MINISTRY OF EDUCATION
SECONDARY ENGAGEMENT PROGRAMME
EASTER TERM 2021
GRADE 11
MATHEMATICS
WEEK 1
LESSON 1
TOPIC: Vectors and Matrices
SUB-TOPIC: Order of Matrices
CONTENT:
Matrix Definition
A matrix is a rectangular array of numbers arranged in rows and columns. The array of numbers
below is an example of a matrix.
3
4
[−1 7 ]
4 −5
The number of rows and columns that a matrix has is called its dimension or its order. By
convention, rows are listed first; and columns, second. Thus, we would say that the dimension
(or order) of the above matrix is 3 x 2, meaning that it has 3 rows and 2 columns.
Numbers that appear in the rows and columns of a matrix are called elements of the matrix. In
the above matrix, the first element in the first column or we can say the first element of the first
row is 3; the element in the second column of the first row is 4; and so on.
1
Example
1. A = [−7]
There is only one row and column in matrix A. The order of matrix A is 1×1.
2. B = [1
5 3]
There is one row and three columns in matrix B. The order of matrix B is 1×3.
3. C = [
6 −2
]
0
3
The elements are arranged in matrix C in 2 rows and 2 columns. The order of matrix C is 2×2.
4 2
4. D = [−1 5]
7 3
Matrix D is a matrix and formed by 3 rows and 2 columns. Therefore, the order of the matrix D
is 3×2.
Exercises:
State the order of the following matrices
1. [15
9
2. [7
−6]
3. [
-5]
−4
]
2
1
4. [7]
3
2
5. [
2 −1 4
]
3
3 5
13
6. [11
10
17
13]
14
3
4
6
7. [−5 −4
−8 ]
−2 −13 −11
8. [
3 8 12
6 9 14
13
]
4
13
16
22
19
9. [ 4
−11
−13
14
1
10. [17
3
6
4
15 13
14
5
23
21
7]
18
9 15
4 13]
9 10
3
WEEK 1
LESSON 2
TOPIC: Vectors and Matrices
SUB-TOPIC: Addition of Matrices
CONTENT: ADDITION OF MATRICES
Two matrices A and B are said to be conformable for addition if they have the same order (i.e., same
number of rows and columns).
To add two matrices: add the numbers in the matching positions:
The two matrices must be the same size, i.e. the rows must match in size, and the columns must
match in size.
Example on Addition of Matrices:
1. If A = [
1 4
2 5
] and B = [
], then
3 7
−1 3
A+B =[
2+1
−1 + 3
=[
3 9
]
2 10
5+4
]
3+7
−1
2
3
3
2. If A = [ 2 −3
1 ], B= [1
3
1 −2
2
−1 2
5
0 3] and M = [
1
−1 0
4
2
], then
4
−1
2
A + B = [ 2 −3
3
1
(−1) + 3
=[ 2+1
3+2
2
=[3
5
3
3 −1
1] + [1 0
−2
2 −1
2
3]
0
2 + (−1)
3+2
(−3) + 0
1+3 ]
1 + (−1) (−2) + 0
1
5
−3
4]
0 −2
A + M is not defined since the order of matrix M is not equal to the order of matrix A.
B + M is also not defined since the order of matrix M is not equal to the order of matrix B.
EXERCICES
1. Find the sum of A and B where A= [
2 3
2. Find A + B when A = [5 6
8 5
3. If A = [
2
−5
3
4
6
] and B = [
]
7
2 −11
4
3 −2 −3
7 ] and B = [5
4
3]
11
1
3
2
2 −1 4
0 −1 2
] and B = [
], compute A + B
−2 3 5
3
0 1
2 3
−2 1
0 4
4. If [
]+[
]=[
], find the value of 𝑥.
−5 7
𝑥 3
−3 9
5. Given A = [
1 4
−4 −1
] and B = [
], compute A + B.
2 3
−3 −2
13 17
22 27
6. Given M = [11 13] and N = [−13 9 ], find M + N.
10 14
−7 14
13 23
9
13
16
14
9
22
19 21
12
11
7. Given P = [ 4 −11 7 ] and Q = [ 14
2
1 ], compute P + Q.
−13
−12 −19 3
14 18
5
11
8. Compute [
12
14
12
]+[
13
22
24
].
13
−6 4
7 14
25
9. Compute [
]+ [
]+ [
−2 3
8 23
22
13
10. Given A = [11
10
17
11
13], B = [ 9
14
11
24
]
23
14
1 0
6 ] and C = [ 0 2], compute A + B + C.
4
−4 1
6
WEEK 1
LESSON 3
TOPIC: Vectors and Matrices
SUB-TOPIC: Subtraction of Matrices
CONTENT:
Two matrices A and B are said to be conformable for subtraction if they have the same order (i.e.
same number of rows and columns) and their difference A - B is defined to be the addition of A
and (-B).
i.e., A – B = A + (-B)
Negative
The negative of a matrix is also simple:
Subtracting
To subtract two matrices: subtract the numbers in the matching positions:
Note: subtracting is actually defined as the addition of a negative matrix: A + (−B)
Example on Subtraction of Matrices:
7
1. If A = [
1 4
2 5
] and B = [
], then
3 7
−1 3
A–B =[
=[
2−1
−1 − 3
5−4
]
3−7
1
1
]
−4 −4
−1
2
3
3
2. If A = [ 2 −3
1 ], B= [1
3
1 −2
2
−1
2
3
3
A – B = [ 2 −3
1 ] – [1
3
1 −2
2
(−1) − 3
=[ 2−1
3−2
−1 2
5
0 3] and M = [
1
−1 0
2
], then
4
−1 2
0 3]
−1 0
2 − (−1)
(−3) − 0
1 − (−1)
3−2
1−3]
(−2) − 0
−4
3
1
=[ 1 −3 −2]
1
2 −2
A – M is not defined since the order of matrix M is not equal to the order of matrix A.
B – M is also not defined since the order of matrix M is not equal to the order of matrix B.
EXERCICES
1. Given the Matrices A= [
2 3
4
6
] and B = [
], evaluate 𝐴 − 𝐵
−5 7
2 −11
2
2. Find A – B when A = [5
8
3 4
3
6 7 ] and B = [5
5 11
1
8
−2 −3
4
3]
3
2
3. If A = [
2 −1 4
0 −1 2
] and B = [
], compute 𝐵 − 𝐴
−2
3 5
3
0 1
2 3
−2 1
0 4
4. If [
] – [
]=[
], find the value of 𝑥.
−5 7
𝑥 3
−3 9
5. Given A = [
1 4
−4 −1
] and B = [
], compute A – B.
2 3
−3 −2
13 17
22
6. Given M = [11 13] and N = [ −13
10 14
−7
16
22
7. Given P = [ 4
−13
11
8. Compute [
12
27
9 ], find M – N.
14
13 23
14
19 21
12
−11 7 ] and Q = [ 14
−12
14 18
9 13
9 11
2
1 ], compute P – Q.
−19 3
14
12 24
] – [
]
13
22 13
−6 4
7 14
25 24
9. Compute [
]– [
]– [
]
−2 3
8 23
22 23
13
10. Given A = [11
10
17
11
],
B
=
[
13
9
14
11
14
1 0
]
and
C
=
[
6
0 2], compute A – B – C.
4
−4 1
9
WEEK 1
LESSON 4
TOPIC: Vectors and Matrices
SUB-TOPIC: Scalar Multiplication
CONTENT:
Multiply by a Constant
We can multiply a matrix by a constant (the value 2 in this case):
We call the constant a scalar, so officially this is called "scalar multiplication"
EXERCISES
Complete the following
1.
2.
3.
4.
10
5.
6.
7.
8.
9.
2
5
10. 5 [
−15
−10 −5
]
10 15
11
WEEK 2
LESSON 1
TOPIC: Vectors and Matrices
SUB-TOPIC: Operations with Matrices
CONTENT: Addition and Subtraction of Matrices and Scalar Multiplication
Example: Given A = (
1
4
2 −1
3 −1
),B=(
) and C = (
), find
0 −6
−4
3
−1
2
1. A + B – 2C
2. A – 3B
Solution:
1
0
1. A + B – 2C = (
=(
4
2 −1
3 −1
)+(
) – 2(
)
−6
−4
3
−1
2
1 + 3 − 2(2)
4 + (−1) − 2(−1)
)
(−1)
0+
− 2(−4)
−6 + 2 − 2(3)
=(
1+3−4
0−1+8
=(
0
7
4−1+2
)
−6 + 2 − 6
5
)
−10
1
4
3 −1
) – 3(
)
0 −6
−1
2
1 − 3(3) 4 − 3(−1)
=(
)
0 − 3(−1) −6 − 3(2)
2. A – 3B = (
1−9 4+3
)
0 + 3 −6 − 6
−8
7
=(
)
3 −12
=(
12
EXERCISE
𝑏
𝑐
1. Given A = (
3 2
7 3
1
), B = (
) and C = (
) and that A+B = C, determine the
𝑐 1
8 10
9
values of 𝑏 and 𝑐.
0 12
𝑎 9
2 −3
), B = (
) and C = (
), find:
8 −6
−3 𝑏
−11
4
2. Given that A = (
(a) 2A
(b) 6B
(c) A + B
(d) 2A – 3B
(e)
1
2
B+A
(f) 𝑎 𝑎𝑛𝑑 𝑏 if C = A + B.
13
WEEK 2
LESSON 2
TOPIC: Vectors and Matrices
SUB-TOPIC: Order of multiplication
CONTENT:
When we are finding the product of two matrices, we first check if they are compatible for
multiplication. The number of columns in the first matrix must be the same as the number of
rows in the second matrix.
9
2 3 1
[
] X [7
0 4 5
2
2X3
8
6]
4
3X2
C1 = R2, hence we can proceed to multiply these matrices
The order of the multiplication will consist of the row of the first matrix and the column of the
second matrix.
[
9
2 3 1
] X [7
0 4 5
2
2X3
8
6]
4
3X2
Order of multiplication (the order of the product)
=2X2
14
EXERCISE
Find the order of matrix multiplication of the following
15
16
WEEK 2
LESSON 3
TOPIC: Vectors and Matrices
SUB-TOPIC: Matrix Multiplication
CONTENT:
When we are finding the product of two matrices, we first check if they are compatible for
multiplication. The number of columns in the first matrix must be the same as the number of
rows in the second matrix.
9
2 3 1
[
] X [7
0 4 5
2
2X3
8
6]
4
3X2
C1 = R2, hence we can proceed to multiply these matrices.
When we multiply two matrices, we use the row of the first with the column of the second.
We know that our product should be a 2 x 2 matrices. Hence,
[
𝑅1𝐶1 𝑅1𝐶2
2
]=[
𝑅2𝐶1 𝑅2𝐶2
0
3 1
] X
4 5
9 8
(2 × 9) + (3 × 7) + (1 × 2)
[7 6 ] = [
𝑅2𝐶1
2 4
=
[
(2 × 9) + (3 × 7) + (1 × 2) (2 × 8) + (3 × 6) + (1 × 4)
]
(0 × 9) + (4 × 7) + (5 × 2) (0 × 8) + (4 × 6) + (5 × 4)
=[
18 + 21 + 2 16 + 18 + 4
]
0 + 28 + 10 0 + 24 + 20
17
𝑅1𝐶2
]
𝑅2𝐶2
=[
41
38
38
]
44
EXERCISES:
Find the following products, using the matrices P = [2
T=[
4
2 −1
].
0 −3 −1
1. PQ
2. QP
3. TS
4. ST
5. SP
6. QS
7. RT
8. R2
18
9
3
5
5
], S = [11] and
−6], Q = [ ], R = [
4
5 4
10
WEEK 2
LESSON 4
TOPIC: Vectors and Matrices
SUB-TOPIC: Multiplying 2X2 matrices
CONTENT:
1 3
2
) and B =(
2 4
4
Given A = (
3
), determine AB.
1
Solution
1
2
AB = (
14
20
AB = (
(1 × 2) + (3 × 4)
3 2 3
)(
)=(
(2 × 2) + (4 × 4)
4 4 1
(1 × 3) + (3 × 1)
2 + 12 3 + 3
) =(
)
(2 × 3) + (4 × 1)
4 + 16 6 + 4
6
)
10
EXERCISES
2 3
3 1
) and B =(
), determine AB.
1 5
4 2
3 −1
−2 4
If C = (
) and D =(
), determine CD.
2
5
−3 1
4 −2
−2
5
If A = (
) and B =(
), determine AB.
3
2
1 −3
4 −2
−2
5
If M = (
) and N =(
), determine MN.
3
2
1 −3
−2 4
3 −1
If P = (
) and Q = (
), determine PQ
−3 1
2
5
1. If A = (
2.
3.
4.
5.
19
WEEK 3
LESSON 1
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: DETERMINANT OF 2X2 MATRICES
CONTENT:
The symbol for determinant is two vertical lines either side.
Example:
|A| means the determinant of the matrix A
Determinant of a 2×2 Matrix
The determinant of matrix A is calculated as
Leading diagonal
Non-leading diagonal
|A| = product of the leading diagonal – product of the non-leading diagonals
|A| = a x d – b x c
Example 1: Find the determinant of the matrix below.
20
Example 2: Find the determinant of the matrix below.
Example 3: Evaluate the determinant of the matrix below.
Example 4: Evaluate the determinant of the matrix below.
Example 5: Find the value of 𝑥 in the matrix below if its determinant is -12.
21
EXERCISE
Evaluate the following determinants:
8
−6
)
−10 9
1. (
2 5
)
1 −3
5.
5 −4
(
)
−3 2
9.
(
2. (
3 −5
)
2 1
6.
0 −4
(
)
−6 −2
10.
(
3. (
−2 4
)
−6 2
7.
−1 1
(
)
−1 4
4. (
−3 −1
)
4 −5
8.
10 −9
(
)
−7 3
22
2 −2
)
7 −7
WEEK 3
LESSON 2
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: SINGULAR AND NON-SINGULAR MATRICES
CONTENT:
𝑎
A matrix is singular if its determinant is 0. Given A =[
𝑐
𝑏
],
𝑑
If |A| = 0, then A is a singular matrix
If |A| ≠ 0, then A is a non-singular matrix.
Example.
Determine the value of 𝑘 if C =[
𝑘
4
−2
] is a singular matrix.
4
Solution
Since C is a singular matrix,
Then |C| = 0
(𝑘 × 4) − −2 × 4 = 0
4𝑘 + 8 = 0
4𝑘 = −8
𝑘=
−8
4
𝑘 = −2
23
EXERCISE
1. M is the matrix(
−4
𝑥
). What value of 𝑥 makes M singular?
−10 −5
2. Determine whether the following is a singular or non-singular matrix
2
5
)
1 −3
(a) (
3 6
)
2 4
(b) (
−2 4
)
−1 2
(c) (
−3
2
)
9 −6
(d) (
5 −4
)
−3
2
(e) (
0 −4
)
−6 −2
(f) (
−1 1
)
−1 1
(g) (
10 −9
)
−7
3
(h) (
−15 −6
)
−10
4
(i) (
2 −2
)
7 −7
(j) (
24
WEEK 3
LESSON 3
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: ADJOINT OF 2X2 MATRICES
CONTENT:
The Adjoint of a 2x2 matrix
If M =
(
𝑎
𝑐
𝑏
)
𝑑
Then M adjoint =
(
𝑑
−𝑐
−𝑏
)
𝑎
Interchange the elements on the leading diagonal.
Multiple the elements on the non-leading diagonals by (-1).
EXAMPLE:
8
5
)?
−1 −3
−3 −5
Solution: B adjoint = (
)
1
8
1. What is the adjoint of B =(
−11 −4
)?
−1 −3
−3
4
Solution: C adjoint = (
)
1 −11
2. What is the adjoint of C =(
25
EXERCISE
State the Adjoint of the following matrices:
1. (
2
5
)
1 −3
2. (
3 −5
)
2
1
3. (
−2 4
)
−6 2
4. (
−3 −1
)
4 −5
5. (
5 −4
)
−3
2
6. (
0 −4
)
−6 −2
7. (
−1 1
)
−1 4
8. (
10
−7
9. (
8
−6
)
−10
9
10. (
2 −2
)
7 −7
−9
)
3
26
WEEK 3
LESSON 4
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: INVERSE OF 2X2 NON SINGULAR MATRICES
CONTENT:
Adjoint
EXAMPLE:
4 5
)?
1 2
Determine the inverse of A =(
A -1 =
A -1 =
A -1 =
A -1 =
1
|𝐴|
× A adjoint
1
2
×(
(4×2)−(5×1)
−1
1
8−5
1
3
× (−12
−5
)
4
−5
)
4
2
(
2 −5
)
−1
4
3
A -1 = (−1
3
27
−5
3
4
3
)
EXERCISE
Determine the inverse of the following matrices:
𝟐
𝟏
𝟓
)
−𝟑
(
𝟑
𝟐
−𝟓
)
𝟏
(
−3 −1
)
4 −5
1.
(
𝟐.
3.
5 −4
)
−3
2
4.
(
5.
(
6.
(
7.
(
8.
(
9.
(
10.
(
0 −4
)
−6 −2
10 −9
)
−7
3
8 −6
)
−10
9
−2 4
)
−6 2
−1 1
)
−1 4
2
7
−2
)
−7
28
WEEK 4
LESSON 1
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: Proving A-1A = I
CONTENT:
4 5
), prove that A-1A = I
1 2
Given A =(
A -1 =
A -1 =
A -1 =
A -1 =
1
|𝐴|
× A adjoint
1
(4×2)−(5×1)
1
8−5
1
× (−12
× (−12
−5
)
4
−5
)
4
2 −5
)
4
3 −1
A-1A =
A-1A =
A-1A =
A-1A =
(
1
2 −5 4 5
)(
)
4 1 2
3 −1
1
(
(2 × 4) + (−5 × 1)
3 (−1 × 4) + (4 × 1)
1
(
(2 × 5) + (−5 × 2)
)
(−1 × 5) + (4 × 2)
8 + (−5) 10 + (−10)
)
3 −4 + 4
−5 + 8
1
(
3 0
)
3 0 3
(
3
A-1A =
(30
3
0
3
3)
3
0
)
1
1 0
Since I =(
), then A-1A = I
0 1
A-1A = (
1
0
29
EXERCISE
3 −5
), prove that A-1A = I
2
1
1. Given A = (
2. Given B =(
−3 −1
), prove that B-1B = I
4 −5
3. Given C = (
5 −4
), prove that C-1C = I
−3
2
−3 2
−1 0
4. The matrices A and B are given as A = [
] and B = [ 1 1].
3 2
4 6
(i)
Determine 𝐴−1 , the inverse of A
(ii) Show that 𝐴−1 𝐴 = 𝐼, the identity matrix.
(iii) Determine the matrix 𝐴2 .
(iv) a)
b)
Explain why the matrix product AB is NOT possible.
Without calculating state the other of the matrix product BA.
30
WEEK 4
LESSON 2
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: SOLVING SIMULTANEOUS EQUATIONS USING MATRIX
METHOD
CONTENT:
Example:
Solve by matrix method the pair of simultaneous equations:
4𝑝 + 3𝑞 = 7 𝑎𝑛𝑑 2𝑝 − 𝑞 = 11
Solution
4𝑝 = 3𝑞 = 7
2𝑝 − 𝑞 = 11
(
7
4
3 𝑝
) (𝑞 ) = ( )
11
2 −1
To solve we will use the inverse of the 2×2 matrix.
4
3
)
2 −1
Let A = (
Then A-1 =
A-1 =
A-1 =
1
−1
(
(4×(−1))−(2×3) −2
1
−3
)
4
−1 −3
)
4
−4−6 −2
(
1
−1
(
−10 −2
𝑝
Hence, (𝑞 ) =
1
−3
)
4
−1
−10 −2
(
−3 7
)( )
4 11
31
𝑝
(𝑞 ) =
𝑝
(𝑞 ) =
𝑝
(𝑞 ) =
1
(−1 × 7) + (−3 × 11)
)
−10 (−2 × 7) + (4 × 11)
1
(
−7 + (−33)
)
−10 (−14) + (44)
1
(
−40
)
30
−10
(
−40
𝑝
−10
(𝑞 ) = ( 30 )
−10
𝑝
4
(𝑞 ) = ( )
−3
Hence, 𝑝 = 4 and 𝑞 = −3.
EXERCISE
Solve by matrix method the pair of simultaneous equations:
1. 𝑥 − 𝑦 = 3 𝑎𝑛𝑑 𝑥 + 𝑦 = 13
2. 2𝑎 + 3𝑏 = 3 𝑎𝑛𝑑 𝑎 + 3𝑏 = 9
3. 3𝑠 − 2𝑡 = 12 𝑎𝑛𝑑 2𝑠 + 𝑡 = 1
4. −3𝑥 + 2𝑦 = −11 𝑎𝑛𝑑 5𝑥 + 4𝑦 = 33
5. 2𝑥 – 2𝑦 – 3 = 0 𝑎𝑛𝑑 8𝑦 = 7𝑥 + 2
32
7.
8.
33
WEEK 4
LESSON 3
TOPIC: Vectors and Matrices
SUB-TOPIC: Representing Vectors
CONTENT:
A vector quantity is defined as one having magnitude (or size) and direction. Examples of
vectors are weight, velocity, acceleration and displacement.
𝒚
𝑥
𝑥
Vector AB can be represented by the column matrix (or column vector), (𝑦). Thus, AB = (𝑦)
34
Example:
6
AB = ( )
7
35
EXERCISE
1. Write the following as column vectors
2. Represent each of the following vectors on graph paper.
4
(a) 𝒘 = ( )
5
(b) 𝒙 = (
−1
)
−5
2
)
−6
(c) 𝒚 = (
36
−3
)
5
(d) 𝒛 = (
WEEK 4
LESSON 4
TOPIC: Vectors and Matrices
SUB-TOPIC: Equal and Inverse Vectors
CONTENT:
1. Two vectors are said to be equal if they both have the same magnitude and direction
2. When two vectors have the same magnitude but opposite directions, then one vector is said to
be the inverse of the other.
37
Example
State the relationship between each of the following pairs of vectors.
Solution
6
−6
(a) 𝒄 = ( ) and 𝒃 = ( )
3
−3
(a) 𝒄 = − 𝒃
5
5
) and q = ( )
12
12
(b) 𝑝 = (
(b) 𝒑 = 𝒒
3
15
) and s =(
)
−4
−20
(c) r = (
(c) 𝒔 = 𝟓𝒓
EXERCISE
State the relationship between each of the following pairs of vectors.
1. 𝒆 = (
−4
−4
) and 𝒇 = ( )
−5
−5
−2
2
) and 𝒎 = ( )
−5
5
2. 𝒍 = (
3. 𝒓 = (
4. 𝒄 = (
−2
−10
) and 𝒔 = (
)
−1.5
−7.5
−2
−18
) and 𝒃 = (
)
3
12
9
−2
−4.5
5. 𝒎 = (
) and 𝒏 = ( 13
)
−6.5
−2
38
MINISTRY OF EDUCATION
SECONDARY ENGAGEMENT PROGRAMME
EASTER TERM 2021
GRADE 11
MATHEMATICS
WEEK 5
LESSON 1
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: VECTOR ALGEBRA
CONTENT: Example:
If 𝒂 = (
−2
4
5
) , 𝒃 = ( ) and 𝒄 = ( ), find
3
0
1
(a) 2𝒂 + 𝒃 + 𝒄
(b) 3𝒂 – 4𝒄
Solution:
(a) 2𝒂 + 𝒃 + 𝒄 = 2 (
(b) 3𝒂 – 4𝒄 = 3(
2 (−2) + 5 + 4
−2
4
−4 + 5 + 4
5
5
)+ ( )+( )=(
)=(
)=( )
2(3) + 1 + 0
3
0
6+1+0
1
7
3(−2) − 4(4)
−2
4
−6 − 16
−22
) – 4( ) = (
)=(
)=(
)
3(3) − 4(0)
3
0
9−0
9
−1
5
2
EXERCISE: Given 𝒂 = ( ) , 𝒃 = ( ) and 𝒄 = ( ), calculate:
6
−4
3
1. 𝒂 + 𝒃
4. 5𝒃 – 4𝒄
7. 2𝒂 + 3𝒃 + 𝒄
2. 𝒃 + 𝒂
5. 4𝒃 – 3𝒄
8. 3𝒂 − 4𝒃 + 5𝒄
3. 𝒂 – 𝒃
6. 𝟐 𝒂 + b
𝟏
9. 5𝒂 − 3𝒃 − 2𝒄
10. 2a + c
WEEK 5
LESSON 2
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: POSITION VECTORS
CONTENT: Position Vectors are vectors drawn from the fixed point O, the origin, to any
point on the Cartesian plane.
Magnitude and direction of position vectors
|OP| = √𝒙𝟐 + 𝒚𝟐
𝒚
Direction = 𝜽 = tan-1 (𝒙)
6
Example: Determine the magnitude and direction of the position vector OA = ( ).
3
Solution
|OA| = √62 + 32 = √36 + 9 = √45 = 6.7 units
𝑦
3
𝜃 = tan-1 (𝑥 ) = tan-1 (6) = tan-1 (0.5) = 26.6°
Thus, OA makes and angle of 26.6° with the positive 𝑥 − 𝑎𝑥𝑖𝑠
EXERCISE.
1. Given the point P (3, 4), determine:
(a) |OP|
(b) the acute angle between OP and the 𝑥 − 𝑎𝑥𝑖𝑠.
2. The point P has the coordinates (4, 6), determine:
(a) |OP|
(b) the acute angle between OP and the 𝑥 − 𝑎𝑥𝑖𝑠.
3. Given the point A (-3, 5), determine:
(a) |OA|
(b) the acute angle between OA and the 𝑥 − 𝑎𝑥𝑖𝑠.
4. Given the point B (-5, 7), determine:
(a) |OB|
(b) the acute angle between OB and the 𝑥 − 𝑎𝑥𝑖𝑠.
5. The point M has the coordinates (-7, 9), determine:
(a) |OM|
(b) the acute angle between OM and the 𝑥 − 𝑎𝑥𝑖𝑠.
WEEK 5
LESSON 3
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: Displacement Vector
CONTENT: Magnitude of a Vector
The magnitude of a vector PQ is the distance between the initial point P and the end point Q. In
symbols the magnitude of PQ is written as |PQ |
Q (𝒙𝟐 , 𝒚𝟐 )
P (𝒙𝟏 , 𝒚𝟏 )
𝑥2 − 𝑥1
PQ = (𝑦 − 𝑦 )
2
1
|PQ| = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
𝑦 −𝑦
𝜃 = tan-1 (𝑥2−𝑥1 )
2
1
Example:
Given the points P (2, 4) and Q (5, 3), determine
(a) PQ
(b) |PQ|
Solution
𝑥2 − 𝑥1
5−2
3
PQ = (𝑦 − 𝑦 ) = (
)=( )
2
1
−1
3−4
|PQ|
= √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
= √(5 − 2)2 + (3 − 4)2 = √32 + (−1)2
= √9 + 1
= √10
= 3.16 units
EXERCISE
1. Given the points P (1, 3) and Q (4, 5), determine
(a) PQ
(b) |PQ|
2. Given the points P (−3, −4) and Q (−6, −7), determine
(a) PQ
(b) |PQ|
3. Given the points M (4, −1) and N (8, −7), determine
(a) MN
(b) |MN|
4. Given the points A (−5, 4) and B (−7, −3), determine
(a) AB
(b) |AB|
5. Given the points C (−4, −6) and D (−3, −6), determine
(a) CD
(b) |CD|
WEEK 5
LESSON 4
TOPIC: Vectors and Matrices
SUB-TOPIC: Triangle Law of vectors
CONTENT:
If two vectors are both going in the same direction (that is, if both are going in a clockwise
direction or both are going in an anticlockwise directions), then we can determine the sum of the
two vectors by completing a triangle. The sum of the two vectors is call the resultant vector.
OP = OA + AP
EXAMPLE:
Determine the resultant vector for the two given vectors.
SOLUTION
EXERCISES
Determine the resultant for each of the following vectors below.
1.
2.
3.
4.
5.
6.
WEEK 6
LESSON 1
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: Parallelogram Law of Vectors
CONTENT:
If two vectors are going in opposite directions, i.e. one clockwise and the other counter
clockwise, then the resultant vector can be obtained by completing a parallelogram.
EXERCISES
Determine the resultant for each of the following vectors below.
1.
2.
3.
4.
5.
6.
WEEK 6
LESSON 2
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: Resultant Vectors
CONTENT:
The resultant of two vectors can be found by either completing a parallelogram (if the vectors are
going in opposite directions) or completing a triangle 9 if the vectors are going in the same
directions).
Example.
−9
5
Given that the position vectors OP = ( ) and OR = ( ) are adjacent sides of a parallelogram
6
3
OPQR, determine:
(a) The resultant of the two vectors, OQ.
(b) The magnitude of OQ , |OQ|.
(c) The acute angle between OQ and the 𝑥 − 𝑎𝑥𝑖𝑠.
Solution:
5 + (−9)
−9
−4
5
(a) The resultant of two vectors, OQ = OP + OR = ( ) + ( ) = (
)=( )
6
9
3
3+6
(b) |OQ| = √(−4)2 + 92 = √16 + 81 = √97 = 9.85units ( 2 dp)
Hence, the magnitude of OQ , |OQ| is 9.85 units
9
(c) 𝜃 = tan-1 (−4) = tan-1(-2.25) = 66.0° ( correct to 1 dp)
Hence, the acute angle between OQ and the x-axis is 66°.
EXERCISE
4
1
6
2
1. Given that the position vectors OP = ( ) and OR = ( ) are adjacent sides of a parallelogram
OPQR, determine:
(a) The resultant of the two vectors, OQ.
(b) The magnitude of OQ , |OQ|.
(c) The acute angle between OQ and the 𝑥 − 𝑎𝑥𝑖𝑠.
2. The points P (3, 5) and R (7, 9) are opposite vertices of a parallelogram OPQR. Determine:
(a) The resultant of the two vectors, OQ.
(b) The magnitude of OQ , |OQ|.
(c) The acute angle between OQ and the 𝑥 − 𝑎𝑥𝑖𝑠.
3. Two opposite sides of a parallelogram OABC are A(−8, 3) and C(−5, 7). Determine:
(a) The resultant of the two vectors, OB.
(b) The magnitude of OB , |OB|.
(c) The acute angle between OB and the 𝑥 − 𝑎𝑥𝑖𝑠.
−6
−2
) and OB = ( ) are adjacent sides of a
5
1
4. Given that the position vectors OA = (
parallelogram OABC, determine:
(a) The resultant of the two vectors, OB.
(b) The magnitude of OB , |OB|.
(c) The acute angle between OB and the 𝑥 − 𝑎𝑥𝑖𝑠.
5. The points K (−3, 8) and M (7, −4) are opposite vertices of a parallelogram OKLM.
Determine:
(a) The resultant of the two vectors, OL.
(b) The magnitude of OL , |OL|.
(c) The acute angle between OL and the 𝑥 − 𝑎𝑥𝑖𝑠.
WEEK 6
LESSON 3
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: Application of Vectors
CONTENT: Example
1. ABCD is a quadrilateral, not drawn to scale, with AB = a, BC= b and AD = 2BC. The point
X divides BD in the ratio 3:2.
(i) Express BD and BX in terms of a and b
1
(ii) Show that XC = 5 (3𝑎 − 𝑏)
Solution:
(i)
𝐵𝐷 = 𝐵𝐴 + 𝐴𝐷
𝐵𝑋 =
𝐵𝐷 = −𝐴𝐵 + 𝐴𝐷
𝐵𝑋 =
𝐵𝐷 = −𝑎 + 2𝑏
(ii)
𝑋𝐶 = 𝑋𝐵 + 𝐵𝐶
𝑋𝐶 = −𝐵𝑋 + 𝐵𝐶
𝑋𝐶 = −
3
(−𝑎 + 2𝑏) + 𝑏
5
6
𝑎 − 𝑏 + 𝑏
5
1
𝑎 − 𝑏
5
3
5
3
𝑋𝐶 =
5
1
𝑋𝐶 = (3𝑎 – 𝑏)
5
𝑋𝐶 =
EXERCISES
1. In the diagram below OP =12p, OQ = 12q. OX = 4p and OY = 4q.
(a) State PQ and XY in terms of p and q
1
(b) Given that XR =4 XQ, write PR and YR in terms of p and q.
(c) Prove by a vector method that the points Y, R and P are collinear.
3
5
3
5
𝐵𝐷
(−𝑎 + 2𝑏)
4
2 6
2. WXYZ is a parallelogram. The position vectors of W, X, Y are: ( ), ( ) 𝑎𝑛𝑑 ( )
7
1 3
respectively.
𝑥
(a) Determine in the form (𝑦), the vectors
(i) WX
(ii) XY
(iii) WZ
(iv) OZ
(b) M is the mid-point of WY and N is the mid-point of XZ. Show by a vector method that:
(i) OM = ON
(ii) |WX| = |XY|
(c) Describe fully the shape of the figure WXYZ.
WEEK 6
LESSON 4
TOPIC: VECTORS AND MATRICES
SUB-TOPIC: Application of Vectors and Matrices
CONTENT:
Answer the question below.
2.
3.
4.
WEEK 7
LESSON 1
TOPIC: Transformation Geometry
SUB-TOPIC: Translation
CONTENT: A translation is a transformation that slides an object to its image position.
OBJECT + TRANSLATION = IMAGE
* TRANSLATION = IMAGE – OBJECT
* OBJECT = IMAGE - TRANSLATION
Example:
1. The vertices of triangle ABC are (−3, 2), (−6, 4) and (−7, 2) respectively. Determine the
7
).
−3
vertices of triangle AʹBʹCʹ under translation 𝑇 = (
Since Object + Translation = Image
Then A
(
B
−3 −6
2
4
C
+
T
=
Aʹ
−3 + 7
−7
7
) + ( )=(
2 + (−3)
2
−3
Bʹ
−6 + 7
4 + (−3)
Cʹ
Aʹ
Bʹ
Cʹ
−7 + 7
4
) =(
2 + (−3)
−1
1
1
0
)
−1
EXERCISE
1. The vertices of triangle ABC are (−4, 3), (−2, 1) and (−3, 5) respectively. Determine the
5
).
−3
vertices of triangle AʹBʹCʹ under translation 𝑇 = (
2. The vertices of quadrilateral ABCD are (−5, 1), (−1, 0), (−3, 4) and (−4, 6) respectively.
−3
).
4
Determine the vertices of triangle AʹBʹCʹDʹ under translation 𝑇 = (
3. The vertices of quadrilateral PQRS are (1, 1), (4, 1), (5, 3) and (2, 3) respectively. Determine
−3
).
−5
the vertices of triangle PʹQʹRʹSʹ under translation 𝑇 = (
𝑥
4. Under the translation T = (𝑦), the point X(3, -4) is mapped onto Xʹ (-5, 2). Determine the
𝑥
translation, T = (𝑦).
5. Rectangle WXYZ is mapped onto rectangle WʹXʹYʹZʹ with vertices Wʹ(2, 1), Xʹ(6, 1), Yʹ(6, 2), Zʹ(2, -2) under the translation 𝑇 = (
−4
). Evaluate the vertices of WXYZ.
5
WEEK 7
LESSON 2
TOPIC: Transformation Geometry
SUB-TOPIC: Reflection in the line 𝒙 = 𝟎 (𝒚 − 𝒂𝒙𝒊𝒔)
CONTENT: A reflection is a transformation in which a figure is flipped, or reflected, over a
line of reflection.
Reflection in the line 𝒙 = 𝟎 (𝒚 − 𝒂𝒙𝒊𝒔)
𝑥
(𝑦 )
−𝑥
(𝑦)
2×2 Matrix representing reflection in the line 𝒙 = 𝟎 (𝒚 − 𝒂𝒙𝒊𝒔)
(
𝟏
𝟎
𝟎
)
𝟏
−𝟏 𝟎
(
)
𝟎 𝟏
Example
Triangle ABC with vertices A(−3, 2), B(1, 4) and C (−1, 1) is reflected in the 𝑦 − 𝑎𝑥𝑖𝑠.
Calculate the coordinates of AʹBʹCʹ, the image of ABC.
𝑀𝑦
(
A
−1 0 −3
)(
0 1
2
B
C
=
1
4
−1
)
1
=
Aʹ
𝐶′
Bʹ
(−1 × (−3)) + (0 × 2)
(
(0 × (−3)) + (1 × 2)
(−1 × 1) + (0 × 4)
(0 × 1) + (1 × 4)
Aʹ
=(
3 + 0 −1 + 0
0+2 0+4
Aʹ
=(
Bʹ
Bʹ
Cʹ
3 −1
2
4
−1
)
1
(−1 × (−1)) + (0 × 1)
)
(0 × (−1)) + (1 × 1)
Cʹ
1+0
)
0+1
EXERCISE
1. Given triangle ONA with coordinates O(-4, 1), N(11, -12) and A(-7, -9), find the image of
ONA after a reflection over the 𝑦 − 𝑎𝑥𝑖𝑠.
2. Draw a quadrilateral ABCD where A (1, 1), B (1, 4), C (3, 3) and D (3, 2). Now reflect the
quadrilateral in 𝑦 − 𝑎𝑥𝑖𝑠. Write down the co-ordinates of A'B'C'D'.
3. Given triangle JBN with coordinates J(4, 5), B(-1, -7), and N(-7, 8), find the image of JBN
after a reflection in the line 𝑥 = 0.
4. Draw a triangle ABC on the graph paper whose co-ordinates are A (-2, 5), B (-3, -1), C (-1, 2)
and reflect this triangle in 𝑦 − 𝑎𝑥𝑖𝑠 to ∆A'B'C'. Find the co-ordinates of the vertices of the
triangle A'B'C'.
5. Given triangle UCJ with coordinates U(-12, 7), C(4, 2), and J(-3, 9), find the image of UCJ
after a reflection over the 𝑦 − 𝑎𝑥𝑖𝑠.
WEEK 7
LESSON 3
TOPIC: Transformation Geometry
SUB-TOPIC: Reflection in the 𝒚 = 𝟎 (𝒙 − 𝒂𝒙𝒊𝒔)
CONTENT:
Reflection in the line 𝒚 = 𝟎 (𝒙 − 𝒂𝒙𝒊𝒔)
𝑥
(𝑦 )
𝑥
(−𝑦)
2×2 Matrix representing reflection in the line 𝒚 = 𝟎 (𝒙 − 𝒂𝒙𝒊𝒔)
𝟏
𝟎
(
𝟎
)
𝟏
𝟏
𝟎
(
𝟎
)
−𝟏
Example
Triangle ABC with vertices A(−3, 2), B(1, 4) and C (−1, 1) is reflected in the 𝑥 − 𝑎𝑥𝑖𝑠.
Calculate the coordinates of AʹBʹCʹ, the image of ABC.
𝑀𝑥
(
A
B
C
=
1 0
−3
)(
0 −1
2
1
4
−1
)
1
=
Aʹ
Bʹ
(1 × (−3)) + (0 × 2)
(
(0 × (−3)) + (−1 × 2)
Bʹ
−3 + 0
0 + (−2)
Aʹ
=(
(1 × (−1)) + (0 × 1)
)
(0 × (−1)) + (−1 × 1)
(1 × 1) + (0 × 4)
(0 × 1) + (−1 × 4)
Aʹ
=(
Cʹ
Bʹ
−3
1
−2 −4
1+0
0 + (−4)
Cʹ
−1 + 0
)
0 + (−1)
Cʹ
−1
)
−1
EXERCISE
1. Given triangle ONA with coordinates O(-4, 1), N(11, -12) and A(-7, -9), find the image of
ONA after a reflection over the 𝑥 − 𝑎𝑥𝑖𝑠.
2. Draw a quadrilateral ABCD where A (1, 1), B (1, 4), C (3, 3) and D (3, 2). Now reflect the
quadrilateral in x-axis. Write down the co-ordinates of A'B'C'D'.
3. Given triangle JBN with coordinates J(4, 5), B(-1, -7), and N(-7, 8), find the image of JBN
after a reflection in the line 𝑦 = 0.
4. Draw a triangle ABC on the graph paper whose co-ordinates are A (-2, 5), B (-3, -1), C (-1, 2)
and reflect this triangle in x-axis to ∆A'B'C'. Find the co-ordinates of the vertices of the triangle
A'B'C'.
5. Given triangle UCJ with coordinates U(-12, 7), C(4, 2), and J(-3, 9), find the image of UCJ
after a reflection over the 𝑥 − 𝑎𝑥𝑖𝑠.
WEEK 7
LESSON 4
TOPIC: Transformation Geometry
SUB-TOPIC: Reflection in the origin
CONTENT:
Reflection in the origin
𝑥
(𝑦 )
−𝑥
(−𝑦)
2×2 Matrix representing reflection in the origin
𝟏
(
𝟎
𝟎
)
𝟏
−𝟏 𝟎
)
𝟎 −𝟏
(
Example
Triangle ABC with vertices A(−3, 2), B(1, 4) and C (−1, 1) is reflected in the line y= 𝑥.
Calculate the coordinates of AʹBʹCʹ, the image of ABC.
𝑀𝑂
(
A
−1 0
−3
)(
0 −1
2
B
C
1 −1
)
4
1
=
=
Aʹ
Bʹ
(−1 × (−3)) + (0 × 2)
(
(0 × (−3)) + (−1 × 2)
(−1 × 1) + (0 × 4)
(0 × 1) + (−1 × 4)
Aʹ
Bʹ
3+0
−1 + 0
=(
0 + (−2) 0 + (−4)
Cʹ
(−1 × (−1)) + (0 × 1)
)
(0 × (−1)) + (−1 × 1)
Cʹ
1+0
)
0 + (−1)
Aʹ
=(
Bʹ
3 −1
−2 −4
Cʹ
1
)
−1)
EXERCISE
1. State the co-ordinates of the following points under reflection in origin:
(i) (−2, −4)
(ii) (−2, 7)
(iii) (0, 0)
2. The point P (𝑎, 𝑏) is mapped onto P’ (−17, 10) on reflection in the origin. Find the values of
𝑎 𝑎𝑛𝑑 𝑏.
3. Write down the coordinates of the image of the point (3, −11) when reflected in the origin.
4. The point P (1, 8) is reflected in the origin to get the image P’. Write down the coordinates of
P’.
5. Given triangle ONA with coordinates O(-4, 1), N(11, -12) and A(-7, -9), find the image of
ONA after a reflection in the origin.
6. Draw a quadrilateral ABCD where A (1, 1), B (1, 4), C (3, 3) and D (3, 2). Now reflect the
quadrilateral in the origin. Write down the co-ordinates of A'B'C'D'.
7. Given triangle JBN with coordinates J(4, 5), B(-1, -7), and N(-7, 8), find the image of JBN
after a reflection in the origin.
8. Draw a triangle ABC on the graph paper whose co-ordinates are A (-2, 5), B (-3, -1), C (-1, 2)
and reflect this triangle in the origin to ∆A'B'C'. Find the co-ordinates of the vertices of the
triangle A'B'C'.
9. Given triangle UCJ with coordinates U(-12, 7), C(4, 2), and J(-3, 9), find the image of UCJ
after a reflection over the in the origin.
10. Plot (1, 0), B(3,2), C(5, 1), D(4, -1). Reflect in the origin.
WEEK 8
LESSON 1
TOPIC: Transformation Geometry
SUB-TOPIC: Reflection in the line 𝒚 = 𝒙
CONTENT:
Reflection in the line 𝒚 = 𝒙
𝑥
(𝑦 )
𝑦
( )
𝑥
2×2 Matrix representing reflection in the line 𝒚 = 𝒙
(
𝟏
𝟎
𝟎
)
𝟏
(
𝟎 𝟏
)
𝟏 𝟎
Example
Triangle ABC with vertices A(−3, 2), B(1, 4) and C (−1, 1) is reflected in the line y= 𝑥.
Calculate the coordinates of AʹBʹCʹ, the image of ABC.
𝑀𝑦=𝑥
A
0 1 −3
(
)(
1 0
2
B
C
1 −1
)
4
1
=
=
Aʹ
(
Bʹ
(0 × (−3)) + (1 × 2)
(1 × (−3)) + (0 × 2)
=(
(0 × 1) + (1 × 4)
(1 × 1) + (0 × 4)
Aʹ
Bʹ
0+2
−3 + 0
0+4
1+0
Aʹ
=(
Cʹ
2
−3
Bʹ
4
1
(0 × (−1)) + (1 × 1)
)
(1 × (−1)) + (0 × 1)
Cʹ
0+1
)
−1 + 0
Cʹ
1
)
−1)
EXERCISE
1. Given triangle ONA with coordinates O(-4, 1), N(11, -12) and A(-7, -9), find the image of
ONA after a reflection in the line 𝑦 = 𝑥.
2. Draw a quadrilateral ABCD where A (1, 1), B (1, 4), C (3, 3) and D (3, 2). Now reflect the
quadrilateral in the line 𝑦 = 𝑥. Write down the co-ordinates of A'B'C'D'.
3. Given triangle JBN with coordinates J(4, 5), B(-1, -7), and N(-7, 8), find the image of JBN
after a reflection in the line 𝑦 = 𝑥.
4. Draw a triangle ABC on the graph paper whose co-ordinates are A (-2, 5), B (-3, -1), C (-1, 2)
and reflect this triangle in the line 𝑦 = 𝑥 to ∆A'B'C'. Find the co-ordinates of the vertices of the
triangle A'B'C'.
5. Given triangle UCJ with coordinates U(-12, 7), C(4, 2), and J(-3, 9), find the image of UCJ
after a reflection over the in the line 𝑦 = 𝑥.
WEEK 8
LESSON 2
TOPIC: Transformation Geometry
SUB-TOPIC: Reflection in the line 𝒚 = −𝒙
CONTENT:
Reflection in the line 𝒚 = −𝒙
𝑥
(𝑦 )
−𝑦
)
−𝑥
(
2×2 Matrix representing reflection in the line 𝒚 = −𝒙
(
𝟏
𝟎
𝟎
)
𝟏
𝟎 −𝟏
(
)
−𝟏
𝟎
Example
Triangle ABC with vertices A(−3, 2), B(1, 4) and C (−1, 1) is reflected in the line y= −𝑥.
State the coordinates of AʹBʹCʹ, the image of ABC.
𝑀𝑦=−𝑥
A
0 −1 −3
(
)(
−1
0
2
B
C
1 −1
)
4
1
=
Aʹ
(
Bʹ
(0 × (−3)) + (−1 × 2)
(−1 × (−3)) + (0 × 2)
Bʹ
0 + (−2)
3+0
Aʹ
=(
(0 × (−1)) + (−1 × 1)
)
(−1 × (−1)) + (0 × 1)
(0 × 1) + (−1 × 4)
(−1 × 1) + (0 × 4)
Aʹ
=(
Cʹ
Cʹ
0 + (−4) 0 + (−1)
)
−1 + 0
1+0
Bʹ
−2 −4
3 −1
Cʹ
−1
)
1
EXERCISE
1. Given triangle ONA with coordinates O(-4, 1), N(11, -12) and A(-7, -9), find the image of
ONA after a reflection in the line 𝑦 = −𝑥.
2. Draw a quadrilateral ABCD where A (1, 1), B (1, 4), C (3, 3) and D (3, 2). Now reflect the
quadrilateral in the line 𝑦 = −𝑥. Write down the co-ordinates of A'B'C'D'.
3. Given triangle JBN with coordinates J(4, 5), B(-1, -7), and N(-7, 8), find the image of JBN
after a reflection in the line 𝑦 = −𝑥.
4. Draw a triangle ABC on the graph paper whose co-ordinates are A (-2, 5), B (-3, -1), C (-1, 2)
and reflect this triangle in the line 𝑦 = −𝑥 to ∆A'B'C'. Find the co-ordinates of the vertices of the
triangle A'B'C'.
5. Given triangle UCJ with coordinates U(-12, 7), C(4, 2), and J(-3, 9), find the image of UCJ
after a reflection over the in the line 𝑦 = −𝑥.
WEEK 8
LESSON 3
TOPIC: Transformation Geometry
SUB-TOPIC: Reflection in the line 𝒙 = 𝒂
CONTENT:
The lines 𝑥 = 𝑎 is parallel to the 𝑦 − 𝑎𝑥𝑖𝑠.
Reflection in the line 𝒙 = 𝒂
𝑥
(𝑦 )
2𝑎 − 𝑥
(
)
𝑦
Example
Triangle ABC with vertices A(−3, 2), B(1, 4) and C (−1, 1) is reflected in the line 𝑥 = 3. State
the coordinates of AʹBʹCʹ, the image of ABC.
Reflection in the line 𝒙 = 𝒂
𝑥
(𝑦 )
2𝑎 − 𝑥
(
)
𝑦
𝑥
(𝑦 )
6−𝑥
(
)
𝑦
A
(
−3
2
B
Reflection in the line 𝒙 = 𝟑
𝑥
(𝑦)
Aʹ
C
6 − (−3)
2
1 −1
)
4
1
(
Bʹ Cʹ
9
2
5
4
(
Aʹ
2(3) − 𝑥
(
)
𝑦
Bʹ
Cʹ
6− 1
4
6 − (−1)
)
1
7
)
1
EXERCISE
1. Given triangle ONA with coordinates O(-4, 1), N(11, -12) and A(-7, -9), find the image of
ONA after a reflection in the line 𝑥 = −1.
2. Draw a quadrilateral ABCD where A (1, 1), B (1, 4), C (3, 3) and D (3, 2). Now reflect the
quadrilateral in the line 𝑥 = 2. Write down the co-ordinates of A'B'C'D'.
3. Given triangle JBN with coordinates J(4, 5), B(-1, -7), and N(-7, 8), find the image of JBN
after a reflection in the line 𝑥 = −2.
4. Draw a triangle ABC on the graph paper whose co-ordinates are A (-2, 5), B (-3, -1), C (-1, 2)
and reflect this triangle in the line 𝑥 = 1.5 to ∆A'B'C'. Find the co-ordinates of the vertices of the
triangle A'B'C'.
5. Given triangle UCJ with coordinates U(-12, 7), C(4, 2), and J(-3, 9), find the image of UCJ
after a reflection over the in the line 𝑥 = −2.5.
WEEK 8
LESSON 4
TOPIC: Transformation Geometry
SUB-TOPIC: Reflection in the line 𝒚 = 𝒃
CONTENT:
The lines 𝑦 = 𝑏 is parallel to the 𝑥 − 𝑎𝑥𝑖𝑠.
Reflection in the line 𝒚 = 𝒃
𝑥
(𝑦)
𝑥
(2𝑏 − 𝑦)
Example
Triangle ABC with vertices A(−3, 2), B(1, 4) and C (−1, 1) is reflected in the line 𝑦 = −2.
State the coordinates of AʹBʹCʹ, the image of ABC.
Reflection in the line𝒚 = 𝒃
Reflection in the line 𝒙 = −𝟐
𝑥
(𝑦 )
𝑥
(2𝑏 − 𝑦)
𝑥
(𝑦 )
𝑥
(−4 − 𝑦)
A
(
−3
2
B
𝑥
(𝑦)
𝑥
(2(−2) − (𝑦)
Aʹ
C
−3
1
−4 − 2 −4 − 4
1 −1
)
4
1
Aʹ
(
Bʹ
(
Bʹ
−3
1
−6 −8
Cʹ
−1)
)
−4 − 1
Cʹ
−1
)
−5
EXERCISE
1. Given triangle ONA with coordinates O(-4, 1), N(11, -12) and A(-7, -9), find the image of
ONA after a reflection in the line 𝑦 = −1.
2. Draw a quadrilateral ABCD where A (1, 1), B (1, 4), C (3, 3) and D (3, 2). Now reflect the
quadrilateral in the line 𝑦 = 2. Write down the co-ordinates of A'B'C'D'.
3. Given triangle JBN with coordinates J(4, 5), B(-1, -7), and N(-7, 8), find the image of JBN
after a reflection in the line 𝑦 = −2.
4. Draw a triangle ABC on the graph paper whose co-ordinates are A (-2, 5), B (-3, -1), C (-1, 2)
and reflect this triangle in the line 𝑦 = 1.5 to ∆A'B'C'. Find the co-ordinates of the vertices of the
triangle A'B'C'.
5. Given triangle UCJ with coordinates U(-12, 7), C(4, 2), and J(-3, 9), find the image of UCJ
after a reflection over the in the line 𝑦 = −2.5.
WEEK 9
LESSON 1
TOPIC: Transformation Geometry
SUB-TOPIC: Rotation
CONTENT: Rotation (When the origin is the centre of rotation)
A rotation is a transformation in which the object is rotated about a fixed point. The direction of
rotation can be clockwise or anticlockwise.
The fixed point in which the rotation takes place is called the center of rotation. The amount of
rotation made is called the angle of rotation.
For any rotation, we need to specify the center, the angle and the direction of rotation.
Anti- Clockwise Rotation about the origin
𝒄𝒐𝒔 𝜽 −𝒔𝒊𝒏𝜽
)
𝒔𝒊𝒏 𝜽 𝒄𝒐𝒔 𝜽
𝑹𝜽 = (
Clockwise Rotation about the origin
𝑹𝜽 = (
𝒄𝒐𝒔 𝜽 𝒔𝒊𝒏𝜽
)
−𝒔𝒊𝒏 𝜽 𝒄𝒐𝒔 𝜽
Example
A parallelogram is define by the points 𝐴(2, 1), 𝐵(3, 3), 𝐶(6, 3) 𝑎𝑛𝑑 𝐷(5, 1). Calculate the
points defining the image when the parallelogram ABCD undergoes an anti-clockwise rotation
about the origin of 180°.
Solution
The anticlockwise rotation of 180° about the origin is:
𝑅𝜃 = (
𝑐𝑜𝑠 180
𝑠𝑖𝑛 180
−𝑠𝑖𝑛180
−1
0
)=(
)
𝑐𝑜𝑠 180
0 −1
A B C D
(
−1
0 2 3
)(
0 −1 1 3
6 5
)
3 1
Aʹ
=(
−1 × 2 + 0 × 1
0 × 2 + (−1 × 1)
Aʹ
=(
Cʹ
Bʹ
−1 × 3 + 0 × 3
−1 × 6 + 0 × 3
−1 × 5 + 0 × 1
)
0 × 3 + (−1 × 3) 0 × 6 + (−1 × 3) 0 × 5 + (−1 × 1)
Bʹ
Cʹ
Dʹ
−2 + 0
−3 + 0
−6 + 0
−5 + 0
)
0 + (−1) 0 + (−3) 0 + (−3) 0 + (−1)
Aʹ
Bʹ
−2 −3
=(
−1 −3
Cʹ
Dʹ
Dʹ
−6 −5
)
−3 −1
EXERCISE
1. A parallelogram ABCD is define by the points 𝐴(3, 3), 𝐵(7, 3), 𝐶(5, 1) 𝑎𝑛𝑑 𝐷(1, 1). Calculate
the points defining the image when the parallelogram ABCD undergoes an anti-clockwise
rotation about the origin of 90°.
2. A trapezium PQRS is define by the points 𝑃(−5, 3), 𝑄(−3, 5), 𝑅(−1, 5) 𝑎𝑛𝑑 𝑆(−1, 3).
Calculate the points defining the image when the trapezium PQRS undergoes a clockwise
rotation about the origin of 90°.
3. A kite KLMN is define by the points 𝐾(−3, 2), 𝐿(−5, 0), 𝑀(−3, −4) 𝑎𝑛𝑑 𝑁(1, 0). Calculate
the points defining the image when the trapezium PQRS undergoes a clockwise rotation about
the origin of 270°.
4. A trapezium PQRS is define by the points 𝑃(−5, 3), 𝑄(−3, 5), 𝑅(−1, 5) 𝑎𝑛𝑑 𝑆(−1, 3).
Calculate the points defining the image when the trapezium PQRS undergoes a clockwise
rotation about the origin of 135°.
5. A parallelogram ABCD is define by the points 𝐴(3, 3), 𝐵(7, 3), 𝐶(5, 1) 𝑎𝑛𝑑 𝐷(1, 1). Calculate
the points defining the image when the parallelogram ABCD undergoes an anti-clockwise
rotation about the origin of 325°.
WEEK 9
LESSON 2
TOPIC: Transformation Geometry
SUB-TOPIC: Rotation
CONTENT: Rotation (When the origin is not the centre of rotation)
When the origin is not the centre of rotation, then the rotation is equivalent to a rotation about the
origin combined with a translation.
𝒑
(
𝒓
𝒙𝟏
𝒒 𝒙
𝒙ʹ
) (𝒚) + (𝒚 ) = ( )
𝒔
𝒚ʹ
𝟏
Rotation
Object
Translation
Image
about
the
origin
Example: A rotation is defined by the matrix equation
(
𝑥ʹ
0 −1 𝑥
5
) (𝑦) + ( ) = ( ). Determine the coordinates of the centre of rotation by calculating
𝑦ʹ
1
0
2
the invariant point of mapping.
Solution: Given that
(
𝑥ʹ
0 −1 𝑥
5
) (𝑦) + ( ) = ( ). Let the centre of rotation be defined by C (a, b). Since the centre of
𝑦ʹ
1
0
2
rotation is an invariant point, then
(
𝑎
0 −1 𝑎
5
) ( ) + ( ) = ( ).
𝑏
1
0 𝑏
2
(
𝑎
0 × 𝑎 + (−1 × 𝑏)
5
)+ ( ) = ( ).
𝑏
2
1×𝑎+0×𝑏
(
𝑎
−𝑏
5
) + ( ) = ( ).
𝑏
𝑎
2
(
𝑎
−𝑏 + 5
) = ( ).
𝑏
𝑎+2
Equating corresponding first elements, we get
−𝑏 + 5 = 𝑎 …… ...eq. 1
𝑎 + 2 = 𝑏 ………eq. 2
Substitute eq. 1 in eq. 2
−𝑏 + 5 + 2 = 𝑏
7= 𝑏+𝑏
7 = 2𝑏
7
2
=b
3.5 = 𝑏
Substitute 𝑏 = 3.5 in eq 1
−𝑏 + 5 = 𝑎
−3.5 + 5 = 𝑎
1.5 = 𝑎
Hence, the centre of rotation is C(1.5, 3.5)
EXERCISE:
1. A rotation is defined by the matrix equation
(
𝑥ʹ
0 −1 𝑥
3
) (𝑦) + ( ) = ( ).
𝑦ʹ
1
0
4
Determine the coordinates of the centre of rotation by calculating the invariant point of mapping.
2. A rotation is defined by the matrix equation
(
𝑥ʹ
−1
0 𝑥
5
) (𝑦) + ( ) = ( ).
𝑦ʹ
0 −1
7
Determine the coordinates of the centre of rotation by calculating the invariant point of mapping.
3. A rotation is defined by the matrix equation
(
𝑥ʹ
0 1 𝑥
4
) (𝑦) + ( ) = ( ).
𝑦ʹ
−1 0
6
Determine the coordinates of the centre of rotation by calculating the invariant point of mapping.
4. A rotation is defined by the matrix equation
𝑥ʹ
( )
𝑦ʹ
𝑥
−1 0
−3
) (𝑦) + ( )
0 −1
−4
(
Determine the coordinates of the centre of rotation by calculating the invariant point of mapping.
5. A rotation is defined by the matrix equation
𝑥ʹ
( )
𝑦ʹ
0 1 𝑥
−4
) (𝑦) + ( )
−1 0
5
(
Determine the coordinates of the centre of rotation by calculating the invariant point of mapping.
WEEK 9
LESSON 3
TOPIC: Transformation Geometry
SUB-TOPIC: Enlargement (When the origin is the centre of enlargement)
CONTENT:
An enlargement is a geometrical transformation of the plane in which the shape is resized using a
scale factor.
The matrix that represents an enlargement with the origin as centre and scale factor k is
𝐸(0,
𝑘)
𝑘
0
0
)
𝑘
= (
Example
Triangle PQR has vertices 𝑃(3, 4), 𝑄(5, 3) 𝑎𝑛𝑑 𝑅(4, 1). Determine the image of triangle PRQ of
scale factor -2, with the origin as the centre of enlargement.
Solution
𝐸(0,−2)
(
P
Q R
−2
0 3 5
)(
0 −2 4 3
Pʹ
4
−2 × 3
) =(
1
−2 × 4
Qʹ
Rʹ
Pʹ
Qʹ
Rʹ
−6 −10 −8
−2 × 5 −2 × 4
) =(
)
−8 −6 −2
−2 × 3 −2 × 1
EXERCISE
1. Triangle PQR has vertices 𝑃(−6, 1), 𝑄(−2, 1) and 𝑅(−3, 4). Determine the image of triangle
PRQ of scale factor 3, with the origin as the centre of enlargement.
2. Triangle ABC has vertices 𝐴(2, 3), 𝐵(4, 5) and 𝐶(5, 2). Determine the image of triangle ABC
of scale factor 2, with the origin as the centre of enlargement.
3. Triangle KLM has vertices 𝐾(−3, −2), 𝐿(−6, −3) and 𝑀(−2, −4). Determine the image of
triangle KLM of scale factor 1.5, with the origin as the centre of enlargement.
4. Triangle XYZ has vertices 𝑋(3, −5), 𝑌(2, −1) and 𝑍(5, −3). Determine the image of triangle
1
XYZ of scale factor 4, with the origin as the centre of enlargement.
5. Triangle KLM has vertices 𝐾(3, 3), 𝐿(3, −6) and 𝑀(6, −6). Determine the image of triangle
KLM of scale factor -2.25, with the origin as the centre of enlargement.
WEEK 9
LESSON 4
TOPIC: Transformation Geometry
SUB-TOPIC: Centre of Enlargement (When the origin is not the centre of
enlargement)
CONTENT:
When the origin is not the centre of enlargement, then the enlargement is equivalent to a
enlargement about the origin combined with a translation.
𝒌
(
𝟎
𝒙𝟏
𝒙ʹ
𝟎 𝒙
) ( ) + (𝒚 ) = ( )
𝒚ʹ
𝒌 𝒚
𝟏
Enlargement Object
Translation
Image
about the
origin
Example: Given that a transformation represented by the matrix equation
(
𝑘
0
𝑥1
𝑥ʹ
2
0 𝑥
) (𝑦) + (𝑦 ) = ( ) undergoes an enlargement of scale factor 6 and translation( ),
𝑦ʹ
1
3
𝑘
determine its centre of enlargement.
𝑘
0
Solution: Given the matrix equation (
𝑥1
𝑥1
𝑥ʹ
2
0 𝑥
) (𝑦) + (𝑦 ) = ( ), 𝑘 = 6 and (𝑦 ) = ( )
𝑦ʹ
1
1
3
𝑘
Let the centre of rotation be defined by C (a, b). Since the centre of rotation is an invariant point,
then
(
𝑎
6 0 𝑎
2
)( ) + ( ) = ( )
𝑏
𝑏
0 6
3
𝑎
6𝑎
2
)+ ( ) = ( )
𝑏
6𝑏
3
So (
(
𝑎
6𝑎 + 2
)=( )
𝑏
6𝑏 + 3
Equating corresponding elements
6𝑎 + 2 = 𝑎
6𝑎 − 𝑎 = −2
5𝑎 = −2
𝑎 =
6𝑏 + 3 = 𝑏
6𝑏 − 𝑏 = −3
5𝑏 = −3
𝑏 =
Centre of Enlargement C(−0.4, −0.6)
−2
5
−3
5
= −0.4
= −0.6
EXERCISE
1. Given that a transformation represented by the matrix equation (
𝑘
0
𝑥1
𝑥ʹ
0 𝑥
) (𝑦) + (𝑦 ) = ( )
𝑦ʹ
1
𝑘
−2
), determine its centre of
3
undergoes an enlargement of scale factor 5 and translation(
enlargement.
2. Given that a transformation represented by the matrix equation (
1
𝑘
0
𝑥1
𝑥ʹ
0 𝑥
) (𝑦) + (𝑦 ) = ( )
𝑦ʹ
1
𝑘
−4
), determine its centre of
−5
undergoes an enlargement of scale factor 3 and translation(
enlargement.
3. Given that a transformation represented by the matrix equation (
undergoes an enlargement of scale factor −3 and translation(
𝑘
0
𝑥1
𝑥ʹ
0 𝑥
) (𝑦) + (𝑦 ) = ( )
𝑦ʹ
1
𝑘
−4
), determine its centre of
7
enlargement.
𝑥ʹ
4. Given that a transformation represented by the matrix equation( )
𝑦ʹ
undergoes an enlargement of scale factor -2 and translation(
𝑘
0
(
𝑥1
0 𝑥
) (𝑦) + (𝑦 )
1
𝑘
−5
), determine its centre of
−3
enlargement.
𝑥ʹ
5. Given that a transformation represented by the matrix equation( )
𝑦ʹ
𝑘
(
0
𝑥1
0 𝑥
) (𝑦) + (𝑦 )
1
𝑘
−4
), determine its centre of
−1
undergoes an enlargement of scale factor 2.5 and translation(
enlargement.
6.
8.
WEEK 10
LESSON 1
TOPIC: Relations, Functions and Graphs
SUB-TOPIC: Image of 𝒙
CONTENT:
Example:
1. If 𝑓(𝑥) = 2𝑥 2 − 3𝑥 + 1, determine the value of each of the following:
(a) 𝑓(−3)
(b) 𝑓(0)
Solution:
(a) Given 𝑓(𝑥) = 2𝑥 2 − 3𝑥 + 1
(b) Given 𝑓(𝑥) = 2𝑥 2 − 3𝑥 + 1
Substitute 𝑥 = −3
Substitute 𝑥 = 0
Then 𝑓(−3) = 2(−3)2 − 3(−3) + 1
Then 𝑓(0) = 2(0)2 − 3(0) + 1
𝑓(−3) = 2(9) + 9 + 1
𝑓(0) = 2(0) − 0 + 1
= 18 + 9 + 1
= 0−0 +1
= 28
= 1
2. If ℎ(𝑥) = 𝑥 2 − 1, for what value of 𝑥 is ℎ(𝑥) = 8?
Solution: Given that ℎ(𝑥) = 𝑥 2 − 1 and ℎ(𝑥) = 8
Then 𝑥 2 − 1 = 8
𝑥2 = 8 + 1
𝑥2 = 9
𝑥 = ±√9
𝑥 = ±3
EXERCISE
1. If 𝑓(𝑥) = 3𝑥 2 + 2𝑥 − 1, determine the value of each of the following:
(a) 𝑓(2)
(b) 𝑓(0)
(c) 𝑓(−1)
1
2. If 𝑓(𝑥) = (4𝑥 − 1), determine the value of each of the following:
2
(a) 𝑓(3)
(b) 𝑓(−2)
(c) 𝑓(0)
1
3. If 𝑓(𝑥) = 𝑥 − 3, determine the value of each of the following:
1
(a) 𝑓(2)
(b) 𝑓(−2)
(c) 𝑓(−1)
1
(d) 𝑓(4)
4. If ℎ(𝑥) = 3(𝑥 − 1) + 2,
for what value of 𝑥 is ℎ(𝑥) = 26?
5. Evaluate 𝑥, if 𝑔(𝑥) = −10 and 𝑔(𝑥) = 3𝑥 − 1.
WEEK 10
LESSON 2
TOPIC: Relations, Functions and Graphs
SUB-TOPIC: Inverse of a function
CONTENT:
Example 1
Here we have the function 𝒇(𝒙) = 𝟐𝒙 + 𝟑, written as a flow diagram:
Algebraically:
𝑓(𝑥) = 2𝑥 + 3
We are given
We can write this as
𝑦 = 2𝑥 + 3
Interchanging 𝑥 and 𝑦
𝑥 = 2𝑦 + 3
Solve for 𝑦
𝑥 – 3 = 2𝑦
𝑥−3
2
= 𝑦
Hence, 𝑓 −1 =
𝑥−3
2
Example 2: Given 𝑓(𝑥) =
3𝑥 + 2
, determine and expression for 𝑓 −1 (𝑥).
2𝑥 − 5
Solution:
Given 𝑓(𝑥) =
3𝑥 + 2
2𝑥 − 5
3𝑥 + 2
Rewriting 𝑦 =
2𝑥 − 5
Interchanging variables
𝑥=
3𝑦 + 2
2𝑦 − 5
Solving for 𝑦
𝑥(2𝑦 − 5) = 3𝑦 + 2
2𝑥𝑦 − 5𝑥 = 3𝑦 + 2
2𝑥𝑦 – 3𝑦 = 2 + 5𝑥
𝑦(2𝑥 − 3) = 5𝑥 + 2
𝑦=
5𝑥+2
2𝑥 − 3
𝑓 −1 =
5𝑥+2
2𝑥 − 3
EXERCISE
1. Given 𝑓(𝑥) = 5𝑥 + 7 determine and expression for 𝑓 −1 (𝑥).
2. Given 𝑔(𝑥) = 7𝑥 − 4 determine and expression for 𝑔−1 (𝑥).
3. Given ℎ(𝑥) = 3𝑥 + 2 determine and expression for ℎ−1 (𝑥).
4. Given 𝑓(𝑥) =
5. Given ℎ(𝑥) =
𝑥−2
3𝑥+1
determine and expression for 𝑓 −1 (𝑥).
5𝑥 + 1
3𝑥 − 1
determine and expression for ℎ−1 (𝑥).
WEEK 10
LESSON 3
TOPIC: Relations, Functions and Graphs
SUB-TOPIC: Composite functions
CONTENT: A composite function is a function that depends on another function. A
composite function is created when one function is substituted into another function.
For example, 𝑓(𝑔(𝑥)) is the composite function that is formed when 𝑔(𝑥) is substituted for 𝑥 in
𝑓(𝑥).
𝑓(𝑔(𝑥)) is read as “𝑓 of 𝑔 of 𝑥”.
𝑓(𝑔(𝑥)) can also be written as (𝑓 ∘ 𝑔)(𝑥) or 𝑓𝑔(𝑥),
Given 𝑓(𝑥) = 𝑎𝑥 + 𝑏 and 𝑔(𝑥) = 𝑐𝑥 + 𝑑, then
𝑓𝑔(𝑥) = 𝑎(𝑐𝑥 + 𝑑) + 𝑏
Example:
1. Given 𝑓(𝑥) = 3𝑥 + 2 𝑎𝑛𝑑 𝑔(𝑥) = 𝑥 + 5, find:
(a) 𝑓𝑔(𝑥)
(b) 𝑔𝑓(𝑥)
Solution
(a) 𝑓(𝑔(𝑥)) = 3(𝑥 + 5) + 2
= 3𝑥 + 15 + 2
= 3𝑥 + 17
2. Given 𝑓(𝑥) = 𝑥 𝟐 + 6 and 𝑔(𝑥) = 2 𝑥 – 1, find
a) (𝑓 ∘ 𝑔)(𝑥)
b) (𝑔 ∘ 𝑓)(𝑥)
(b) 𝑔 (𝑓(𝑥)) = (3𝑥 + 2) + 5
= 3𝑥 + 7
Solution:
a) (𝑓 ∘ 𝑔)(x)
= (2x – 1)2 + 6
= 4x2 – 4x + 1 + 6
= 4x2 – 4x + 7
b) (𝑔 ∘ 𝑓)(x)
= 2(x2 + 6) – 1
= 2x2 + 12 – 1
= 2x2 + 11
EXERCISE:
2
1. Given 𝑓(𝑥) = 𝑥 + 𝑥 and 𝑔(𝑥) = 4 − 𝑥
Find
(a) (𝑓 ∘ 𝑔)(𝑥)
(b) (𝑔 ∘ 𝑓)(𝑥)
2. Given the functions, determine the value of each composite function.
𝑓(𝑥) = 4𝑥 + 1,
2
𝑔(𝑥) = 𝑥 − 𝑥 + 5
(a) (𝑓 ∘ 𝑔)(𝑥)
(b) (𝑔 ∘ 𝑓)(𝑥)
3. Given the functions, determine the value of each composite function.
𝑓(𝑥) = 2𝑥 − 1,
𝑔(𝑥) = 𝑥 3 − 5,
ℎ(𝑥) = 5 − 𝑥 2
(a) 𝑓𝑔(𝑥)
(b) 𝑔𝑓(𝑥)
(c) ℎ𝑔(𝑥)
4. 𝑓(𝑥) = 2𝑥 4 + 𝑥 2 + 1, 𝑔(𝑥) = √𝑥. Find 𝑓(𝑔(𝑥))
WEEK 10
LESSON 4
TOPIC: Relations, Functions and Graphs
SUB-TOPIC: Inverse of a composite functions
CONTENT:
Example: Given 𝑓: 𝑥
3𝑥 + 2 and
2𝑥 − 1, determine (𝑓𝑔)−1 (𝑥)
𝑔: 𝑥
Solution
Step 1: find 𝑓𝑔(𝑥)
Since 𝑓(𝑥) = 3𝑥 + 2 and 𝑔(𝑥) = 2𝑥 − 1
Then 𝑓𝑔(𝑥) = 3(2𝑥 − 1) + 2
= 6𝑥 − 3 + 2
= 6𝑥 – 1
Step 2: determine the (𝑓𝑔)−1 (𝑥)
𝑓𝑔(𝑥) = 6𝑥 – 1
Since
Rewriting:
𝑦 = 6𝑥 − 1
Interchanging the variables
𝑥 = 6𝑦 − 1
Make 𝑦 the subject of the equation
𝑥 + 1 = 6𝑦
Hence, (𝑓𝑔)−1 (𝑥) =
𝑥+1
6
𝑥+1
6
= 𝑦
EXERCISE
2
1. Given 𝑓(𝑥) = 𝑥 + 𝑥 and 𝑔(𝑥) = 4 − 𝑥
Find
(a) (𝑓𝑔)−1 (𝑥)
(b) (𝑔𝑓)−1 (𝑥)
2. Given the functions, determine the value of each composite function.
𝑓(𝑥) = 4𝑥 + 1,
2
𝑔(𝑥) = 𝑥 − 𝑥 + 5
(a) 𝑓𝑔(2)
(b) (𝑓𝑔)−1 (𝑥)
(c) (𝑔𝑓)−1 (−2)
3. Given the functions, determine the value of each composite function.
𝑓(𝑥) = 2𝑥 − 1,
𝑔(𝑥) = 𝑥 3 − 5,
ℎ(𝑥) = 5 − 𝑥 2
(a) (𝑓 𝑔)(𝑥)
(b) (𝑓𝑔)−1 (𝑥)
(c) (ℎ𝑔)−1 (−1)
4. 𝑓(𝑥) = 2𝑥 4 + 𝑥 2 + 1, 𝑔(𝑥) = √𝑥. Find:
(a) 𝑓𝑔(𝑥)
(b) (𝑓𝑔)−1 (−3)
5. Given that 𝑓(𝑥) = 𝑥 2 , 𝑔(𝑥) = 3 + 𝑥 and ℎ(𝑥) = 𝑥 − 1, determine:
(a) 𝑓𝑔ℎ(𝑥)
(b) (𝑓𝑔ℎ)−1(2)
MINISTRY OF EDUCATION
SECONDARY ENGAGEMENT PROGRAMME
EASTER TERM 2021
GRADE 11
MATHEMATICS
WORKSHEET
1
WEEK 11
LESSON 1
TOPIC: Trigonometry
SUB-TOPIC: Bearings
CONTENT:
Bearings are a way of expressing the angle between two objects. There is a specific set of rules
about how bearings should be calculated and expressed.
1. Always measure bearings from the North line.
2. Always express your answers as three-figure bearings (so 60° would be 060°).
3. Always draw and measure bearings clockwise.
For example:
(a) N
(b)
N
(c)
N
B
065°
A
Q 150°
K 210°
The bearing of B from A is 065°.
P
The bearing of P from Q is 150°
2
L
The bearing of L from K is 210°
EXERCISE:
1. State the bearing of A from B
N
A
47°
B
2. Draw a rough sketch to illustrate each of the following bearings. Mark the angle in your
sketch.
(a) From a point P, the bearing of a point Q is 30°.
(b) From a point A, the bearing of a point B is 140°.
(c) The bearing of a point K from a point L is 250°.
(d) The bearing of a place M from a place N is 330°.
3. The bearing of C from A is 190°, and C is 5km from A. The bearing of D from A is 270°, and
D is 7km from A.
Draw a rough sketch to illustrate the information above.
4. George needs to take the route as follows. 6km on a bearing of 080° from A to B, 5km on a
bearing of 160° from B to C. Draw a rough sketch to illustrate George’s route.
3
5. Starting at point A, Freya makes a journey as follows. 8km on a bearing of 135° to B, 4km on
a bearing of 180° to C, 6km on a bearing of 315° to D. Draw a suitable diagram to represent her
journey.
4
WEEK 11
LESSON 2
TOPIC: Trigonometry
SUB-TOPIC: Bearings
CONTENT: Example:
(a) The bearing of B from A is 065°.
(b) The bearing of P from Q is 150°
State the bearing of A from B.
State the bearing of Q from P.
N
N
𝛼 B
N
065°
A
The bearing of A from B is 360° − (180° – 065°)
The bearing of Q from P is 360° −
(180° – 150°)
= 360° − 115° = 245°
= 360° − 30° = 330°
5
EXERCISE:
1. The bearing of a point P from a point Q is 70°. Calculate the bearing of Q from P.
2. The bearing of a point A from a point B is 145°. Calculate the bearing of B from A.
3. The bearing of a ship S from a yacht Y is 220°. Calculate the bearing of the yacht from the
ship.
4. The bearing of an airport A from a plane P is 310°. Calculate the bearing of P from A.
5. The bearing of a ship S from a harbor H is 339°. Calculate the bearing of H from S.
6.
7.
8.
6
WEEK 11
LESSON 3
TOPIC: Trigonometry
SUB-TOPIC: Trigonometrical Ratios
CONTENT: Use for any right-angled triangle.
Example: In the triangle shown below, find the value of x, accurate to three decimal places.
Solution
Given opp = 65
𝜃 = 200
Hypotenuse
Opposite
find hyp = x
Using sin 𝜃 =
Then sin 20° =
7
65
𝑥
𝑥=
65
sin 20°
𝑜𝑝𝑝
ℎ𝑦𝑝
𝑥 = 90.047 units
EXERCISE
1. A hawk sitting on top a tree branch spots a mouse on the ground 15 feet from the base of
the tree. The hawk swoops down toward the mouse at an angle of 30 degrees. What is
the distance from the top of tree branch to the mouse?
2. A yacht is anchored 90 feet offshore from the base of a lighthouse. The angle of
elevation from the boat to the top of the lighthouse is 26°. The distance between the yacht
and the top of the lighthouse is about 100 feet. Round answer to the nearest WHOLE
NUMBER.
3. A person 100 meters from the base of a tree, observes that the angle between the ground
and the top of the tree is 18 degrees. Estimate the height h of the tree to the nearest tenth
of a meter.
4. The angle of elevation of a hot air balloon, climbing vertically, changes from 25 degrees
at 10:00 am to 60 degrees at 10:02 am. The point of observation of the angle of elevation
is situated 300 meters away from the take off point. What is the upward speed, assumed
constant, of the balloon? Give the answer in meters per second and round to two decimal
places.
5. When the top T of a mountain is viewed from point A, 2000 m from ground, the angle of
depression a is equal to 15o and when it is viewed from point B on the ground the angle
8
of elevation b is equal to 10o. If points A and B are on the same vertical line, find the
height h of the mountain. (round answer to one decimal place).
6. A tourist views a deer from a height of 45 feet. The horizontal distance between the
tourist and the deer is 130 feet. At what angle (x) should the tourist hold his camera to
photograph the deer? Round your answer to the nearest degree.
7.
Solve for 𝑎
2
8. If tan A = 3, then find all the other trigonometric ratios.
9. A 70 foot ramp rises from the first floor to the second floor of a parking garage. The
ramp makes an angle with the ground. How high above the first floor is the second floor?
9
10. You see Mr. Singh flying a kite in the park. The kite string is 65 meters long. What angle
does the string need to form with the ground so that the kite is 30 feet from the ground?
10
WEEK 11
LESSON 4
TOPIC: Trigonometry
SUB-TOPIC: Application of Trigonometrical Ratios
CONTENT:
Example:
Calculate the distance travelled south and the distance travelled east by a ship sailing on a
bearing of 150° for 90km.
Considering the right-angled ∆ OAB
Angle AOB = 180° - 150° = 30°
So cos 30°
=
𝑂𝐵
𝑂𝐴
=
𝑂𝐵
90 𝑘𝑚
𝑂𝐵 = 90𝑘𝑚 × 𝑐𝑜𝑠 30° = 90𝑘𝑚 × 0.866 = 77.94 𝑘𝑚
Hence, the distance travelled south by the ship is 77.9km
11
Now sin 30°
=
𝐴𝐵
𝑂𝐴
=
𝐴𝐵
90 𝑘𝑚
𝐴𝐵 = 90 𝑘𝑚 × 𝑠𝑖𝑛 30° = 90𝑘𝑚 × 0.5 = 45𝑘𝑚
Hence, the distance travelled east by the ship is 45km
EXERCISE:
1. By drawing a diagram, determine the distance travelled north and the distance travelled east
by a plane flying on a bearing of 50° for 100km.
2. By drawing a diagram, determine the distance travelled south and the distance travelled east
by a ship sailing on a bearing of 140° for 90km.
3. By drawing a diagram, determine the distance travelled south and the distance travelled west
by a yacht sailing on a bearing of 220° for 85km.
4. By drawing a diagram, determine the distance travelled north and the distance travelled west
by a plane flying on a bearing of 300° for 65km.
5. By drawing a diagram, determine the distance travelled east and the distance travelled south
by a ship sailing on a bearing of 158° for 95km.
12
WEEK 12
LESSON 1
TOPIC: Trigonometry
SUB-TOPIC: Sine Rule
CONTENT:
Or
Example:
1. Calculate the length of c.
13
𝑎
Law of Sines:
Then:
𝑆𝑖𝑛 𝐴
7
𝑆𝑖𝑛 35°
c=
c=
=
𝑏
𝑆𝑖𝑛 𝐵
=
𝑐
𝑆𝑖𝑛 𝐶
𝑐
=
𝑆𝑖𝑛 105°
7 × 𝑆𝑖𝑛 105°
sin 35°
7 ×0.574
0.966
c = 11.8 units (1dp)
2. Find the value of angle B
Law of Sines:
Then:
𝑆𝑖𝑛 𝐵
4.7
𝑆𝑖𝑛 𝐴
𝑎
=
=
𝑆𝑖𝑛 𝐵
𝑆𝑖𝑛 63°
5.5
𝑏
=
𝑆𝑖𝑛 𝐶
𝑐
Sin B =
4.7 × 𝑆𝑖𝑛 63°
5.5
B = 𝑠𝑖𝑛−1 (0.7614)
B = 49.6° (1dp)
14
= 0.7614
EXERCISE:
1. Calculate angle R
2. Calculate the length of side BC of the triangle shown below.
3. Calculate the missing lengths of the following triangle.
4. Calculate the angles of the triangle shown below.
15
5. Work out the length of x in the diagram below:
16
WEEK 12
LESSON 2
TOPIC: Trigonometry
SUB-TOPIC: Application of Sine Rule
CONTENT:
Example:
A man leaves a point A walking at 6.5 km/h in a direction of 0700. A cyclist leaves the same
point at the same time in a direction 1300 travelling at a constant speed. Find the average speed
of the cyclist if the walker and the cyclist are 80km apart after 5 hours.
Solution:
Speed =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑇𝑖𝑚𝑒
Distance = Speed × time
After 5 hours, the walker travelled 5 × 6.5 = 32. 5 km as shown in the diagram below.
If AC is the distance the cyclist travels in 5 hours then BC = 80km.
17
In the triangle ABC, two sides are known and angle BAC is also known.
18
EXERCISE:
1. Two ships S and T leave a port O at the same time. Ship S moves at 22 km/h on a bearing
of 1250 and T on a bearing of 2000. After 1 1/2 hours, the bearing of ship T from S is
2400. Calculate, correct to one decimal place,
a) The initial speed of the ship T:
b) The distance between the positions of the ships at that instance.
2. Two vehicles P and Q set out from a station at the same time. P moves up north and Q on
a bearing of 0470. When Q had travelled 10 km, the vehicles were 8 km apart. How far
was P from the station?
3. Three towns A, B and C are all at sea level. The bearings of towns B and C from A are
360 and 2470 If B is 120 km from A and C is 234 km from A, calculate the distance BC
and bearing of town C from B.
4. From the lighthouse L an aircraft carrier A is 15 km away on a bearing of 1120 and a
submarine S is 26 km away on a bearing of 2000. Find:
a) The distance between A and S.
b) The bearing of A from S.
5. Two honey bees A and B leave the hive H at the same time; A flies 27 m due south and B
flies 9 m on a bearing 1110. How far apart are they?
6. From A, B lies 11 km away on a bearing of 0410 and C lies 8 km away on a bearing of
3410. Find
a) The distance between B and C.
b) The bearing of B from C.
7. An aircraft flies from its base 200 km on a bearing of 1620, then 350 km on a bearing of
2600, and then returns directly to base. Calculate the length of the shortest route and the
bearing of the return journey from the base.
8. A destroyer D and a cruiser C leave port P at the same time. The destroyer sails 25 km on
a bearing of 0400 and the cruiser sails 30 km on a bearing of 3200. How far apart are the
ships?
9. A vertical flagstaff 15 m high stands on a horizontal plane. Find the angles of elevation of
the top and middle point of the flagstaff from a point on the horizontal plane 4.5 m from
the foot of the flagstaff.
19
WEEK 12
LESSON 3
TOPIC: Trigonometry
SUB-TOPIC: Cosine Rule
CONTENT:
The Cosine Rule can be used in any triangle where you are trying to relate all three sides to one
angle.
Finding Sides
If you need to find the length of a side, you need to know the other two sides and the opposite
angle.
You need to use the version of the Cosine Rule where a2 is the subject of the formula:
a2 = b2 + c2 – 2bc cos(A)
Side a is the one you are trying to find. Sides b and c are the other two sides, and angle A is
the angle opposite side a. (two given sides and the angle between them)
Finding Sides Example
Work out the length of x in the diagram below:
Step 1: Start by writing out the Cosine Rule formula for finding sides:
a2 = b2 + c2 – 2bc cos(A)
Step 2: Fill in the values you know, and the unknown length:
x2 = 222 + 282 – 2×22×28×cos(97°)
Step 3: Evaluate the right-hand-side and then square-root to find the length:
20
x2 = 222 + 282 – 2×22×28×cos(97°)
(evaluate the right hand side)
x2 = 1418.143.....
(square-root both sides)
x = 37.7 (accurate to 3 significant figures)
2. Work out angle P° in the diagram below:
Step 1: Start by writing out the Cosine Rule formula for finding angles: cos(A) =
Step 2: Fill in the values you know, and the unknown length: cos(P) =
𝑏2 + 𝑐 2 – 𝑎2
52 + 82 – 7
2𝑏𝑐
2×5×8
Step 3: Evaluate the right-hand-side and then use inverse-cosine (cos–1) to find the angle:
cos(P) =
52 + 82 – 7
2×5×8
cos(P) = 0.5
P = cos–1(0.5)
P = 60°
21
EXERCISE:
1. a) Find the missing side in the diagram below:
2. Find the missing angle in the diagram below:
3. Find the unknown angle in each of the following diagrams:
4. Find the unknown angle in each of the following diagrams:
5. Solve the triangle ABC in which AC = 105cm, AB = 76cm and A = 29°.
22
WEEK 12
LESSON 4
TOPIC: Trigonometry
SUB-TOPIC: Application of Cosine Rule
CONTENT:
Example:
A tunnel is to be built through a mountain. To estimate the length of the tunnel, a surveyor makes
the measurements shown in Figure below. Use the surveyor’s data to approximate the length of
the tunnel.
Solution:
Using the cosine rule: 𝑎2 = 𝑏 2 + 𝑐 2 − 2𝑎𝑏 𝑐𝑜𝑠𝐴
𝑎 = √3882 + 2122 − 2(388)(212) 𝐶𝑜𝑠 82.40
𝑎 = 416.8 𝑓𝑡
23
EXERCISE:
1. A farmer has a triangular field with sides 120 yards, 170 yards, and 220 yards. Find the
area of the field in square yards. Then find the number of acres if 1 acre = 4840 square
yards.
2. Three scientists are out setting up equipment to gather data on a local mountain. Person 1
is 131.5 yards away from Person 2, who is 67.8 yards away from Person 3. Person 1 is
72.6 yards away from the mountain. The mountains forms a 103∘ angle with Person 1
and Person 3, while Person 2 forms a 92.7∘ angle with Person 1 and Person 3. Find the
angle formed by Person 3 with Person 1 and the mountain.
3. A support at a construction site is being used to hold up a board so that it makes a
triangle, like this:
If the angle between the support and the ground is 17∘, the length of the support is
2.5 meters, and the distance between where the board touches the ground and the bottom
of the support is 3 meters, how far along the board is the support touching?
24
4. Dirk wants to find the length of a long building from one side (point A) to the other
(point B). He stands outside of the building (at point C), where he is 500 ft from point A
and 220 ft from point B. The angle at C is 94∘. Find the length of the building.
5. While hiking one day you walk for 5 miles due east, then turn to the left and walk 3 more
miles 30∘ west of north. At this point you want to return home. How far are you from
home if you were to walk in a straight line?
25
WEEK 13
LESSON 1
TOPIC: Trigonometry
SUB-TOPIC: Trigonometrical Formula for Area of parallelogram
CONTENT:
Example:
Find the area of a parallelogram if its two adjacent sides are 80cm and 40cm and the angle
between them is 56°.
Solution:
Let a = 80 cm and b = 40 cm.
The angle between a and b = 56°.
Area = ab sine (α)
Substitute.
A = 80 × 40 sine (56)
A = 3,200 sine 56
A = 2,652.9 𝑐𝑚2
26
EXERCISE:
Calculate the area of the shapes below:
1.
2.
3.
27
WEEK 13
LESSON 2
TOPIC: Trigonometry
SUB-TOPIC: Trigonometrical Formula for Area of Triangle
CONTENT:
Area =
𝟏
𝟐
𝒂𝒃 𝑺𝒊𝒏 𝜽
Example:
To find the area of the triangle
Use the formula
inserting the values that you know.
Solve for the value of the area.
28
𝐴 = 8660.25 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
The area is about 8,660 square units.
EXERCISE:
1. Find the area of ΔPQR
2. Find the area of triangle ABC.
3. Calculate the area of ABC
4. Calculate the area of ABC
29
5. Calculate the area of DEF
30
WEEK 13
LESSON 3
TOPIC: Trigonometry
SUB-TOPIC: Area of Segment of a circle
CONTENT:
Example: Determine the area of the shaded region in each figure below.
In circle A, the radius is 2.6 units, ∠CAB = 79°, and BC = 3.3 units.
31
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐴𝐵𝐶 =
1
2
𝑏𝑐 𝑠𝑖𝑛𝐴
1
𝐴 = 2 (2.6)(2.6) sin 79
𝐴 = 3.3 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠
𝜃
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟 = 𝜋𝑟 2 × 360
𝐴 = 3.142(2.6)2 ×
79
360
A = 4.66 Square units
Area of Segment = Area of sector – Area of Triangle
A = 4.66 Square units – 3.3 Square units
Alternatively
A = 1.36 Square units
32
EXERCISE:
1. A lawn sprinkler waters a circular portion of a yard. The radius of the circular potion is 7
ft. Calculate the area of the yard watered by the sprinkler to the nearest square foot.
2.
3.
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WEEK 13
LESSON 4
TOPIC: Trigonometry
SUB-TOPIC: CXC Questions on the Trigonometry
CONTENT:
1.
2.
34
3.
35
4.
5.
36
6.
7.
37
WEEK 14
LESSON 1
TOPIC: Geometry 2
SUB-TOPIC: Length of a straight line
CONTENT:
Example: Find the distance between A(2, 3) and B(6, 8).
𝑥1 = 2, 𝑥2 = 6, 𝑦1 = 3, 𝑦2 = 8
|AB| =√(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
|AB| =√(6 − 2)2 + (8 − 3)2
|AB| =√42 + 52
|AB| =√16 + 25
|AB| =√41
|AB| = 6.4 units
Exercise:
1. What would be the length of a line with endpoints at (1,6) and (9,9)?
2. What is the length of a line segment with end points (0,1) 𝑎𝑛𝑑 (5,7)?
3. What line goes through the points (−1,5) 𝑎𝑛𝑑 (5,15)?
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4. What is the length of a line with endpoints at (4,10) 𝑎𝑛𝑑 (8,13)?
5. What is the length of a line with endpoints at (6,13) 𝑎𝑛𝑑 (9,17)?
6. Find the distance between points (2, −4) 𝑎𝑛𝑑 (14, −2).
7. What is the length of a line with endpoints of (2, −4) 𝑎𝑛𝑑 (7,9)?
8. What is the distance between (1,5) 𝑎𝑛𝑑 (6,17)?
9. A line segment has endpoints at (2,3) 𝑎𝑛𝑑 (−7,3). What is the distance of this segment?
10. If a line has a length of 5, and the endpoints are (0,0) 𝑎𝑛𝑑 (4, 𝑦), what is the value of y?
39
WEEK 14
LESSON 2
TOPIC: Geometry 2
SUB-TOPIC: Gradient of straight line
CONTENT: The gradient of a straight line is the rate at which the line rises (or falls)
vertically for every unit across to the right. That is:
In general, if we have two coordinates (𝑥1 , 𝑦1 ) ) and (𝑥2 , 𝑦2 ) then the gradient of the line that
passes through them is,
Gradient =
𝑟𝑖𝑠𝑒
=
=
𝑟𝑢𝑛
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑦
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑥
𝑦2 −𝑦1
𝑥2 −𝑥1
The gradient of a straight line is denoted by m where:
40
Example:
1. Find the gradient of the straight line joining the points P(– 4, 5) and Q(4, 17).
Solution:
So, the gradient of the line PQ is 1.5.
3. A horse gallops for 20 minutes and covers a distance of 15 km, as shown in the diagram.
Find the gradient of the line and describe its meaning.
41
Solution:
In the above example, we notice that the gradient of the distance-time graph gives the speed (in
kilometres per minute); and the distance covered by the horse can be represented by the
equation:
42
EXERCISE
1. Find the gradient of the straight line joining the points 𝐴(6, 0) 𝑎𝑛𝑑 𝐵(0, 3).
2. Find the gradient of the line joining the points 𝐴(5, 8) 𝑎𝑛𝑑 𝐵(3, 10).
3. Given two points, 𝑃 = (0, – 1) 𝑎𝑛𝑑 𝑄 = (4,1), calculate the slope of the line.
4. Find the slope of the line containing the points (−10, −4) 𝑎𝑛𝑑 (−15, −6).
5. The cost of transporting documents by courier is given by the line segment drawn in the
diagram. Find the gradient of the line segment; and describe its meaning.
43
WEEK 14
LESSON 3
TOPIC: Geometry 2
SUB-TOPIC: Midpoint of a straight line
CONTENT:
Example: Determine the midpoint of the points A(1,4) and B(5,6) is
MAB = (
𝑥1 +𝑥2 𝑦1 +𝑦2
2
,
2
1+5 4+6
)= (
2
,
2
6 10
)= ( ,
2
2
)=
(3, 5)
Example: M(3, 8) is the midpoint of the line AB. A has the coordinates (-2, 3), Find the
coordinates of B.
Solution:
Let the coordinates of B be (x, y)
Coordinates of B = (8, 13)
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EXERCISE:
1. Find the midpoint of the two points A(1, -3) and B(4, 5).
2. Find the midpoint of the two points (5, 8) and (-5, -6).
3. Find the midpoint of the two points (-4, -8) and (-2, -10).
2 1
11 7
3 4
6
4. Determine the midpoint of the two points ( , ) and ( , ).
9
5. For a line segment DE, one endpoint is D(6, 5) and the midpoint M(4, 2). Find the coordinates
of the other endpoint, E.
45
WEEK 14
LESSON 4
TOPIC: Geometry 2
SUB-TOPIC: Equation of a straight line given two points
CONTENT: Equation of a straight line is 𝒚 = 𝒎𝒙 + 𝒄
Where m is the gradient and c is the y-intercept
Example: Find the equation of the straight line passing through the points (2, 3) and (6, − 5).
𝑚𝐴𝐵
=
=
𝑦2 −𝑦1
𝑥2 −𝑥1
==
−5 −3
6 −2
===
−8
4
= −2
Using 𝑚𝐴𝐵 = −2 and (2, 3); 𝑥1 = 2 and 𝑦1 = 3
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
Alternatively
𝑦 − 3 = −2(𝑥 − 2)
𝑦 = 𝑚𝑥 + 𝑐
𝑦 − 3 = −2𝑥 + 4
𝑐 = 𝑦 − 𝑚𝑥
𝑦 = −2𝑥 + 4 + 3
𝑐 = 3 − (−2)(2)
𝑦 = −2𝑥 + 7
𝑐 =3+4
𝑐=7
Now the equation of a straight line is written in the form 𝑦 = 𝑚𝑥 + 𝑐 substitute -2 for m and
7 for c the equation is 𝑦 = −2𝑥 + 7.
46
EXERCISE
1. Find the equation of the straight line joining the points (- 3, 4) and (5, - 2).
2. Find the equation of the straight line joining the points (4, 6) and (8, 26).
3. Find the equation of the straight line passing through (1, 1) and (4, −8).
4. Find the equation of the straight line passing through (3, 4) and (5, 4).
5. Find the equation of the straight line joining the points (0, 2) and (4, 0).
47
WEEK 15
LESSON 1
TOPIC: Geometry 2
SUB-TOPIC: Equation of a straight line given the two intercept point
CONTENT:
The 𝒙 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕 is the point 𝑨(𝒂, 𝟎) where the line cuts the 𝒙 − 𝒂𝒙𝒊𝒔.
The 𝒚 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕 is the point 𝑩(𝟎, 𝒃) where the line cuts the 𝒚 − 𝒂𝒙𝒊𝒔.
Example:
Given the intercepts 𝐴( −2, 0) and B(0, 8). Determine the equation of the line AB.
48
Solution:
𝐴( −2, 0) and B(0, 8)
𝑚𝐴𝐵
=
=
𝑦2 −𝑦1
𝑥2 −𝑥1
==
𝑥1 = −2, 𝑥2 = 0, 𝑦1 = 0, 𝑦2 = 8
8−0
0−(−2)
8
=== =4
2
Using 𝑚𝐴𝐵 = 4 and (−2, 0); 𝑥1 = −2 and 𝑦1 = 0
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑦 − 0 = 4(𝑥 − (−2))
𝑦 = 4(𝑥 + 2)
𝑦 = 4𝑥 + 8
EXERCISE
1. Given the intercepts 𝐴( −4, 0) and B(0, 12). Determine the equation of the line AB.
2. Given the intercepts 𝑀(2, 0) and N(0, 2). Determine the equation of the line MN.
3. State the equation of the line passing through the intercepts 𝑋( −16, 0) and Y(0, 24).
4. State the equation of the line passing through the intercepts 𝑃( 6, 0) and Q(0, −15).
5. Given the intercepts 𝐴( −3, 0) and B(0, −9). Determine the equation of the line AB.
49
WEEK 15
LESSON 2
TOPIC: Geometry 2
SUB-TOPIC: Parallel Lines
CONTENT:
Parallel lines are a fixed distance apart and will never meet, no matter how long they are
extended. Lines that are parallel have the same gradient.
Example
1. State the equation of a line that is parallel to 𝑦 = 3𝑥 + 7 and passes through the point (1, −3)
50
Solution:
Given 𝑦 = 3𝑥 + 7,
Written in the form 𝑦 = 𝑚𝑥 + 𝑐
the gradient, 𝑚 = 3
To be parallel, two lines must have the same gradient.
Using 𝑚 = 3 and (1, −3) where 𝑥1 = 1 and 𝑦1 = −3
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑦 − (−3) = 3(𝑥 − 1)
𝑦 + 3 = 3𝑥 − 3
𝑦 = 3𝑥 − 3 − 3
𝑦 = 3𝑥 − 6
EXERCISE
1. Find the equation of a line passing through the point (– 6, 5) parallel to the line 3𝑥 – 5𝑦 = 9.
2. What is the equation of a line parallel to y = −4x + 5 and passing through the point
(6, −3)?
3. Find the equation of a line passing through the point (3, – 4) parallel to the line 8𝑥 + 6𝑦 =
15.
4. Find the equation of a line passing through the point (6, 3) parallel to the line 4𝑥 – 5𝑦 =
– 10.
5. Find the equation of a line passing through the point (−8, −2) perpendicular to the line 𝑦 =
−2𝑥 + 6
51
WEEK 15
LESSON 3
TOPIC: Geometry 2
SUB-TOPIC: Perpendicular Lines
CONTENT: Two lines are perpendicular lines if their slopes are opposite reciprocals of each
other, or the product of their slopes equals - 1.

Slopes are Opposite Reciprocals:

Product of Slopes:

Graph:
52
Example: Find the equation of a line that is perpendicular to =
1
5
𝑥 − 2 , and passing through
the point (1, −3).
This is the graph of the given line and the point.
Given 𝒚 =
1
5
1
𝑥−2
𝑚1 = 5
Since the lines are perpendicular, then 𝑚1 × 𝑚2 = −1
1
𝑚2 = − 𝑚
1
The opposite reciprocal of 𝑚1 = −5. Hence, 𝑚2 = −5
Using 𝑚2 = −5 and (1, −3) where 𝑥1 = 1 and 𝑦1 = −3
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑦 − (−3) = −5(𝑥 − 1)
𝑦 + 3 = −5𝑥 + 5
𝑦 = −5𝑥 + 5 − 3
53
𝑦 = −5𝑥 + 2
EXERCISE
1. Find the equation of a line passing through the point (3, –4) perpendicular to the line 8𝑥 +
6𝑦 = 15.
2. Find the equation of a line passing through the point (6, 3) perpendicular to the line
4𝑥 – 5𝑦 = – 10.
3. One line passes through the points (0, –4) and (–1, –7); another line passes through the
points (3, 0) and (–3, 2). Are these lines parallel, perpendicular, or neither?
4. Find the slope of a line perpendicular to the line y = – 4x + 9.
5. What is the equation of a line perpendicular to 2y = x - 4 and passing through the point (−4,
1)?
54
WEEK 15
LESSON 4
TOPIC: Geometry 2
SUB-TOPIC: Equation of Perpendicular Bisector
CONTENT:
The perpendicular bisector is a line that divides a line segment into two equal sizes and makes a
right angle with the line segment it cuts through.
Explanation:
The vertical line would be the perpendicular bisector to segment AB. Note the two dashes on
each side of the bisected segment show congruence.
Example: Find the equation of the perpendicular bisector of the points (1, 4) and (5, −2).
Explanation:
 a perpendicular bisector, bisects a line segment at right angles
 to obtain the equation we require slope and a point on it
 find the midpoint and slope of the given points
55
𝑥1 +𝑥2 𝑦1 +𝑦2
M=(
,
2
2
1+5 4−2
)= (
2
,
2
6 2
)= (2 , 2) = (3, 1)←point on bisector
calculate the slope m using the gradient formula
(1, 4) and (5, −2)
𝑚=
𝑦2 −𝑦1
𝑥2 −𝑥1
=
−2 − 4
5 −1
=
−6
4
=
−3
2
given a line with slope m then the slope of a line perpendicular to it is
−1
𝑚𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 =
𝑚
−1
2
⇒𝑚𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 = −3 = ←slope of bisector
3
2
2
using 𝑚 = 3 and (𝑥1 , 𝑦1) = (3,1) then
2
𝑦 − 1 = 3 (𝑥 − 3)←in point-slope form
2
⇒𝑦 − 1 = 3 𝑥 − 2
2
⇒𝑦 = 3 𝑥 − 1←in slope-intercept form
56
EXERCISE
1. Find the equation of the perpendicular bisector of the points (- 3, 4) and (5, - 2).
2. Find the equation of the perpendicular bisector of the points (4, 6) and (8, 26).
3. Find the equation of the perpendicular bisector of the points (1, 1) and (4, −8).
4. Find the equation of the perpendicular bisector of the points (3, 4) and (5, 4).
5. The coordinates of the point P and Q are (−2, −3) and (4, 5) respectively. 𝑋 is the mid-point
of 𝑃𝑄.
(a) Calculate:
(i) the length of 𝑃𝑄
(ii) the gradient of 𝑃𝑄
(iii) the coordinates of 𝑋
(iv) the intercept of 𝑃𝑄 on the y-axis
(b) Determine the equation of the perpendicular bisector of 𝑃𝑄.
6.
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7.
8.
9.
58
59