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Thread: Point distance from straight

  1. #1
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    [Solved] Point distance from straight

    Hi,
    I have done a few tasks, that are the same type, for example: "Calculate the point a = [1, 2] distance from y = 3x + 2. These are easy and simple to do for me, but now I came across to a same subject but a different kind, that I don't really understand

    The question is:
    "Calculate the point (2, 6) distance from straight r = r0 + tv, when r0 = [1, -1] and v = [2, 10]

    I don't understand how to start doing it. Hope for some help.
    Thank you.
    Last edited by Koppara; Dec 19th 2018 at 09:05 PM.
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  2. #2
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    Re: Point distance from straight

    $r =(1,-1)+t(2,10)$

    can be written

    $y+1 = 5(x-1)$

    $y = 5x-6$

    can you finish?
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  3. #3
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    Re: Point distance from straight

    Quote Originally Posted by romsek View Post
    $r =(1,-1)+t(2,10)$

    can be written

    $y+1 = 5(x-1)$

    $y = 5x-6$

    can you finish?
    Ah, you can think it like that too.
    Thank you very much!
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  4. #4
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    Re: Point distance from straight

    Quote Originally Posted by Koppara View Post
    Ah, you can think it like that too.
    Thank you very much!
    also see this page
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  5. #5
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    Re: Point distance from straight

    Quote Originally Posted by romsek View Post
    $r =(1,-1)+t(2,10)$ can be written
    $y+1 = 5(x-1)$
    $y = 5x-6$
    To clarify: $(1,-1)$ is a point $(x_0,y_0)$ on the line and $(2,10)$ is the direction or slope of the line - how $x$ and $y$ change together.

    Thus the gradient $m=\frac{\Delta y}{\Delta x}=\frac{10}2=5$ and the the equation of the line is \begin{align}y-y_0 &= m(x-x_0) \\ y+1 &= 5(x-1)\end{align} as per romsek.
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  6. #6
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    Re: Point distance from straight

    As an alternative approach, you can look for the vector perpendicular to the direction of the line (i.e. $\langle2,10\rangle$) that passes through the point $P=(2,6)$.

    The points on the line are $Q_t=\langle1,-1\rangle + t\langle 2,10\rangle$. We just need to find the one for which the vector $\vec{PQ_t}$ joining it to $P$ is perpendicular to $\langle2,10\rangle$.

    $$\vec{PQ_t}=\vec{OQ_t} - \vec{OP} = \langle1,-1\rangle + t\langle 2,10\rangle - \langle 2,6 \rangle = \langle 2t-1, 10t-7 \rangle$$

    For this to be perpendicular to $\langle2,10\rangle$, we require that the scalar (dot) product of $\langle2,10\rangle$ and $\vec{PQ_t}$ be equal to zero.

    That is \begin{align}\vec{PQ_t} \cdot \langle2,10\rangle = \langle 2t-1, 10t-7 \rangle \cdot \langle2,10\rangle &= 0 \\ 2(2t-1) + 10(10t-7) &= 0 \\ 104t - 72 &= 0 \\ t &= \tfrac9{13} \end{align}

    Thus the vector $\vec{PQ_t}=\langle 2t-1, 10t-7 \rangle$ with $t=\frac9{13}$ is the perpendicular that joins the line and the point $P$. It's length is the perpendicular distance between the line and $P$.
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