #7
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    Quote Originally Posted by Walkman View Post
    Just as i thought. Can't get it right...

    Keep getting wierd answers all the time...

    help please..?
    You have been shown excatly how to do this question and given all the information required for doing it. Your troubles will be caused by poor algebra. So please show all your working so that your mistake(s) can be explained to you.
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