# SS1 Further Mathematics Third Term Lesson Notes

### Further Mathematics Third Term SSS1 Revision

WEEK 2

Reference Materials: New Further Mathematics project 1, by Adigun et al. Page 178

Previous Knowledge:  Students can identify calculating devices.

Instructional Materials:  Charts showing flowcharts.

### ContentFLOWCHART

A flowchart is a diagrammatical representation of a solution to a problem.

Example: The perimeter of a rectangle is 2(l+b). Draw a flowchart to determine and represent the information.

1. A flowchart plays a very important role in computer programming.
2. It facilitates the interpretation and solution of problems.
3. It can be easily understood.
4. It can help in planning and development of algorithm for solving problems.

WEEK 3

Topic: Gradients of straight lines and curves

Duration: 40 minutes

Learning Objectives: By the end of the lesson, students should be able to calculate the gradient of a straight line.

Reference Materials: i. New General Mathematics for SSS 2, by M.F Macrae et al. Pages 184 – 192.

Previous Knowledge: Students can draw the graph of a linear equation (straight-line graph).

Instructional Materials: Graph board and graph book.

Content: GRADIENT OF A STRAIGHT LINE

The gradient of a straight line is the rate of change of y compared with x.

For example, if the gradient is 2, then for any increase in x, y increases two times as much.

Gradient of AB = Increase in y from A to B  =  MB

Increase in x from A to B AM

Example

Find the gradient of the line joining P(7, -2) and Q(-1, 2)

Gradient of PQ = increase in y   =    –    AQ

Increase in x                PA

-48 = -12

Example 2

Find the gradient of the line 7x + 4y – 8 = 0

Re-arrange the equation: 4y = – 7x + 8

y = -74 + 2

Therefore, gradient (m) = -74 , y – intercept (c) = 2

SKETCHING GRAPHS OF STRAIGHT LINES

Given the equation

y = 3x – 2 , gradient = 3, y – intercept(c) = -2

2x + 3y = 6, gradient = -23 , y – intercept(c) = 2

Example

Sketch the graph of the line whose equation is 4x – 3y = 12

Solution

When x = 0 ,- 3y = 12

y = – 4

The line crosses the y – axis at (0, – 4).

When y = 0 , 4x = 12

x = 3

The line crosses the x – axis at (3, 0).

From the graph:

Gradient m = y2 –  y1x2 –  x1

= 0 +  43 +  0   = 43

y – intercept = – 4

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Lines parallel to axes

Any line parallel to the x – axis has a gradient of zero. The equation of such lines is always in the form

y = c, where c may be any number.

The figure below shows the graph of y = 5 and y = – 3.

Notice that the equation of the x – axis is y = 0

————————————

————————————-

The gradient of a line that is parallel to the y – axis is undefined. The equations of such a lines are always in the form x = a , where a may be any number.

The figure below shows the graph of line x = 2 and x = – 4.

Notice that the equation of the y – axis is x = 0

EQUATION OF A STRAIGHT LINE

Equation of a straight line is of the form y = mx + c, where m is the gradient and c is the y – intercept.

Example 1

Determine the equation of a straight line whose gradient is 13 and passes through the point (- 3, 2).

Solution

Using the formula y – y1 = m(x – x1)

Where (x1, y1) = (- 3, 2) and m = 13

y – 2 = 13 (x + 3)

3y – 6 = – x – 3

x + 3y = 3

Example 2

Find the equation of the straight line passing through the points (1, 4) and (- 2, 6).

Using the formula

y –  y1y2 –  y1x –  x1x2 –  x1

Where = (x1, y1) = ( 1, 4) and (x2, y2) = (- 2, 6), the equation is

y –  46 –  4 = x –  1-2 –  1

cross multiply

– 3y + 12 = 2x – 2

2x + 3y = 14

Example

Draw the graph of y = 14x2 for values of x from –2 to 3. Find the gradient of the curve at the point where x has the value (a) (b) – 2

Solution

 x -2 -1 0 1 2 3 y 1 ¼ 0 ¼ 1 2¼

(a) Gradient of the curve where x = 3

= MP = 2.25     =       =  9      =  11/2

TM                   1.5         1.5             6

(b) Gradient of curve where x = – 2

= – QN      = –  1 = – 1

NR              1

WEEK 4

Topic: Straight line

Sub-topic: Angle of slope and angle between lines

Duration: 80 minutes

Learning Objectives: By the end of the lesson, students should be able to calculate the angle of slope and angle between two lines.

Reference Materials:  New Further Mathematics Project 2 by M. R Tuttuh Adegun

Previous Knowledge: Students can draw the graph of a linear equation (straight-line graph).

Instructional Materials: Graph board and graph book.

Content: ANGLE OF SLOPE

Example: Find the gradient of the line joining (3, 2) and (7, 10) and the angle of slope of the line.

Solution

Let m be the gradient of the line, then

m = 10-27-3=84=2

Let be the angle of slope of the line; then:

tanθ=2

θ=63.43°

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ANGLE BETWEEN TWO LINES

Condition for Parallelism

If two lines are parallel, the angle between them is zero, hence  tanθ=0

Example: Determine if AB is parallel to PQ in each of the following.

1. A(3, 1); B(4, 3)  and  P(4,6); Q(5, 8)
2. A(-1, -2); B(2, -3)  and   P(5, 4) ; Q(6, 7)

Solution

1. Let m1 be the gradient joining A and B and m2 be the gradient joining P and Q.

m1=3-14-3=2

m2=8-65-4=2

Since  m1=m2 ; AB||PQ

1. Let m1 be the gradient joining A and B and m2 be the gradient joining P and Q.

m1=-3-(-2)2-(-1)=-13

m2=7-46-5=3

Since  m1m2 ; AB is not parallel to PQ

### CONDITION FOR PERPENDICULARITY

If the lines are perpendicular, α=90° and  tanα=∞; therefore:

1 + m1m2=0

m1m2=-1

m1=-1m2

Example: Determine if AB is parallel to PQ in each of the following.

1. A(5, -1); B(3, 2)  and  P(2, 4); Q(5, 6)
2. A(-1, -2); B(2, -3)  and   P(5, 4) ; Q(6, 7)

Solution

1. Let m1 be the gradient joining A and B and m2 be the gradient joining P and Q.

m1=2-(-1)3-5=-32

m2=6-45-2=23

Since  m1m2=-1AB is perpendicular to PQ

1. Let m1 be the gradient joining A and B and m2 be the gradient joining P and Q.

m1=-3-(-2)2-(-1)=-13

m2=7-46-5=3

Since  m1m2=-1; AB is perpendicular to PQ

EQUATION OF A LINE

The equation of a straight line is given by: y =mx + c

Example: Find the gradient and intercept on the y-axis of the following lines:

1. y = 3x – 4
2. y = – ½x – 3

Solution:

1. Compare y = 3x – 4 with y = mx + c ; Hence the gradient is 3, intercept on y-axis is -4
2. Gradient is – ½ , intercept on y-axis

Example: Find the equation of a straight line of slope 2, if it passes through the point (3, -2)

y – y1=m(x-x1)

m = 2; x1=3 and y1=-2

Hence the equation of the straight line is:

y – (-2) = 2(x – 3)

y + 2 = 2x – 6

y = 2x -6 -2 = 2x – 8

y = 2x – 8

WEEK 5

### Topic:Vectors

Sub-topic: Modulus of a vector

Duration: 80 minutes

Learning Objectives: By the end of the lesson, students should be able to perform simple operations on vectors.

Reference Materials:  New Further Mathematics Project 2 by M. R Tuttuh Adegun

Previous Knowledge: Students can perform arithmetic operations on vectors

Instructional Materials:  Mathematical set.

Content: MAGNITUDE OF A VECTOR

The magnitude of a vector a, sometimes called the modulus of the vector is represented by |a|.

Zero Vector: The zero vector is a vector with zero magnitude.

Unit Vector: The unit vector is the vector represented by a and is such that a = |a| a

Negative Vector: The negative vector of a is written as – a

Equality of vector: Two vectors are equal when they have same magnitude and direction.

Example: Find the modulus of each of the following vectors

1. 3i + 4j
2. -2i – 5j
3. isinθ-jcosθ

Solution

1. Let r = 3i + 4j ; then |r| = 32+42=25=5
2. Let r = -2i – 5j ; then |r| = 2252=4+25=29
3. Let r = isinθ-jcosθ; then |r| = sinθ2+(-cos⁡)2=1=1

Example: If r1=7i+3j, r2=2i-5j ; find the modulus and direction cosines of:

1. r1+r2
2. r2r1

Solution

r1=7i+3j, r2=2i-5j

1. r1+r2=9i-2j

|r1 + r2| = 92+(-2)2=81+4=85

Let cos1and cos1 be the direction cosines of r1+r2

cos1=985

cos1=-285

1. r2r1=-5i-8j

|r2r1| = (-5)2+(-8)2=25+64=89

Let cos2and cos2 be the direction cosines of r2r1

cos α2=-589

cos β2=-889

UNIT VECTOR

Example: Find the unit vectors in the directions of the following vectors

1. r = 21 + 3j
2. q = 4i – 5j
3. p = 7i + 2j – 3k
4. t = 3i -5j -3k

Solution

1. Let r be the unit vector in the direction of r; then

r=rr=1132i+3j

1. Let q be the unit vector in the direction of q; then

q=qq=1414i-5j

1. Let p be the unit vector in the direction of p; then

p=pp=1627i+2j-3k

1. Let t be the unit vector in the direction of t; then

t=tt=1433i-5j-3k

ARITHMETIC OPERATIONS ON VECTORS

Example: If p = 2i –  3j; q =  3i + 5j and r = i + j; Find the values of

1. 2p + q + 3r
2. 3p – 2q

Solution

1. 2p = 2(2i – 3j ) = 4i – 6j

3r = 3( i + j ) = 3i + 3j

Therefore; 2p + q + 3r = (4i – 6j) + (3i + 5j) + (3i + 3j)

= 10i + 2j

1. 3p = 3(3i – 3j) = 9i – 9j

2q = 2(3i + 5j) =  6i + 10j

Therefore 3p – 2q = (9i – 9j) – (6i + 10j) =3i – 19j

Example: Given that OC= a – b and OD= 2a + 3b, where a = 2i + 3j and b = 3i – 2j, find CD

CD=CO+OD=ODOC

= (2a + 3b) – (a – b)

= 2a + 3b – a + b = a + 4b

= (2i + 3j) + 4(3i – 2j)   = 14i – 5j

Evaluation: New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 5

Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.

Assignment: New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 6

WEEK 6

### MAGNITUDE OF A VECTOR

The magnitude of a vector a, sometimes called the modulus of the vector is represented by |a|.

Zero Vector: The zero vector is a vector with zero magnitude.

Unit Vector: The unit vector is the vector represented by a and is such that a = |a| a

Negative Vector: The negative vector of a is written as – a

Equality of vector: Two vectors are equal when they have same magnitude and direction.

Example: Find the modulus of each of the following vectors

1. 3i + 4j
2. -2i – 5j
3. isinθ-jcosθ

Solution

1. Let r = 3i + 4j ; then |r| = 32+42=25=5
2. Let r = -2i – 5j ; then |r| = 2252=4+25=29
3. Let r = isinθ-jcosθ; then |r| = sinθ2+(-cos⁡)2=1=1

Example: If r1=7i+3j, r2=2i-5j ; find the modulus and direction cosines of:

1. r1+r2
2. r2r1

Solution

r1=7i+3j, r2=2i-5j

1. r1+r2=9i-2j

|r1 + r2| = 92+(-2)2=81+4=85

Let cos1and cos1 be the direction cosines of r1+r2

cos1=985

cos1=-285

1. r2r1=-5i-8j

|r2r1| = (-5)2+(-8)2=25+64=89

Let cos2and cos2 be the direction cosines of r2r1

cos α2=-589

cos β2=-889

Evaluation: New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 10

Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.

Assignment: New Further Mathematics Project 2, by M.R Tuttuh Adegun

UNIT VECTOR

Example: Find the unit vectors in the directions of the following vectors

1. r = 21 + 3j
2. q = 4i – 5j
3. p = 7i + 2j – 3k
4. t = 3i -5j -3k

Solution

1. Let r be the unit vector in the direction of r; then

r=rr=1132i+3j

1. Let q be the unit vector in the direction of q; then

q=qq=1414i-5j

1. Let p be the unit vector in the direction of p; then

p=pp=1627i+2j-3k

1. Let t be the unit vector in the direction of t; then

t=tt=1433i-5j-3k

Evaluation: New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 10

Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.

Assignment: New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 12

ARITHMETIC OPERATIONS ON VECTORS

Example: If p = 2i –  3j; q =  3i + 5j and r = i + j; Find the values of

1. 2p + q + 3r
2. 3p – 2q

Solution

1. 2p = 2(2i – 3j ) = 4i – 6j

3r = 3( i + j ) = 3i + 3j

Therefore; 2p + q + 3r = (4i – 6j) + (3i + 5j) + (3i + 3j)

= 10i + 2j

1. 3p = 3(3i – 3j) = 9i – 9j

2q = 2(3i + 5j) =  6i + 10j

Therefore 3p – 2q = (9i – 9j) – (6i + 10j) =3i – 19j

Example: Given that OC= a – b and OD= 2a + 3b, where a = 2i + 3j and b = 3i – 2j, find CD

CD=CO+OD=ODOC

= (2a + 3b) – (a – b)

= 2a + 3b – a + b = a + 4b

= (2i + 3j) + 4(3i – 2j)   = 14i – 5j

Evaluation: New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 5

Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.

Assignment: New Further Mathematics Project 2, by M.R Tuttuh Adegun et al. Page 262, Exercise 14, no 6

WEEK 8

### Topic:Straight line

Sub-topic: Angle between lines

Duration: 80 minutes

Learning Objectives: By the end of the lesson, students should be able to calculate the angle between two lines.

Reference Materials:  New Further Mathematics Project 2 by M. R Tuttuh Adegun

Previous Knowledge: Students can draw the graph of a linear equation (straight-line graph).

Instructional Materials: Graph board and graph book.

Content: ANGLE  BETWEEN TWO LINES

The acute angle between lines of gradient m1 and m2

tanα=m2m11+m1m2

Example: Find the acute angle between the lines x + 4y = 12 and y – 2x + 6 =0.

Solution

The gradients are -1/4 and 2

tanα=14-21+(-14.2)= |4.5| =77.47o

The gradient intercept form of the equation of a line is y = mx + c

Example: Determine the equation of the line whose gradient is -2 and y-intercept is 3..

Solution

Let the equation of the line be y = mx + c, where m = -2 and c = 3

Hence, the equation of the line is y = -2x + 3

Presentation:

Step I: Teacher revises the last topic with the students and does necessary corrections.

Step II: Teacher introduces the new topic to the students and explains by giving illustrative examples.

Step III: Teacher welcomes and answers questions from the students.

Step IV: Teacher gives notes to the students and ensures they copy correctly.

Step V: Teacher evaluates the students on topic discussed.

Evaluation: 1. Determine the equation of the line whose gradient is 3 and y-intercept is -4.

1. Find the acute angle between the lines 2y = 3x – 8 and 5y = x + 7.

Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.

Assignment: 1. Determine the equation of the line whose gradient is 3¼  and y-intercept is -6.

1. Find the acute angle between the lines 2y = – 5x + 8 and y = 3x – 7.

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PERIOD 3

Topic: Equation of a straight line

Sub-topic: Gradient and one point form and two point form

Duration: 40 minutes

Learning Objectives: By the end of the lesson, students should be able to determine the equation of a line in different forms.

Reference Materials: i. New Further Mathematics for SSS 2 Project 2.

Previous Knowledge: Students can calculate angle between two lines.

Instructional Materials: Graph book.

Content: GRADIENT AND ONE POINT FORM

Equation of a line through (x, y) with gradient m is y – y1 = m(x – x1)

Example: A straight line has a gradient of -3/2 and passes through the point (1, 4). Find its equation and its intercept on the y-axis

Solution

In this case (x, y) = (1, 4) and m = – 3/2

So the equation is

y – 4 = – 32(x – 1)

• 2y – 8 = -3(x – 1)
• 2y + 3x =11

So y = – 32x+112 ; Hence the intercept on y-axis is 5½

Equation of a line through the two points (x1, y1) and (x2, y2)is y – y1y2y1 = x – x1x2x1

Example: Find the equation of a line AB which passes through the points (1, -1) and (-2, -13)

Solution

Using y – y1y2y1 = x – x1x2x1; y--1x-1=-13--1-2-1

Therefore y + 1 = 4x – 4

• y = 4x – 5.

Thus the gradient of AB is 4.

Evaluation:  Find the equation of a line AB which passes through the points (-2, -3) and (-2, -13)

Conclusion: Teacher summarizes the topic, marks the students’ notes, does correction and allows the students to copy.

Assignment: Find the equation of a line AB which passes through the points (-1, 2) and (3, 0)

Assignment: New General Mathematics for SSS 2, by M.F Macrae et al. Page 190, Exercise 16d, no 2a, 2c

MID TERM TEST FIRST TERM FURTHER MATHS SS 1

Further Mathematics Third Term SSS1 Revision

### SCHEME OF WORK FURTHER MATHEMATICS SSS 1 THIRD TERM

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