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Determine all subsets of a given set of points which lie on the same line

Consider the set of points

    \begin{align*}   A = (2,1,1), \quad B = (6,-1,1), \quad C = (-6,5,1), \quad D = (-2,3,1), \\  E = (1,1,1), \quad F = (-4,4,1), \quad G = (-13,9,1), \quad H = (14,-6,1). \end{align*}

The points A,B,C lie on a line. Determine all other subsets of three or more points which lie on the same line.


First, we notice that the third coordinate of all of the given points is 1. So, if we check that the x,y coordinates are on the same line, then the points will be on the same line. Next, we look at which other points are on the same line as A,B,C. This line is given by

    \[ L(A; A-B) = \{ (2,1,1) + t(-4,2,0) \mid t \in \mathbb{R} \} = \{ (2-4t,1+2t,1) \mid t \in \mathbb{R} \}. \]

Checking the remaining points we have
The point D = (-2,3,1) is on L since 2-4t = -2 implies t = 1 and 1+2t = 3. Thus, D = (2-4t,1+2t,1) for t = 1.
The point E = (1,1,1) is not on L since 2-4t = 1 implies t = \frac{1}{4}, but then 1 + 2t = \frac{3}{2} \neq 1.
The point F = (-4,4,1) is on L since 2-4t = -4 implies t = \frac{3}{2} and then 1+2t = 4. Thus, F = 2-4t,1+2t,1) for t = \frac{3}{2} so F is on L.
The point G = (-13,9,1) is not on L since 2-4t = -13 implies t = \frac{15}{4}, but then 1+2t = \frac{17}{2} \neq 9. SO G is not on L.
The point H = (14,-6,1) is not on L since 2-4t = 14 implies t = -3, but then 1+2t = -5 \neq -6. So H is not on L.

Therefore, we have that \{ A,B,C,D,F \} are all on the same line (hence, every subset of these with three or more elements is a subset with three or more points on the same line).

None of the other points E,G,H are on the same line since H is not on the unique line containing E,G.

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