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Thread: projecting lines

  1. #1
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    projecting lines

    I have a scheme on projecting lines on another line, but I don't know how to calculate the points that are projected on the lines.

    I want to calculate them using the coordinates of the rectangle corners.

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  2. #2
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    Projecting lines

    Hello moment
    Quote Originally Posted by moment View Post
    I have a scheme on projecting lines on another line, but I don't know how to calculate the points that are projected on the lines.

    I want to calculate them using the coordinates of the rectangle corners.
    As I understand it, you need to find, using coordinate geometry, the positions of the vertices of rectangles when they are projected onto a series of lines. So, to be specific, I think you're saying that, given a point $\displaystyle P$ (representing a typical vertex of a rectangle) you need to find the position of the projection of $\displaystyle P$ onto a line $\displaystyle l$.

    If that's so, then the following may be helpful.

    Suppose that $\displaystyle P$ has coordinates $\displaystyle (x_1,y_1)$ and that the equation of the line $\displaystyle l$ is $\displaystyle y = mx + c$. Suppose further that $\displaystyle Q$ is the projection of $\displaystyle P$ onto $\displaystyle l$, and that the coordinates of $\displaystyle Q$ are $\displaystyle (x_2, y_2)$.

    Then, because $\displaystyle Q$ lies on $\displaystyle l$:

    $\displaystyle y_2 = mx_2 + c$ (1)

    and because $\displaystyle PQ \bot l$:

    $\displaystyle \frac{y_1-y_2}{x_1-x_2} = -\frac{1}{m}$ (2)

    Substitute from (1) into (2):

    $\displaystyle \frac{y_1-(mx_2+c)}{x_1-x_2}=-\frac{1}{m}$

    $\displaystyle \Rightarrow my_1-m^2x_2-mc=x_2-x_1$

    $\displaystyle \Rightarrow my_1 - mc +x_1=x_2(1+m^2)$

    $\displaystyle \Rightarrow x_2=\frac{x_1+my_1 - mc}{1+m^2}$ (3)

    Substitute into (1):

    $\displaystyle y_2 = \frac{mx_1 + m^2y_1 - m^2c + c + cm^2}{1+m^2}$

    $\displaystyle \Rightarrow y_2 = \frac{mx_1 + m^2y_1 +c}{1+m^2}$ (4)

    So equations (3) and (4) give us the coordinates of Q in terms of $\displaystyle x_1, y_1, m$ and $\displaystyle c$.

    Of course, you may be able to simplify these results if you are able to choose a particular origin of coordinates. For example, if the origin can be chosen to lie on the line $\displaystyle l$, then $\displaystyle c = 0$, and equations (3) and (4) simplify to:

    $\displaystyle \Rightarrow x_2=\frac{x_1+my_1}{1+m^2}$

    $\displaystyle \Rightarrow y_2 = \frac{m(x_1 + my_1)}{1+m^2}$

    Or, if you could also choose two lines $\displaystyle l$ at $\displaystyle 45^o$ to the axes, then you can make $\displaystyle m = \pm 1$, and the equations would simplify even further to:

    $\displaystyle \Rightarrow x_2=\frac{x_1\pm y_1}{2}$

    $\displaystyle \Rightarrow y_2 = \frac{y_1 \pm x_1}{2}$

    I hope this gives you some ideas.

    Grandad
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