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# 5.3: Uniqueness of a perpendicular

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## Theorem $$\PageIndex{1}$$

There is one and only one line that passes thru a given point $$P$$ and is perpendicular to a given line $$\ell$$.

According to the above theorem, there is a unique point $$Q \in \ell$$ such that $$(QP) \perp \ell$$. This point $$Q$$ is called the foot point of $$P$$ on $$\ell$$.

Proof

If $$P \in \ell$$, then both existence and uniqueness follow from Axiom III.

Existence for $$P \not\in \ell$$. Let $$A$$ and $$B$$ be two distinct points of $$\ell$$. Choose $$P'$$ so that $$AP' = AP$$ and $$\meauredangle BAP' \equiv - \measuredangle BAP$$. According to Axiom IV, $$\triangle AP'B \cong \triangle APB$$. In particular, $$AP = AP'$$ and $$BP = BP'$$.

According to Theorem 5.2.1, $$A$$ and $$B$$ lie on the perpendicular bisector to $$[PP']$$. In particular, $$(PP') \perp (AB) = \ell$$.

Uniqueness for $$P \not\in \ell$$. From above we can choose a point $$P'$$ in such a way that $$\ell$$ forms the perpendicular bisector to $$[PP']$$.

Assume $$m \perp \ell$$ and $$P \in m$$. Then $$m$$ is a perpendicular bisector to some segment $$[QQ']$$ of $$\ell$$; in particular, $$PQ = PQ'$$.

Since $$\ell$$ is the perpendicular bisector to $$[PP']$$, we get that $$PQ = P'Q$$ and $$PQ' = P'Q'$$. Therefore,

$$P'Q = PQ = PQ' = P'Q'$$.

By Theorem 5.2.1, $$P'$$ lies on the perpendicular bisector to $$[QQ']$$, which is $$m$$. By Axiom II, $$m = (PP')$$.

This page titled 5.3: Uniqueness of a perpendicular 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.

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