# 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.

*Content* **FLOWCHART**

*Content*

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.

**Advantages of flowcharts**

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

**WEEK 3**

** Topic:** Gradients of straight lines and curves

** Sub-topic:** Gradients of straight lines

** 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

**may be any number.**

*c*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

**may be any number.**

*a*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 – y1 = x – 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

**GRADIENT OF A CURVE**

**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

= gradient of tangent PT

= MP = 2.25 = 2¼ = 9 = 11/2

TM 1.5 1.5 6

(b) Gradient of curve where x = – 2

= gradient of tangent QR

= – 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.

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

**Solution**

- 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**

- 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.

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

**Solution**

- 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=-1; **AB is perpendicular to PQ**

- 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:

- y = 3x – 4
- y = – ½x – 3

**Solution:**

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

**GRADIENT AND ONE POINT FORM**

**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

*Topic:*

** 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

- 3i + 4j
- -2i – 5j
- isinθ-jcosθ

**Solution**

- Let r = 3i + 4j ; then |r| = 32+42=25=5
- Let r = -2i – 5j ; then |r| = –22–52=4+25=29
- 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:

- r1+r2
- r2–r1

**Solution**

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

- 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

- r2–r1=-5i-8j

|r2–r1| = (-5)2+(-8)2=25+64=89

Let cos2and cos2 be the direction cosines of r2–r1

cos α2=-589

cos β2=-889

**UNIT VECTOR**

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

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

**Solution**

- Let r be the unit vector in the direction of r; then

r=rr=1132i+3j

- Let q be the unit vector in the direction of q; then

q=qq=1414i-5j

- Let p be the unit vector in the direction of p; then

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

- 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

- 2p + q + 3r
- 3p – 2q

**Solution**

- 2p = 2(2i – 3j ) = 4i – 6j

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

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

= **10i + 2j**

- 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=OD–OC

= (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

- 3i + 4j
- -2i – 5j
- isinθ-jcosθ

**Solution**

- Let r = 3i + 4j ; then |r| = 32+42=25=5
- Let r = -2i – 5j ; then |r| = –22–52=4+25=29
- 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:

- r1+r2
- r2–r1

**Solution**

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

- 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

- r2–r1=-5i-8j

|r2–r1| = (-5)2+(-8)2=25+64=89

Let cos2and cos2 be the direction cosines of r2–r1

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

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

**Solution**

- Let r be the unit vector in the direction of r; then

r=rr=1132i+3j

- Let q be the unit vector in the direction of q; then

q=qq=1414i-5j

- Let p be the unit vector in the direction of p; then

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

- 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

- 2p + q + 3r
- 3p – 2q

**Solution**

- 2p = 2(2i – 3j ) = 4i – 6j

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

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

= **10i + 2j**

- 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=OD–OC

= (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

*Topic:*

** 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=m2–m11+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

**GRADIENT INTERCEPT FORM**

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.

- 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.

- 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 – y****1**** = m(x – x****1****)**

**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

**GRADIENT AND TWO POINT FORM**

Equation of a line through the two points (x1, y1) and (x2, y2)is y – y1y2–y1** = **x – x1x2–x1

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

**Solution**

Using y – y1y2–y1** = **x – x1x2–x1**; **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