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Suppose, $AB$ and $CD$ are two line segments. And, they have slopes $m_1$ and $m_2$ respectively. They will intersect with each other if, $m_1 - m_2 \ne 0.$ No, this is incorrect. They are parallel if $m_1 = m_2$. They may intersect each other if $m_1 \neq m_2$. For instance, consider the line segment through $(0,0)$ and $(1,1)$ (which has slope ...

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It is impossible to calculate your figure, because the information you have does not completely determine the figure. Your fear about "multiple solutions" is correct--in fact, infinitely many of them. You need more information to get a unique solution--the "unknown's" in your drawing would do. Here are two diagrams, both satisfying your conditions and ...

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Assuming you don't do anything too crazy with a line (like moving it along a space-filling curve), you will get a (part of) ruled surface. Two examples from Wikipedia: helicoid and hyperboloid.

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Perhaps along the lines of what you are thinking, if you subtract one point from the rest so that one point is $(0,0)$, then all the points are on a common line if and only if every point $(x_i,y_i)$ is an exact multiple of every other point. As long as the line is not horizontal, you can test this for example by computing $x_i/y_i$ for each remaining point ...

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As noted in the comments, we can cut to the chase by simply tabulating where the various superimposed vertices accumulate at the end of the edge-dropping process. For a regular $n$-gon, we can assign integer weights $p_i$ (with $i=0, 1, \dots n-1$) that count the vertices that land on vertex $P_i$. Of course, we must have $\sum_i p_i = n$. In what follows, ...

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Here is a way to do it without using quaternions. Refer to https://en.wikipedia.org/wiki/Rotation_matrix#In_three_dimensions for rotation matrix about an axis and an angle. The axis $\vec{u}$ in your case is $(2,1,2)$. You also need to normalize it. Say the angle is $\theta$. You can then represent the original $(x,y,z)$ by the new $(x',y',z')$ using the ...

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