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I don't know how quite to phrase this, but I'll try. Because two point are co-linear and two lines cannot always be used to define a plane and aren't always in the same plane, are two lines always co-something? 2 0 dimensional shapes make one 1 dimensional shape. 3 0 dimensional shapes make 1 two dimensional shape. x-1 0 dimensional shapes make a x dimensional shape. What do x-1 1 dimensional shapes make?

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If you allow one of your points to be the origin (or one of your lines to pass through the origin), then you can formulate and answer this question using linear algebra. For instance, given two points (the origin $O$ and one other point $P$), you have a single vector $\overrightarrow{OP}$, and the span of this vector is $1$-dimensional vector space, a line. Given three points (the origin, $P$, and $Q$), they are contained in the vector space spanned by the two vectors $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ which is at most a $2$-dimensional vector space, a plane.

Now, given two lines, the first through the origin and another point $P$, and the second through $Q$ and $R$, these lines are contained in the vector space spanned by the vectors $\overrightarrow{OP}$, $\overrightarrow{OQ}$, and $\overrightarrow{QR}$, which is at most $3$-dimensional. So two lines are always contained in a $3$-dimensional space.

In the same way, if we have $x-1$ lines, then we can choose that the first goes through the origin and hence is determined by a single vector (one other point). The other lines can each be determined by two vectors (a point on the line and the vector from that point to another on the line). So they will be contained in a space of dimension $2x-3$.

To remove the hypothesis that the origin is involved, you can use affine combinations and the related notion of span, but it is the same idea.

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    $\begingroup$ Using the same kind of counting, it will span a subspace that is $x(y+1)-1$ dimensional (at most). For each of the $x$ objects (except one), you get $y+1$ vectors, and $y$ for the remaining one. $\endgroup$
    – cws
    Commented Oct 24, 2014 at 6:36

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