# 1.6: Half-lines and segments

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Assume there is a line $$l$$ passing thru two distinct points $$P$$ and $$Q$$. In this case we might denote $$l$$ as $$(PQ)$$. There might be more than one line thru $$P$$ and $$Q$$, but if we write $$(PQ)$$ we assume that we made a choice of such line.

We will denote by $$[PQ)$$ the half-line that starts at $$P$$ and contains $$Q$$. Formally speaking, $$[PQ)$$ is a subset of $$(PQ)$$ which corresponds to $$[0,\infty)$$ under an isometry $$f: (PQ) \to \mathbb{R}$$ such that $$f(P) = 0$$ and $$f(Q) > 0$$.

The subset of line $$(PQ)$$ between $$P$$ and $$Q$$ is called the segment between $$P$$ and $$Q$$ and denoted by $$[PQ]$$. Formally, the segment can be defined as the intersection of two half-lines: $$[PQ] = [PQ) \cap [QP)$$.

Exercise $$\PageIndex{1}$$

Show that

(a) if $$X \in [PQ)$$, then $$QX = |PX - PQ|$$;

(b) if $$X \in [PQ]$$, then $$QX + XQ = PQ$$.

Hint

Fix an isometry $$f: (PQ) \to \mathbb{R}$$ such that $$f(P) = 0$$ and $$f(Q) = q > 0$$.

Assume that $$f(X) = x$$. By the definition of the half-line $$X \in [PQ)$$ if and only if $$x \ge 0$$. Show that the latter holds if and only if $$|x - q| = ||x| - |q||$$. Hence (a) follows.

To prove (b), observe that $$X \in [PQ]$$ if and only if $$0 \le x \le q$$. Show that the latter holds if and only if $$|x - q| + |x| = |q|$$.

This page titled 1.6: Half-lines and segments is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.