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See Fig. 11-11. (a) The locus is a line parallel to the two given lines and midway between them. (b) The locus is the perpendicular bisector of the line joining the two points. (c) The locus is a circle concentric with the earth and of radius 100 mi greater than that of the earth. (d) The locus is a circle of radius 10 mi with its center at the gun.
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Fig. 11-11
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11.2 Determining the locus of the center of a circle
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Determine the locus of the center of a circular disk (a) moving so that it touches each of two parallel lines; (b) moving tangentially to two concentric circles; (c) moving so that its rim passes through a fixed point; (d) rolling along a large fixed circular hoop.
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See Fig. 11-12.
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Fig. 11-12
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CHAPTER 11 Locus
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(a) The locus is a line parallel to the two given lines and midway between them. (b) The locus is a circle concentric with the given circles and midway between them. (c) The locus is a circle whose center is the given point and whose radius is the radius of the circular disk. (d) The locus is a circle outside the given circle and concentric to it.
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11.3 Constructing loci Construct (a) the locus of points equidistant from two given points; (b) the locus of points equidistant from two given parallel lines; (c) the locus of points at a given distance from a given circle whose radius is less than that distance.
Solutions
See Fig. 11-13.
Fig. 11-13
11.2 Locating Points by Means of Intersecting Loci
A point or points which satisfy two conditions may be found by drawing the locus for each condition. The required points are the points of intersection of the two loci.
SOLVED PROBLEM
11.4 Locating points that satisfy two conditions On a map locate buried treasure that is 3 ft from a tree (T ) and equidistant from two points (A and B) in Fig. 11-14.
Fig. 11-14
CHAPTER 11 Locus
Solution
The required loci are (1) the perpendicular bisector of AB and (2) a circle with its center at T and radius 3 ft. As shown, these meet in P1 and P2, which are the locations of the treasure. Note: The diagram shows the two loci intersecting at P1 and P2. However, there are three possible kinds of solutions, depending on the location of T with respect to A and B: 1. The solution has two points if the loci intersect. 2. The solution has one point if the perpendicular bisector is tangent to the circle. 3. The solution has no points if the perpendicular bisector does not meet the circle.
11.3 Proving a Locus
To prove that a locus satisfies a given condition, it is necessary to prove the locus theorem and its converse or its inverse. Thus to prove that a circle A of radius 2 in is the locus of points 2 in from A, it is necessary to prove either that 1. Any point on circle A is 2 in from A. 2. Any point 2 in from A is on circle A (converse of statement 1). or that 1. Any point on circle A is 2 in from A. 2. Any point not on circle A is not 2 in from A (inverse of statement 1). These statements are easily proved using the principle that a point is outside, on, or inside a circle according as its distance from the center is greater than, equal to, or less than the radius of the circle.
SOLVED PROBLEM
11.5 Proving a locus theorem Prove that the locus of points equidistant from two given points is the perpendicular bisector of the segment joining the two points.
Solution
First prove that any point on the locus satisfies the condition: Given: Points A and B. CD is the ' bisector of AB. To Prove: Any point P on CD is equidistant from A and B; that is, PA > PB. Plan: Prove ^PEA > ^PEB to obtain PA > PB. PROOF:
Statements 1. CD is the ' bisector of AB. 2. /PEA > /PEB 3. 4. 5. 6. AE > EB PE > PE ^PEA > ^PEB PA > PB
Reasons 1. Given 2. Perpendiculars form right angles; all right angles are congruent. 3. To bisect is to divide into congruent parts. 4. Reflexive property 5. SAS 6. Corresponding parts of > triangles are >.
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