compass and straightedge construction of perpendicular
Let $P$ be a point and $\mathrm{\ell}$ be a line in the Euclidean plane^{}. One can construct a line $m$ perpendicular^{} to $\mathrm{\ell}$ and passing through $P$. The construction given here yields $m$ in any circumstance: Whether $P\in \mathrm{\ell}$ or $P\notin \mathrm{\ell}$ does not matter. On the other hand, the construction looks quite different in these two cases. Thus, the sequence of pictures on the left (in which $\mathrm{\ell}$ is in red) is for the case that $P\notin \mathrm{\ell}$, and the sequence of pictures on the right (in which $\mathrm{\ell}$ is in green) is for the case that $P\in \mathrm{\ell}$.

1.
With one point of the compass on $P$, draw an arc that intersects $\mathrm{\ell}$ at two points. Label these as $Q$ and $R$.

2.
Construct the perpendicular bisector^{} of $\overline{QR}$. This line is $m$.
This construction is justified because $Q$ and $R$ are constructed so that $P$ is equidistant from them and thus lies on the perpendicular bisector of $\overline{QR}$.
In the case that $P\notin \mathrm{\ell}$, this construction is referred to as dropping the perpendicular from $P$ to $\mathrm{\ell}$. In the case that $P\in \mathrm{\ell}$, this construction is referred to as erecting the perpendicular to $\mathrm{\ell}$ at $P$.
If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.
Title  compass and straightedge construction of perpendicular 
Canonical name  CompassAndStraightedgeConstructionOfPerpendicular 
Date of creation  20130322 17:14:01 
Last modified on  20130322 17:14:01 
Owner  Wkbj79 (1863) 
Last modified by  Wkbj79 (1863) 
Numerical id  5 
Author  Wkbj79 (1863) 
Entry type  Algorithm 
Classification  msc 51M15 
Classification  msc 5100 
Related topic  ProjectionOfPoint 
Defines  drop the perpendicular 
Defines  dropping the perpendicular 
Defines  erect the perpendicular 
Defines  erecting the perpendicular 