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Thread: Parallel and Perpendicular lines...?

  1. May 1st 2008, 12:51 AM #1
    mooch
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    Parallel and Perpendicular lines...?

    Hi folks, yr 10 student dilemna here!
    need to know this 4 test tomoro, just some pointers for finding the equations of straight lines which pass through a given point and perpendicular to the line whose equation is given by:...... and so on,, same with parallel.
    how to show that lines are perpendicular, with 2 equations???
    any help welcome,
    thanks a lot!
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  2. May 1st 2008, 12:55 AM #2
    Danshader
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    the lines are perpendicular when the gradient of the first line multiplied by the gradient of the second line gives you -1

    <br /> 1^{st} Line, y = m_1 x+ c_1<br />

    <br /> 2^{nd} Line, y = m_2 x+ c_2<br />

    <br /> m_1 m_2 = -1<br />
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  3. May 1st 2008, 01:04 AM #3
    earboth
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    Quote Originally Posted by mooch View Post
    Hi folks, yr 10 student dilemna here!
    need to know this 4 test tomoro, just some pointers for finding the equations of straight lines which pass through a given point and perpendicular to the line whose equation is given by:...... and so on,, same with parallel.
    how to show that lines are perpendicular, with 2 equations???
    any help welcome,
    thanks a lot!
    The general equation of a straight line is:

    y = \underbrace{m}_{direction} \cdot x + \underbrace{c}_{y-intercept}

    Two lines are parallel if they have the same direction (=slope, gradient):

    l_1: y = m_1 \cdot x + c_1 ... and ... l_2: y = m_2 \cdot x + c_2

    l_1\parallel l_2~\implies~m_1=m_2


    Two lines are perpendicular if

    m_1 \cdot m_2 = -1

    l_1\perp l_2~\implies~m_2=-\frac1{m_1}


    If you have the slope (=direction) of a straight line and a point P(x_P, y_P) which is placed on this line (or the line passes through the point P) then the coordinates of P must satisfy the equation:

    y-y_P = m(x-x_P) ... This is the point-slope-formula of a straight line.
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