Similarly it may be shown that each of QR, RS,

SB = AP.

By this method a st. line may be divided into any number of equal parts.

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66. Example 1.-A is a point and from A straight lines are drawn in different directions in the same plane.

On each line a distance of one inch is measured from A and the resulting points are B, C, D, etc.

Is there any one line that contains all of the points in the plane that are at a distance of one inch

from A?

To answer this question describe a circle with centre A and radius one inch. The circumference of this circle is a line that passes through all the points.

Mark any other point P on the

circumference.

What is the distance of P from A? From the definition of a circle the answer to this question is one inch.

If any point Q be taken within the circle, its distance from A is less than one inch, and if any point R be taken without the circle, its distance from A is greater than one inch.

Thus every point in the circumference satisfies the condition of being just one inch from A, and no point, in the plane, that is not on the circumference does satisfy this condition.

This circumference is called the locus of all points in the plane that are at a distance of one inch from A.

Example 2: AB is a straight line of indefinite length, to which any number of perpendiculars are drawn.

E

-B

H P K *R

On each of these perpendiculars a distance of one centimetre is measured from AB, and the resulting points are C, D, E, etc.

Are there any lines that contain all of the points, such as C, D, etc., that are at a distance of one centimetre from AB?

Draw two straight lines parallel to AB, each at a distance of one centimetre from AB, and one or other of these lines will pass through each of the points.

Any point P in CF, or in GK, is at a distance of one centimetre from AB; any point Q in the space between CF and GK is less than one centimetre from AB, and any point R in the plane and neither between CF and GK nor in one of these lines is more than one centimetre from AB.

Thus every point in CF and GK satisfies the condition of being just one centimetre from AB, and no point outside of these lines and in the plane does satisfy this condition.

The two lines GF, GK make up the locus of all points in the plane that are at a distance of one centimetre from AB.

Definition. When a figure consisting of a line or lines contains all the points that satisfy a given condition, and no others, this figure is called the locus of these points.

67. In place of speaking of the "locus of the points which satisfy a given condition," the alternative expression "locus of the point which satisfies a given condition" may be used.

Suppose a point to move in a plane so that it traces out a continuous line, but its distance from a fixed point A in the plane is always one inch; then it must move on the circumference of the circle of centre A and radius one inch, and the locus of the point in its different positions is that circumference.

The following definition of a locus may thus be given as an alternative to that in § 66.

Definition. If a point moves on a line, or on lines, so that it constantly satisfies a given condition, the figure consisting of the line, or lines, is the locus of the point.

THEOREM 22

The locus of a point which is equidistant from two given points is the right bisector of the straight line joining the two given points.

Hypothesis. P is a point equidistant from A and B. To prove that P is on the right bisector of AB Construction.-Bisect AB at C.

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May 8

THEOREM 23

5/48

Do same as

The locus of a point which is equidistant from two given intersecting straight lines is the pair of straight lines which bisect the angles between the two straight lines.

22

E

B

Hypothesis. AB, CD are two st. lines cutting at
E; GF, HK are the bisectors of Ls made by AB, CD.

To prove that the locus of a point equidistant from
AB, and CD consists of GF and HK.

Construction. Take any point P in GF. Draw
PXL AB, PY CD.

Proof.

< PEX = < PEY,

In As PEX, PEY,

< PXE = / PYE,

PE is common,

PX - PY. (I-14, p. 54.)

.. every point in GF is equidistant from AB and CD.

Similarly it may be shown that every point in
HK is equidistant from AB and CD.

.. the locus of points equidistant from AB, CD consists.
of GF and HK.

There may

well as P.

be other points as