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	2013-03-22 17:14:01
Last modified on	2013-03-22 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 51-00
Related topic	ProjectionOfPoint
Defines	drop the perpendicular
Defines	dropping the perpendicular
Defines	erect the perpendicular
Defines	erecting the perpendicular