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The locus of points equidistant from two given points is a straight line that is called the perpendicular bisector of the line segment connecting the points. The perpendicular bisectors of any two sides of a triangle intersect in exactly one point. This point must be equidistant from the vertices of the triangle.
Line DE bisects line AB at D, line EF is a perpendicular bisector of segment AD at C, and line EF is the interior bisector of right angle AED. In geometry, bisection is the division of something into two equal or congruent parts (having the same shape and size). Usually it involves a bisecting line, also called a bisector.
The line with equation ax + by + c = 0 has slope -a/b, so any line perpendicular to it will have slope b/a (the negative reciprocal). Let (m, n) be the point of intersection of the line ax + by + c = 0 and the line perpendicular to it which passes through the point (x 0, y 0). The line through these two points is perpendicular to the original ...
The set of points equidistant from two points is a perpendicular bisector to the line segment connecting the two points. [8] The set of points equidistant from two intersecting lines is the union of their two angle bisectors. All conic sections are loci: [9] Circle: the set of points at constant distance (the radius) from a fixed point (the ...
Constructing the perpendicular bisector from a segment; Finding the midpoint of a segment. Drawing a perpendicular line from a point to a line. Bisecting an angle; Mirroring a point in a line; Constructing a line through a point tangent to a circle; Constructing a circle through 3 noncollinear points; Drawing a line through a given point ...
To make the perpendicular to the line AB through the point P using compass-and-straightedge construction, proceed as follows (see figure left): Step 1 (red): construct a circle with center at P to create points A' and B' on the line AB, which are equidistant from P. Step 2 (green): construct circles centered at A' and B' having equal radius.
Three lines, each formed by drawing an external equilateral triangle on one of the sides of a given triangle and connecting the new vertex to the original triangle's opposite vertex, are concurrent at a point called the first isogonal center. In the case in which the original triangle has no angle greater than 120°, this point is also the ...
Through every pair of points there are two horocycles. The centres of the horocycles are the ideal points of the perpendicular bisector of the line-segment between them. Given any three distinct points, they all lie on either a line, hypercycle, horocycle, or circle. The length of a line-segment is the shortest length between two points.