Patterns and Relationships

What are Relations?

A relation is a rule which connects one set to another. We can express relations as an ordered pair (2, 6), in a diagram form, a rule form or graphical form. Relations could be a connection between the first set known as the domain and the second set called the co-domain.

Types of Relations

There are four types of relations and they are;

1. One -to-one relation: This is a relation in which each element in the domain (first set) has exactly one image in the co-domain (second set).

Example:

2. One-to-many relation: This is a relation in which one element in the domain has more than one (many) in the co-domain.

3. Many- to-one relation: This is a relation in which more than one element in the domain has only one element in the co-domain.

4. Many-to-many relation: This is a relation in which many elements in the domain have many images in the co-domain.

Activity 4

Can you draw a diagram to show many-to-many relations?

It could look something like this, but you could have a better one.

Functions Functions are relations where each element in the domain has only one image in the co-domain. One-to-one and many-to-one relations are functions because each element in the domain has exactly one image in the co-domain. Thus, every member of the domain has only one image in the co-domain.

Mapping Mapping is a relation such that each element in the domain is associated with an element in the co-domain.

A subset of the co-domain which is actually used by the function is called the range of the function. This is illustrated in figure below.

A subset of the co-domain, the range of the function as shown Rules of Mappings There are two rules for mappings

1. Linear mapping A mapping is said to be linear if the difference between the consecutive elements in both the domain and the co-domain is constant.

The rule for linear mapping is of the form y = mx + c Where, m = constant difference of the co − domain____________________________ constant difference of the domain

Example 9

What is the rule of the mapping?

Solution

The rule of the mapping is of the form y = mx + c m = 8 − 4/2 − 0 m = 2 Put the value of m into the equation y = 2x + c…(1) Now take any coordinate say (0, 4) and put it into equation 1(i.e. x = 0 and y = 4), 4 = 2(0) + c c = 4 Put c back into equation 1 to give the rule for the mapping above.

∴ y = 2x + 4

2. Exponential mapping A mapping is said to be an exponential mapping if the ratio between the consecutive elements in the co-domain is constant. The rule for exponential mapping is given as; y = brˣ−a

Example 10

i.

Common ratio (r) 4/2 = 8/4 = 16/8 = 2 The first term of x = a = 1 The first term of y = b = 2 Y can be expressed in terms of x by y = brˣ−a Hence the rule for the mapping is:

y = 2 × 2ˣ⁻¹ii.

common ratio (r) 1 / 9/1 / 3 = 1 / 27/1 / 9 = 1 / 81/1 / 27 = 1/3 the first term of x = a = 1 the first term of y = b = 1/3 y can be expressed in terms of x by y = brˣ−a Hence the rule for the mapping is:y = y = 1/3 × ( ¹__ 3)ˣ⁻¹ Application of Concepts in Real-world

Example 11

Adowa has three jumpers and two skirts, combine these in six possible ways using arrow diagram.

Solution

Let the jumpers be A, B, and C Let the skirts be D and E.

Graphs of Linear Functions

Let’s look at graphs of linear functions and their interpretations.

Figure 2: Straight line graph Identifying Linear Graphs

Example 12

Mr. Benyah asks Yakubu to identify whether the given equation 3x − 7y = 16 forms a linear graph or not without plotting its values.

Solution

First, Yakubu needs to identify the type of equation. Next, he needs to remember that any linear equation in two variables always represent a straight line if it can be rearranged to be in the form y = mx + c.

3x − 7y = 16 can be rearranged to 7y = 3x − 16 and then further rearranged to y = 3/7 x − 16/7 . Therefore, the above equation represents a straight line.

Example 13

Draw a graph of a straight line with a gradient of 1 and explain your answer.

Figure 3: Straight line graph This graph with a gradient 1 pass through the origin and slopes from left to right.

For every one unit we go to the right, we go up one unit.

Now draw a straight-line graph with a gradient of -1 and explain your answer.

Figure 4: Straight line graph This graph with a gradient of -1 slopes downwards and it passes through the origin. For every one unit we go to the right, we go down one unit.

Example 14

Draw the graph of y = 4x + 6 and explain what happens if the constant 6 is changed to 1.

Solution

The graph’s gradient is 4, so for every unit we move to the right, we move up 4 units. The graph intersects the y axis at (0,6). This means the graph looks like this:

Figure 5: Straight line graph If the equation is changed to y = 4x + 1 the gradient remains the same, but the intersection with the y axis is now at (0,1), so the moves “down” in order to cut y-axis at (0, 1).

Example 15

Draw the graph of y = 4x + 6 and explain what happens if the coefficient 4 is changed to 2, 1, 0 and -1 respectively.

Solution

As the co-efficient decreases, the gradient changes as the lines are rotated clockwise about (0, 6)

Figure 6: Straight line graph

Activity 5

Family Relations

Relation: Consider a family where each person is related to others through various roles (parent, child, sibling).

Function: If we focus on the relation between parents and children, each parent (father or mother) typically has a set of children. In this case, the relation from parent to child is a function because each parent (input) is associated with a unique child or set of children (output). For example, Mr. Kwame has children Ama, Kofi, and Kwesi; Mrs. Akua has children Yaa and Kojo. This demonstrates a function where each parent has a specific set of children.

Hello Learner! Now that you have studied the above scenario, discuss with your friend (either at home or in school) how functions can be identified from the following relations:

i) Academic adviser to student

ii) Employer to employees

iii) Market vendor to products

Activity 6

Ama has three Defenders and two Strikers, combine these in six possible ways using arrow diagram.

Solution

Step-by-step solution:

1. List all Defenders and Strikers:

- Defenders: A, B, C

- Strikers: D, E

2. Pair each Defender with each Striker:

- A with D

- A with E

- B with D

- B with E

- C with D

- C with E

3. An arrow diagram below is a visual representation that combines all.

Each arrow from a Defender to a Striker represents a specific outfit combination.DEABC You can see from the arrow diagram that the solution is complete with six possible combinations shown by the six arrows.

Activity 7

Example 16

Consider a linear mapping having the differences between consecutive images being constant. E.g. Given two sets X = {0 , 1 , 2 , 3 , 4} and Y = {2, 5, 8, 11, 14}

• Let the first term of set X be a

• Let the first term of set Y be b

• Denote each element in set X be x and that of set Y be y as shown below.X01234 2Y 5 8 11 14

• Let the common difference between consecutive image in Y be d

• Let the common difference between consecutive image in X be k

• Calculate the common difference between consecutive image in Y d = 5 − 2 = 8 − 5 = 11 − 8 = 14 − 11 = 3

• Calculate the common difference between consecutive image in X k = 5 − 4 = 6 − 5 = 7 − 6 = 8 − 7 = 1

• The first term of the set x = a = 0

• The first term of the set y = b = 2

• Express y in terms of x by y = b + ᵈ_ ₖ(x − a)

• Hence you arrived at the rule for the above mapping as y = 2 + 3/1(x − 0) y = 2 + 3x

Example 17

Find the rule for the mapping

• Cross check both the substitutions and the answer if they are correct or not

• Check the substitutions a = 0, b = 2, d = 4 and k = 2

• Check the rule for the mapping as given by y = 2 + 4/2(x − 0)

• Check the final answer y = 2 + 2x

Activity 8

Given an Exponential mapping having a common ratio between consecutive images being constant. E.g. Consider the two sets X = {0, 1, 2, 3, } and Y = {3, 6, 12, 24,}

• Let the first term in the set X = a and that of set Y = b

• Represent each element of set X be x and that of Y be y Using the diagramY361224X0123

• The common ratio of the element Y = r = 6/3 = 12/6 = 24/12 = 2

• Let the first term of the set X = a = 0

• let the first term of the set Y = b = 3

• Express y in terms of x by the standard relation y = b rˣ−a and substitute all the values as mentioned above and we have y = 3 × 2ˣ

Facts about gradient 1 The gradient of a line is the measure of the steepness of a straight line.

2 The gradient of a line can be either positive (uphill) or negative (downhill) or 0 (horizontal) and does not need to be a whole number.

3 The gradient of a line is the measure of the steepness of a straight line.

How to Understand the Gradient of a Line

Imagine walking up a set of stairs. Each step has the same height and you can only take one step forward each time you move. If the steps are taller, you will reach the top of the stairs quicker, if each step is shorter, you will reach the top of the stairs more slowly.

Let’s look at sets of stairs, The blue steps are taller than the red steps and so the gradient is steeper (notice the blue arrow is steeper than the red arrow).

The green steps are not as tall as the red steps so the gradient is shallower (the green arrow is shallower than the red arrow).

Gradients can be positive or negative but are always observed from left to right.

Figure 7: Straight line graph Finding gradients: In placing a ladder against a wall or tree, a change in the position of the top of the ladder will be because of a change in position of the foot of the same ladder.

Figure 8: Gradient of a line Gradient = Rise____ Run = Change in y_________ Change in x = y₂– y₁_____ x₂– x₁ The gradient of a straight line is denoted by m where: m = y₂ − y₁_____ x₂ − x₁ Now that we have understood the concept of gradient and learned the formula to find the gradient of a straight line joining two points, it’s time to put our knowledge into practice. Let’s undertake the two activities below on finding gradient between two points using the formulae.

Activity 9

Find the gradient of the straight line joining the points P(– 4, 5) and Q(4, 17).

Step 1: Identify the coordinates of the points P(–4, 5) and Q(4, 17)

Step 2: Write the formula for gradient (m) m = y₂ − y₁_____ x₂ − x₁

Step 3: Plug in the coordinates into the formula m = 17 − 5/4 − ( − 4)

Step 4: Simplify the expression m = 12/8 = 3/2

Step 5: Calculate the gradient m = 1.5

Figure 9: Gradient of a line The gradient of the straight line joining points P(–4, 5) and Q(4, 17) is 1.5.

Note: If the gradient of a line is positive, then the line slopes upward as the value of x increases.

Activity 10

Find the gradient of the straight line joining the points A(6, 0) and B(0, 3).

Step 1: Identify the coordinates of the points A(6, 0) and B(0, 3)

Step 2: Write the formula for gradient (m) m = y₂ − y₁_ x₂ − x₁

Step 3: Plug in the coordinates into the formula m = 3− 0/0 − 6

Step 4: Simplify the numerator 3 – 0 = 3

Step 5: Simplify the denominator 0 – 6 = – 6

Step 6: Write the simplified formula m = 3__

–6

Step 7: Calculate the gradient m = − 1/2

Figure 10: Gradient of a line The gradient of the straight line joining points A(6, 0) and B(0, 3) is −1/2

Note: When the gradient of the line is negative. It indicates that the line slopes downward from the left to right. As the value of (x ) increases, the corresponding (y ) values decreases, resulting in a reduction along the line.

Now that you have mastered the skill of finding the gradient of a line when given two points, let’s explore how gradients are applied in real-world scenarios Applications of Gradients Gradients are an important part of life. The roof of a house is built with a gradient to enable rain water to run down the roof. An aeroplane ascends at a particular gradient after take-off, flies at a different gradient and descends at another gradient to safely land. Tennis courts, roads, football and cricket grounds are made with a gradient to assist drainage.

Activity 11

Example 18

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

Figure 11: Gradient of a line Finding the Gradient of a Horse’s Gallop

Solution:

Let (t₁, d₁) = (0, 0) and (t₂, d₂) = (20, 15)

Step 1: Identify the coordinates

- Time (x-axis): 0 minutes to 20 minutes

- Distance (y-axis): 0 km to 15 km

Step 2: Write the formula m = d₂₋ t₂_____ d₁ − t₁

Step 3: Plug in the values m = (15 km − 0 km)__________________ (20 minutes − 0 minutes)

Step 4: Simplify and calculate m = 15 km/20 minutes m = 3/4 km / minute The gradient of the horse’s gallop is m = 3/4 km / minute Meaning: So, the gradient of the line is 3/4 km/min. 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: d = 3__ 4t (∴ Distance = Speed × Time) Interpretation: The gradient is a measure of the horse’s speed. A steeper gradient would indicate a faster speed, while a shallower gradient would indicate a slower speed.

Hello learner, now that you have completed the task of determining the gradient of a horse’s gallop, let’s move on to another example.

Example 19

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.

Figure 12: Gradient of a line

Solution

Let (d₁, c₁) = (0, 5) and (d₂, c₂) = (6, 23) Now, m = d₂₋ c₂______ d₁ − c₁ = 23 − 5/6 − 0 = 18/6 = 3 So, the gradient of the line is 3. This means that the cost of transporting documents is GH¢ 3 per km plus a fixed charge of GH¢ 5, i.e. it costs GH¢5 for the courier to arrive and GH¢ 3 for every kilometre travelled to deliver the documents.

Hello learners, Now that we have explored the applications of gradients, let’s take the next step and learn how to find the gradient of a straight line when given its equation.

Finding the Gradient of a Straight Line Given the Equation We can determine the gradient from a given equation of a straight line when the equations are given in the form or can be rearranged to be in the form y = mx + c, where m is the gradient.

Example 20

Given the equation 2y − 6x = 12 , first rewrite the equation in the general form y = mx + c, where m is the gradient.

Therefore, 2y − 6x = 12 can be written as 2y = 6x + 12 Now, making one y the subject we have y = 3x + 6 .

Since our new equation is in the general form, we compare and identify the gradient. Hence, the gradient of the equation 2y − 6x = 12 is 3.

Now that you have grasped the example above, please proceed with this activity

Activity 12

Find the gradient of the equation 3y + 2x = 7.

Let’s solve it with the steps below:

Step 1: Write the equation in slope-intercept form (y = mx + c ) To find the gradient (m), we need to rewrite the equation in slope-intercept form.

Step 2: Subtract 2x from both sides 3y + 2x − 2x = 7 − 2x

Step 3: Simplify

3y = 7 − 2x

Step 4: Divide both sides by 3 3y/3 = 7 − 2x/3

Step 5: Simplify

y = 7/3 − 2/3 x or y = −2/3 x + 7/3

Step 6: Identify the gradient (m) The gradient (m) is the coefficient of x, which is −2/3.

Therefore, the gradient of the equation 3y + 2x = 7 is −2/3.

Now that we’ve learned how to determine the gradient of a given equation, let’s move on to finding the equation of a straight line.

Finding the Equation of a Straight Line

Example 21

Find the equation of the line with gradient −2 that passes through the point (3, −4).

Solution

Put m = −2, x₁ = 3 and y₁ =−4 into the formula y − y₁ = m(x − x₁ ) y − y₁ = m(x − x₁ ) y − − 4 = − 2(x − 3) Expand the brackets and simplify, y + 4 = − 2x + 6 y = −2x + 2

Figure 13: Gradient of a line

Perpendicular Lines

Definition: Perpendicular lines are two lines that intersect at a 90° angle, forming right angles.

Figure 14: Perpendicular lines Real-Life Example: Consider a door frame and the floor. The door frame (vertical line) and the floor (horizontal line) intersect to form a right angle, making them perpendicular.

Figure 15: Perpendicular lines Parallel Lines Definition: Parallel lines are two or more lines that never intersect. They have the same slope and are equidistant from each other.

Figure 16: Parallel lines Real-Life Example: Look at railroad tracks. The two tracks run alongside each other, never converging or intersecting. This demonstrates parallel lines in real life.

Figure 17: Railway lines showing Parallel lines

Example 22

Identify which of the lines are parallel and which perpendicular.

i. y = 3x + 1

ii. y = 3x + 12

iii. y = 1/4 x − 5

iv. y = − 4x + 1

Solution

Parallel lines have the same slope. Since the functions y = 3x + 1 and y = 3x + 12 each have the same gradient (= 3), they represent parallel lines.

Figure 18: Parallel lines Perpendicular lines have negative reciprocal slopes. Since − 4 and 1/4 are negative reciprocals the equations y = 1/4 x − 5 and y = − 4x + 1, they represent perpendicular lines.

Figure 19: Perpendicular lines

Activity 13

Example 23

Find the equation of a line that is perpendicular to the line y = 2x − 2 and goes through the point (1,3).

Solution

We know that the general equation of a straight line is y = mx + c.

Firstly, we need to find the gradient of the line y = 2x – 2 .

If we label the gradient of our line as m₁and the gradient of the line that is perpendicular with our line m₂ then we know that the product of those two gradients should be –1.

The gradient of our line is m₁ = 2 Meanwhile the gradient of the line perpendicular to our line is:

m₂ = − 1/m₁ m₂ = −1_ 2 After finding the gradient the equation of the line we want to find takes the form y = −1/2 x + c To find the value of c, we substitute the point (1,3) on our equation, since the graph of this line passes through this point.

y = −1/2 x + c 3 = −1/2 × 1 + c 3 = −1/2 + c c = 3 + 1/2 c = 6/2 + 1/2 = 7/2 The final form of our line that is perpendicular with the given line is: y = −1/2 x + 7/2

Example 24

Find the equation of a line that is parallel to the line x + y – 1 = 0 and goes through the point (-1, 1).

Solution

Firstly, we rewrite our line x + y − 1 = 0 in the correct form y = − x + 1 Then we find the gradient of our line that is m₁= –1 We know that parallel lines have the same gradient so the gradient of the line we are going to find is m₁= m₂= – 1 The line takes the form y = − x + c To find the value of c, we substitute the point (-1,1) into our equation, since the graph passes through this point.

y = − x + c 1 = -(-1) + c 1 = 1 + c c = 1 – 1 = 0 The final form of our parallel line is y = − x .

Finding the Magnitude of Line Segment

The magnitude of a line also known as its “length”, “distance” or “modulus” describes the length of a line linking two points. Relating length of objects discussed from activities, if P and Q have coordinates (x₁, y₁) and (x₂, y₂). From the figure below:

Figure 20: Magnitude of Line Segment

∆ x = x₂ − x₁ ∆ y = y₂ − y₁ Where ∆ means a change By Pythagoras theorem, |^(PQ)|²= ∆x²+ ∆y²|^(PQ)| = √∆x²+ ∆y²|^(PQ)| = √________________ (x₂ − x₁)²+ (y₂ − y₁)²So, if P (x₁, y₁) and Q (x₂, y₂) are two points in the x-y plane, then the distance between P and Q is |^(PQ)| = √________________ (x₂ − x₁)²+ (y₂ − y₁)²

Note: The symbol ( D ) called delta as used here implies change in x₂, x₁ and y₂, y₁ or the differences in their values.

Example 25

Determine the distance between the points

(a) P(2, 1) and Q(5, 5)

(b) A(7, -3) and B(-1, 5)

(c) D(4, 1) and E(-3, -5)

Solution

The distance between two points given by √________________ (x₂ − x₁)²+ (y₂ − y₁)²(a) P(2, 1) and Q(5, 5) |PQ| = √______________ (5 − 2)²+ (5 − 1)²= √3²+ 4²|PQ| = √25 = 5 units

(b) A (7, -3) and B(-1, 5) ⇒ |AB| = √_______________ (− 1 − 7)²+ (5 + 3)²= √( − 8)²+ 8²|AB| = √128 units

(c) D(4, 1) and E(-3, -5) |DE| = √________________ (− 3 − 4)²+ (− 5 − 1)²= √____________ ( − 7)²+ ( − 6)²|DE| = √49 + 36 = √85 units Determine the Midpoint of a Line What is a midpoint?

In geometry, the midpoint is the middle point of a line segment. It is equidistant from both endpoints, and it is the midpoint of both the segment and the endpoints.

It bisects the segment.

For example: Consider the following line segment and the points A (3, 4) and B (5, 10).

We can determine the mid-point.

HOW?

If you add both x co-ordinates and then divide by two you get (3 + 5)______ 2 = 8/2 = 4 If you add both y co-ordinates and then divide by two you get (4 + 10)______ 2 = 14/2 = 7 This gives a new point with co-ordinates (4, 7). This point is exactly halfway between A and B.

Activity 14

Tom and Alex are planning a road trip from City A to City B. City A is located at (30, 40) on a map, and City B is located at (60, 80). If they stop for lunch at the midpoint of their journey, what are the coordinates of the lunch spot?

Step 1: Find the midpoint formula The midpoint formula is:, ( x₁ + x₂_____ 2 ), ( y₁ + y₂_____ 2 )

Step 2: Identify the coordinates City A: (x₁, y₁) = (30, 40) City B (x₂, y₂) = (60, 80)

Step 3: Plug in the values Midpoint = ( 30 + 60/2 , 40 + 80/2 )

Step 4: Calculate the coordinates Midpoint = (90/2 , 120/2 ) Midpoint = (45, 60) The coordinates of the lunch spot are (45, 60).