compass and straightedge construction of perpendicular


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

  1. 1.

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

    ....PQRQPR
  2. 2.

    Construct the perpendicular bisectorMathworldPlanetmath of QR¯. This line is m.

    ....PQRQPR

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 QR¯.

In the case that P, this construction is referred to as dropping the perpendicular from P to . In the case that P, this construction is referred to as erecting the perpendicular to 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