(which relates to the property of vectors stating that and have opposite direction and same modulus).
where represents any segment of the form . If it is stated that we will work with directed segments, it’s customary to omit the overlining and to just write , convention we will follow now.
It does not make sense to compare signs of non-collinear segments. So if are not on the same line (or parallel lines) we cannot relate the signs of and .
It can be proved considering cases that no matter the relative position of three points on a line, the following equality holds:
In the above picture . Notice that goes to the left since is the segment that starts at and ends at . Also, taking gives which is consistent with the earlier remarks.
Just like undirected segments in Euclidean geometry (and unlike vectors), directed segments can be divided to obtain a ratio. Such ratio is the number obtained dividing the undirected segments, but taking signs int oaccount (ratio of two segments with the same direction is positive, and negative otherwise).
Given two points on a line, we can locate any other point on the line considering the ratio . In other words, if and only if . Moreover, to each point corresponds an extended 11We use extended reals to avoid dealing with separate cases, allowing us to deal also with points at infinity. real and to each extended real corresponds a point such that .
Notice that is positive when and have the same direction, which happens if and only if is between and . If lies outside , then and have negative signs and so the ratio will be negative.
|Date of creation||2013-03-22 14:56:59|
|Last modified on||2013-03-22 14:56:59|
|Last modified by||drini (3)|