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 \(\ell\).
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,
\(P'Q = PQ = PQ' = P'Q'\).