Loci | Mathematics Form 4

KCSE ONLINE

Esoma Online Revision Resources

Loci - Mathematics Form 4


Segment
Area bounded by an arc and a chord


Sector
This is the area bounded by two radii and an arc.


2.Angle properties of circle
Angle in a semicircle:

A diameter subtends an angle of 90O
at the circumference.
Angle at the center and angle subtends on the circumference by the same arc.


The angle subtends at the centre is equal to the angle subtended on the circumference by the same arc.


3.Drawing a straight line
I. Draw the line y= 2x+3 hence show the region y=2x+3.



2.Show the region (R) represented by the inequalities.
y > 2x + 3, x > - 1.5, y < 3

 



OBJECTIVES:


By the end of the lesson you should be able to


a)Define locus

b)Describe common types of loci

c)Construct loci involving

I.Inequalities

II.Chords

III.Points under given conditions

IV.Intersecting loci.




LOCI

COMMON TYPES OF LOCI

In the life situations movement of objects and bodies creates different paths, regions like sectors chords etc. this knowledge is appreciated in constructions of buildings, flights, navigations, migration of wildebeests, flamingos, etc.

The secondhand of a clock occupies a series of different positions in the course of each minute. If all these positions are combined the circumference of a circle is obtained, which the tip of the second hand traces out within the course of each minute.



Similarly, a raindrop which hits an umbrella on its way to the ground traces out a path as shown in the picture below.


In the above illustration such a patch traced out by a moving object is called a locus. The tip of the second hand in a clock traces out a path which is the circumference of a circle.


A locus (plural of loci) of a point is the set of all possible positions occupied by a point which vanes its position according to some given law. It can also be defined further as the path, area or volume traced out by a point, line or region as it moves according to some given laws.





The perpendicular bisector loci.

Draw a line of length 6cm.Using the same measurements make arcs on both sides of the line at point A and similarly at point B.
Then join points C and D on the intersecting arcs to form a line CD.

Measure AP, and PB, what do you notice?
We notice that AP is equal to PB.


Measure the angle APC and angle BPC.
What do you notice?
We notice that angle APC is 90
o
and angle BPC is 90o
.


The line CD divides the AB into two equal parts and it is called a perpendicular
bisector, locus.
Notice that any point on the line CD is equidistant to the point A and B.

The locus of a point at a given distance from a fixed point which is 4 cm.

Using the same measurements from point A, make other positions for p. Using A as the centre join the various positions of P. What is the resulting path? The resulting path is a circumference of a circle centre A, with radius 4cm.

All points on a circle describe a locus of a point at constant distance from a fixed point.

The locus of point moving at a fixed distance from a given line.


Draw a line PQ of length 8cm.
Mark a point N such that is 3 cm from the line PQ. Draw perpendicular lines at points P and Q on both sides of the line PQ. Mark points on the perpendicular line 3cm from points P and Q and join them.
The set of points 3cm from line PQ is a parallel line on both sides of the line. The locus of a point which is d units from a fixed line PQ is a pair parallel lines d units away either side of the line PQ.

Angle bisector locus
Draw lines AB and BC such that they intersect to form angle ABC.

Standing at point B mark an arc on the line BA and BC 2cm away from point B.

Using the arcs make two arcs to intersect at point D.
Join B and D.
Measure angle ABD and angle DBC.
What do you notice?
We notice angle ABD and angle DBC are equal.
The line BD which bisects the angle ABC is called an angle bisector.
Note also that a point which lies on a bisector of a given angle is equidistant from the lines including that angle.


Example
Construct an equilateral triangle PQR of side 8cm. on the diagram:
i) Construct the locus x which is equidistant from the side PQ and QR.
ii) Construct the locus of a point Y equidistant from P and Q.

Solution.
i) Draw triangle PQR


Locus of x is the angle bisector.

Locus of angle PQR.



Construction of the locus of a point equidistant from two other points


Example:

Construct an equilateral triangle PQR of sides 8cm. On the diagram construct the locus of a point Y equidistant from points P and R.

Solution:
The locus of a point (Y) equidistant from P and R is a perpendicular bisect of line PR.The locus of Y is the angle bisector locus of angle PQR.


Construct an angle locus.

Angles in the segment are equal


Draw a circle radius 3cm, mark chord AB to form a minor and a major segment. Chord AB subtends angles ao,
bo,
co
and do
on the circumference. Measure the angles.
What do you notice? Angles ao,
bo,
co
and do
are all equal. The angles are in the same segment.

 


Example 1.

A line AB is 6 cm long. Construct the locus of all points P such that angle APB is equal to 70o.

 

Solution.


The locus of all points P such that angle APB is equal to 70O
is called constant angle locus. If angle APB = 70O.
Line AB is a chord that subtends 70O
on the circumference and 140O
at the centre of a circle centre O. If angle APB = 70O,
then for any point P on the circumference. APB = 70O
. Line AB is a chord that subtends 70O
on the circumference and 140O
at the centre of a circle centre O.
Draw a line AB of length 6cm. Measure 20O
at points A and B on both sides.




Example 2.

P and Q are fixed point 5cm. construct the locus point k such that angle PKQ is equal to 60o
. This is a constant angle locus such that PQ is a chord of circles centre O1
and O2
, with chord PQ subtending angles 60o
at the circumferences of the circles and 120o
at the centres of O1
and O2
the circles. Notice that angle O1
PQ = angle O1
QP = angle PO2
Q= angle QO2
P which is 30o.
Angle P, K, Q is equal to 60o
and angle PK2
Q is also equal to 60o.


Construct the locus of a point Y equidistant from P and Q
Example:

Construct an equilateral triangle PQR of sides 8cm and on the diagram construct the locus of a point (X) which is equidistant from the sides PR and QR.

Solution

The locus of a point (X) equidistant from P and Q is a perpendicular bisect of line PQ.The locus of X is the angle bisector locus of angle PRQ.

Construct angle locus.
Angles in the segment are equal.
Draw a circle radius 3cm, mark chord AB to form a minor and a major segment.
Chord AB subtends angles a
o
,bo
, co
and do
on the circumference. Measure the angles.What do you notice?
Angles aO,bO,cO and dO are all equal. The angles are in the same segment.

Example 1.
A line AB is 6 cm long. Construct the locus of all points P such that angle APB is equal to 70O
.
Solution.
The locus of all points P such that angle APB is equal to 70O
is called constant angle locus.
If <APB = 70
O

Line AB is a chord that subtends 70
O
on the circumference and 140O
at the centre of a circle centre O.

If <APB = 70O, then for any point P on the circumference. APB = 70O.
Line AB is a chord that subtends 70Oon the circumference and 140O at the centre of a circle centre O.
Draw a line AB of length 6cm. Measure 20O at points A and B on both sides.


The lines intersect at the centre (O) of a circle of radius OA.
i. Using OA as the radius draw major segment on both sides of the line AB.
ii. Measure angles A, P, B and AP
2
B subtends by the chord AB on the circumference of circle centre O1
and O2
. What do you notice?



We notice that angle A, P1
, B is equal to 70O
. Notice that the angle subtends by chord AB at the centre is equal to 140O.


Example 2.
2) P and Q are fixed point 5cm. construct the locus point k such that <PKQ is equal to 60O
.
This is a constant angle locus such that PQ is a chord of circles centre O1
and O2
, with chord PQ subtending angles 60O
at the circumferences of the circles and 120O
at the centres of O1
and O2
the circles.
Notice that angle O
1
PQ =angle O1
QP =angle PO2
Q= angle QO2
P which is 30O


Angle P, K, Q is equal to 60O and angle PK2Q is also equal to 60O.

Order this CD Today to Experience the Full Multimedia State of the Art Technology!

For Best results INSTALL Adobe Flash Player Version 16 to play the interactive content in your computer. Test the Sample e-Content link below to find out if you have Adobe Flash in your computer.

Sample Coursework e-Content CD

Other Goodies for KCSE ONLINE Members!

Coursework e-Content CD covers all the topics for a particular class per year and costs 1200/- ( Per Subject per Class ).

Purchase Online and have the CD sent to your nearest Parcel Service. Pay the amount to Patrick 0721806317 by M-PESA then provide your address for delivery of the Parcel. Alternatively, you can use BUY GOODS TILL NUMBER 827208 Ask for clarification if you get stuck.

Install ADOBE Flash Player for Best Results

For Best results INSTALL Adobe Flash Player Version 16 to play the interactive content in your computer. Test the link below to find out if you have Adobe Flash in your computer.

Search

Subject Menu