# compass and straightedge construction of perpendicular

Let $P$ be a point and $\ell$ be a line in the Euclidean plane  . One can construct a line $m$ perpendicular    to $\ell$ and passing through $P$. The construction given here yields $m$ in any circumstance: Whether $P\in\ell$ or $P\notin\ell$ 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 $\ell$ is in red) is for the case that $P\notin\ell$, and the sequence of pictures on the right (in which $\ell$ is in green) is for the case that $P\in\ell$.

1. 1.

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

2. 2.

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 $P\notin\ell$, this construction is referred to as dropping the perpendicular from $P$ to $\ell$. In the case that $P\in\ell$, this construction is referred to as erecting the perpendicular to $\ell$ 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