The line through them (operation 1) is the perpendicular bisector.

If p and q are distinct, choose their perpendicular bisector as mirror.

Circles tangent to two given points must lie on the perpendicular bisector.

Construct the perpendicular bisector of the line segment.

Its perpendicular bisector is another chord, which is a diameter of the circle.

One diagonal is the perpendicular bisector of the other diagonal.

A chord's perpendicular bisector passes through the center of the circle and bisects the angle.

The Perpendicular bisector construction can be reversed via isogonal conjugation.

The center of the solution circle is equally distant from all three points, and therefore must lie on the perpendicular bisector line of any two.

Particularly important is the perpendicular bisector of a segment, which, according to its name, meets the segment at right angles.

Working out the math, the perpendicular bisector of the side of the pyramid comes out to 612 feet.

The centers of the circles in this pencil lie on the perpendicular bisector of CD.

Therefore O must lie on the perpendicular bisector of the line segment PQ.

The perpendicular bisector of a segment also has the property that each of its points is equidistant from the segment's endpoints.

The line DD' is the perpendicular bisector of AB.

This is equivalent to finding the perpendicular bisector of the line segment p'p. This can be done in four steps:

In two-dimensional Euclidian geometry the locus of points equidistant from two given (different) points is their perpendicular bisector.

G. C. Shephard, The perpendicular bisector construction, Geom.

It is easy to see that A is a circling point with a center that is on the perpendicular bisector of the segment A A2.

(A perpendicular bisector is a line that forms a right angle with one of the triangle's sides and intersects that side at its midpoint.)

The line through these five points is one of the two principal axes of the hyperbola, the other being the perpendicular bisector of the transverse axis.

The perpendicular bisector of line segment is the Lemoine line, which contains the three centers of the circles of Apollonius.

The 2-wise intersection point therefore is the midpoint of a side of the anticomplementary triangle, and H lies on the perpendicular bisector of this side.

The perpendicular bisector construction arises naturally in an attempt to find a replacement for the circumcenter of a quadrilateral in the case that it is noncyclic.

The line determined by the points of intersection of the two circles is the perpendicular bisector of the segment, since it crosses the segment at its center.