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Math Help - Parametric representation of lines.. (simple)

  1. #1
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    Parametric representation of lines.. (simple)

    I just can't get my head around this, what am I doing wrong.

    for example, given that,
    r(t) = (1-t)r0 + t(r1)

    and say we have a line segment: (-2,-1) to (1,2)

    How is it that x = -2 + 3t and y = -1 +3t
    r(t) = (1-t)<-2,-1> + t<1,2>
    I am not seeing how this expression becomes: <-2+3t, -1+3t>


    If you multiply it out, don't you get: -2 + 2t - 1 + t + t + 2t = -3 + 6t
    How are you supposed to go from
    r(t) = (1-t)<-2,-1> + t<1,2>
    to
    <-2+3t, -1+3t>


    Same for a line segment from (-5,3) to (0,2)

    Thanks
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  2. #2
    Junior Member Bradyns's Avatar
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    Re: Parametric representation of lines.. (simple)

    The vector equation for a line when you have two points in R^2 is.

    v = (p_1, p_2) + t(q_1 - p_1, q_2 - p_2), where t \in \mathbb{R}

    So, for your points (-2,-1), (1,2), we could write.
    v = (-2,-1) + t(1 - (-2),2 - (-1))

    Which, when simplified, is:
    v = (-2,-1) + t(3,3)

    When parameterized, you get,
    x = -2 + 3t
    y = -1 + 3t
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  3. #3
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    Re: Parametric representation of lines.. (simple)

    To understand parametric forms of lines, you must first understand what a line IS. A line is defined by its direction, is infinitely long, and is positioned somewhere.

    Notice that vectors have a similar definition, by their direction and their magnitude.

    That would mean that vectors and lines should be related, and they are.

    If you know two points that lie on a line, you can get a vector that has the same direction, but it will only be as long as the line segment. It can be made infinitely long by multiplying by a large parameter.

    Now you almost have a line, the only problem is that vectors aren't defined as having any position, and so are usually positioned as starting at the origin. But the points you know the line goes through actually tell us how much of a numerical adjustment we need in the direction of each axis. So by adding the components of one of the points to the components of your now-infinitely long vector, you position it where you need.

    To illustrate with your example, we know that the line has to go through \displaystyle \begin{align*} (-2, -1) \end{align*} and \displaystyle \begin{align*} (1, 2) \end{align*}. We can get the direction of the line by the direction vector \displaystyle \begin{align*} < 1 - (-2) , 2 - (-1) > \, = \, < 3, 3 > \end{align*}.

    Now to make it infinitely long, we multiply by a parameter \displaystyle \begin{align*} t \end{align*}, giving \displaystyle \begin{align*} t < 3, 3 > \, = \, < 3t, 3t > \end{align*}.

    Finally, we position it where we need by making the numerical adjustment implied by either of the points. Say we use \displaystyle \begin{align*} ( -2, -1 ) \end{align*}, then adding these components to our infinitely long vector gives \displaystyle \begin{align*} < 3t - 2 , 3t - 1 > \end{align*}.

    So the equation of our line is \displaystyle \begin{align*} <3t - 2 , 3t - 1>  \end{align*}, and if we wanted to we could write it as

    \displaystyle \begin{align*} x &= 3t - 2 \\ y &= 3t - 1 \end{align*}


    Now attempt your second problem.
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