4) Linear Relationships. yr 9 - Notes

Maths in Context: Animation, Cartoon, and Game Developers

• Animation uses algebra, coordinate geometry, linear relations, trigonometry, and advanced mathematics.

• Animators code a subdivision process using the midpoint formula to morph 2D polygons into curves.

• Repeatedly subdividing lengths and slicing off corners creates a virtual curve illusion.

• A similar process builds 3D shapes from polyhedral objects with flat polygon faces.

• Repeated subdivision and reduction result in finished shapes with countless flat surfaces, appearing as smooth curves on screen.

• Linear algebra controls position and movement of virtual objects.

• Linear equations transform virtual objects by altering vertex coordinates.

• Movement is achieved by adding or subtracting coordinates.

• Enlargement or shrinking uses multiplication or division of coordinates.

• Rotation is achieved using trigonometry.

• Example: The cartoon Bluey uses these techniques.

Chapter Contents

• 4A Introducing linear relationships (Consolidating)

• 4B Graphing straight lines using intercepts

• 4C Lines with one intercept

• 4D Gradient

• 4E Gradient and direct proportion

• 4F Gradient – intercept form

• 4G Finding the equation of a line using y = mx + c

• 4H Midpoint and length of a line segment from diagrams

• 4I Perpendicular lines and parallel lines

• 4J Linear modelling

• 4K Graphical solutions to simultaneous equations

NSW Syllabus

• Develops mathematical understanding by exploring and connecting concepts, applying techniques, and communicating reasoning.

• Determines midpoint, gradient, and length of an interval; graphs linear relationships with and without digital tools.

• Graphs and interprets linear relationships using gradient/slope-intercept form.

• Describes and applies transformations, midpoint, gradient, distance formulas, and line equations to solve problems.

Online Resources

• Interactive Textbook, including HOTmaths content, video demonstrations, auto-marked quizzes.

Linear Relationships

• Mathematical rules describe relationships between variables.

• Linear relationships can be illustrated with a straight line graph.

• Example: Volume of petrol increasing at a constant rate.

Lesson Starter: Is it Linear?

• Three rules linking $x$ and $y$:

  1. $y_1 = \frac{1}{2}x + 1$

  2. $y_2 = x^2 - 1$

  3. $y_3 = 3x - 4$

• Task: Complete a table and graph for these rules.

• Determine which rule is linear based on the table and graph.

4A Introducing Linear Relationships (Consolidating)

• Learning Intentions

  • Review the Cartesian plane and plot points using coordinates.

  • Understand that a linear relationship forms a straight line.

  • Complete a table of values and plot points to form a linear graph.

  • Arrange the equation of a line in various forms, including general form.

  • Determine if a point is on a line by substituting into the equation.

  • Identify the x-intercept and y-intercept.

• Past, Present, and Future Learning

  • Consolidates Stage 4 concepts used in Stages 5 and 6.

  • Some questions are more challenging than in Year 8.

  • Concepts will be revised and extended in Year 10.

  • Expertise is assumed knowledge for Stage 6 Advanced.

• Example: A car's annual running costs are $4000$ p.a., and the car cost $17000$. Total cost $C$ for $n$ years is expressed as: $C = 4000n + 17000$

4A Introducing Linear Relationships

• Building Understanding

  1. Refer to the Cartesian plane:

     • Give the coordinates of points A − L.

     • State which points lie on the x-axis and y-axis.

     • State which points lie inside the 2nd and 4th quadrants.

  2. For the rule $y = 3x − 4$, find the value of $y$ for given $x$-values.

• Key Ideas

  • Coordinate geometry links geometry and algebra.

  • The Cartesian plane consists of two axes dividing the plane into four quadrants.

    • Horizontal x-axis and vertical y-axis intersect at right angles at the origin $O(0, 0)$.

    • A point's position is given by the coordinate pair $(x, y)$, indicating horizontal and vertical positions from the origin.

  • A linear relationship is a set of ordered pairs $(x, y)$ that form a straight line when graphed.

  • Linear relationships have rules of the form:

    • $y = mx + c$ (e.g., $y = 2x + 1$).

    • $ax + by = d$ or $ax + by + c = 0$ (e.g., $2x − 3y = 4$ or $2x − 3y − 4 = 0$).

  • The x-intercept is the point where the graph meets the x-axis.

  • The y-intercept is the point where the graph meets the y-axis.

Example: For the line $2x + 3y = 6$
* Y Intercept : 2
* X Intercept : 3

4A Introducing Linear Relationships

• Example 1: Plotting points to graph straight lines

  • Using $-3 gtrslant x gtrslant 3$, construct a table of values and plot a graph for these linear relationships:

    • a) $y = x + 2$

    • b) $y = -2x + 2$

• Solution:

  • a) Substitute each value of $x$ into the rule $y = x + 2$

  • b) Substitute each value of $x$ into the rule $y = -2x + 2$

4A Introducing Linear Relationships

• Example 2: Reading off the x-intercept and y-intercept

  • Write down the coordinates of the x-intercept and y-intercept from this table and graph

    • The x interecept is the point where $y = 0$

    • The y intercept is the point where $x = 0$

• Example 3: Decide if a point is on a line

  • Decide if the point $(-2, 4)$ is on the line with the given rules:

    • a) $y = 2x + 10$

    • b) $y = -x + 2$

• Solution:

  • a) Substitute $x = -2$ into the equation:
    $y = 2x + 10$
    $y = 2(-2) + 10 = 6$
    Since $y = 6 \neq 4$, the point $(-2, 4)$ is not on the line

  • b) Substitute $x = -2$ into the equation:
    $y = -x + 2$
    $y = -(-2) + 2 = 4$
    Since $y = 4$, the point $(-2, 4)$ is on the line

Exercise 4A

• Fluency questions 1-3 (1/2)\n 1-4 (1/2)\n 1-4 (1/2)

• Problem Solving Questions 5, 6

• Reasoning Questions 7, 8, 9

4B Graphing Straight Lines Using Intercepts

• When linear rules are graphed, all points lie on a straight line.

• Graphing a straight line requires only two points.

• Two critical points are the x-intercept and y-intercept.

Lesson starter: Two key points

Consider the realtion $y = \frac{1}{2}x + 1$ and complete this table and graph.

• What are coordinates of the pount where the line crosses the y-axis? That is, state the coordinates of the y-intercept.

• What are coordinates of the pount where the line crosses the x-axis? That is, state the coordinates of the x-intercept.

• Discuss how you might find the coordinates of the x- and y-intercepts without drawing a table and plotting pounts. Explain your method.

4B Graphing Straight Lines Using Intercepts

• Learning Intentions

  • To understand that only two points are required to graph a straight line

  • To be able to graph a line by finding the x-intercept and y-intercept

• Past, present and future learning

  • This section addresses the Stage 5 Path Topic called Linear relationships C

  • These concepts will be revised, briefly, in Chapter 1 of our Year 10 book and also in Stage 6 Advanced

• Businesses can tax deduct the annual decrease in equipment values over n years. The straight line joining intercepts (0, V) and (n, 0) shows that the rate of decrease per year is $\frac{V}{R}$

4B Graphing Straight Lines Using Intercepts

• Key Ideas

  • Two points are required to sketch a straight line graph. Often these points are the axes intercepts.

  • The y-intercept is the point where the line intersects y-axis (where $x = 0$
    Substitute $x = 0$ to find the y-intercept

  • The x-intercept is the point where the line intersects x-axis (where $y = 0$
    Substitute $y = 0$ to find the x-intercept
    Example: $2x + 3y = 6$. Then y-int is (0,2) and x-int (3,0)

Example 4: Sketching with intercepts

• Sketch the graph of the following, showing the x- and y-intercepts

  • a) $2x + 3y = 6$
    To Find y-intercept, let $x = 0$
    $2(0) + 3y = 6$
    $3y = 6$
    $y = 2$
    Therefore, the y-intercept is at $(0, 2)$
    To Find x-intercept, let $y = 0$
    $2x + 3(0) = 6$
    $2x = 6$
    $x = 3$
    Therefore, the x-intercept is at $(3, 0)$

  • b) $y = 2x - 6$
    To Find y-intercept, let $x = 0$
    $y = 2(0) - 6$
    $y = -6$
    Therefore, the y-intercept is at $(0, -6)$
    To Find x-intercept, let $y = 0$
    $0 = 2x - 6$
    $6 = 2x$
    $x = 3$
    Therefore, the x-intercept is at $(3, 0)$

Exercise 4B:

• Fluency questions 1-2 (1/2)\n 1-3 (1/2)\n 1-3 (1/3)

• Problem Solving Questions 4, 5

• Reasoning Questions 7, 8, 9

4C Lines with One Intercept

Lesson starter: What rule satisfies all points?

• Here is one vertical and one horizontal line.

• For the vertical line shown, write down the corrdinates of all the points shown as dots.

• What is always true for each coordinate pair?

• What simple equation describes every pount on the line?

• For the horizontal line shown, write down the coordinates of all the points shown as dots.
  What is always true for each coordinate pair?

• What simple equation describes every pount on the line?

• Where do the two lines Intersect?

4C Lines with One Intercept

• Learning intentions

  • To be able to use equations to graph vertical lines and horizontal lines

  • To know that every line of the form y = mx passes through the origin

  • To be able to graph lines of the form y = mx usung the origin and one other pount

• Past, present and future learning

  • This section addresses the Stage 5 Core Topic called Linear relationshups A

  • These concepts will be revised, briefly, in Chapter 1 of our Year 10 book and also in Stage 6 Advanced

• Lines with one intercept include vertical lines, horizontal lines, and lines that pass through the origin
  Linear relationships with graphs through (0, 0) include:
  Fertiliser in kg = usage rate in kg/acre x number of acres
  Annual interest = interest rate p.a. x amount invested
  Weekly pay = pay rate in $/h x number of hours

Key Ideas:

• Vertical line: x = k

  • Parallel to the y-axis

  • Equation of the form x = k, where k is a constant

  • x-intercept is at (k, 0)

• Horizontal line: y = c

  • Parallel to the x-axis

  • Equation of the form y = c, where c is a constant

  • y-intercept is at (0, c)

• Lines through the origin (0, 0): y = mx

  • y-intercept is at (0, 0)

  • x-intercept is at (0, 0)

  • Substitute x = 1 or any other value of x to find a second point

4C Lines with One Intercept

Graphing vertical and horizontal Lines

• Sketch the graph of the following horizontal and vertical lines

  • a) $y = 3$
    Sketch a horizontal line through all points where y = 3, the y-intercept is at (0,3)

  • b) $x = -4$
    Sketch a horizontal line through all points where x = -4, the x-intercept is at (-4, 0)

Sketching lines that pass through the origin

• Sketch the graph of $y = 3x$

  • The x- and the y-intercept are both at (0, 0).

  • Another point, let $x = 1$, $y = 3 \times (1)$
    $y = 3$
    Another point is at (1, 3), then use the other two points to sketch the graph

Exercise 4C:

• Fluency questions 1-3 (1/2)\n 1-4 (1/2)\n 1-4 (1/2)

• Problem Solving Questions 5, 6

• Reasoning Questions 10, 11, 12
  A

4D: Gradient

Lesson starter: Which line is the steepest?

• The three lines shown Connect the points A-H

• Calculate the rise and run (working from left to right) and also the fraction $\frac{rise}{run}$ for these segments
  a) AB b) BC c) BD d) EF e) GH

• What do you notice about the fraction $\frac{rise}{run}$ for parts a, b, and c?

• How does the $\frac{rise}{run}$ for EF compare with the $\frac{rise}{run}$ for parts a, b, and c? Which of the two lines is the steeper?

• Your $\frac{rise}{run}$ for GH should be negative. Why is this the case?

• Discuss whether or not GH is steeper than AD

4D Gradient Learning Intentions:

* to understand what is meant by the gradient of the line
* to know that the gradient can be positive, negative, zero, or undefined
* to be able to the find the gradient of aline from a garph or using two given points

4D Gradient Past, Present, Future Learnings:

* core topic learnings from stage 5 A
* brief review in stage 6 advanced

4D Gradient Key Ideas:

* Gradient (m) = rise/run or the ratio between two points on the line
* Run is horizontal and often positive
* rise can be negative(Negative gradient) or positive(Positive gradient)
* gradient can be zero for horizontal line or undefined for vertical line

4D Gradient Building Understanding:

* Calculate the gradient using rise/run for these lines, Remember to negative if line slopes downward, from left to right
* use words, positive, negative, zero, or undefined to complete each sentence.

Finding the gradient of a line:

Example 7: to work gradient as being positive, negative, zero or undefined

Finding the gradient given gradient and rise or run:

#### 4D Example 8: Finding the gradient between two points:
* rise/run method to find m(gradient) by subtracting 2 points.

4D Gradient Exercise 4D: with varying range of all questions provided previously:

4D Finding the gradient: problem-solving 4,5,6

• Find: the gradient corresponding to the following slopes with all values having to follow gradient(M) rise/run or $\frac {y2-y1}{x2-x1}$

4D Gradients reasoning: 7,8

* Give reasons why is a line with a gradient is steeper than one of $\frac {7}{11}$ or $\frac {3}{5}$

4D Gradient Enrichment:

Enrichment question regarding a theoretical gradients being a slope between varying given point values and having to be able to use formulas to find the right answer and use gradient methods of solving/finding values.

4D Gradient Key understanding for formulas:

* (run = x2/x1) * (rise = y2/y1) * (Gradient method/solving or using =$\frac {y2-y1} {x2-x1}$)or rise/run (Y axes over X axes )

4E Gradient and Direct Proportion Key information relating to the chapter.

4E Lesson starter: Average Speed:

Over 5 hours, Sandy travels 420 km:
* what is Average speed for the trip?
* speed is a rate and discuss about speed as a rate.
* draw a graph distance vs time, assuming a constant speed for the trip sandy made.
what does your graph intersect at and why?
* find the gradient of your graph and what do you notice about the rate.
* find a rule linking distance (d) and time (t)

4E Gradient and Direct Proportion Learning intentions:

To understand:
* that a gradient is the rate of change for the variable in regards to another variable
* what it means for two variables to be directly proportional to the information
* to be able to form equation and work with equations involving the variation.

4E gradient and Direct Proportion:

*Past, present, and future learning’s on topic:

Address learning stage 5:

• core topics called “Linear relationships A and Variation A”
  Core concepts will be revised briefly into the early stages of year 10

4E gradient and Direct Proportion

Bicycle’s speed “V” m/min, is in direct propertion
to the number of pedal R ”p”

Key Ideas to gradient and direct proportion

 * Direct variation occurs with two varibles are directly proportional in relation to their situation

and can vary in variable relationship based on their gradients relationship to a certain formula; the equations from y=/+mx+c

4E Gradient and Direct Proportion Building Understanding: graph shows cyclist travels for 4 hours

* speed related change of distance travel and notice of what the answers do with gradient.
* missing values for H with given table with numbers for solving
* finding the correct graph that matches the graph the information from the previous slide and learning intentions.
* find the gradient of the graph with given graph/s and solve.

**4E Gradients Direct Proportions Example 9 exploring direct proportion.

* Water is poured into an empty tank at a constant rate and this and its time and solutions.
* draw a graph volume and time with in variable situations and gradient to the formulas.

4E Gradient Direct Porportions EXCERSISE 4E and problem solving questions.

4E Reasoning questions numbers: 9,10 to be solved

4E Enrichment Rate challenge numbers: 13, 14 extra hard problems.

Start of 4F Gradient – intercept form Lesson 4F

4 f gradients intercept form Lesson starter” Family Traits “
Lesson: relating gradients to there given rule for either each variable gradient.

4F key gradients and the value of what slope intercept:

 means to be used for questions or values to be solved on paper or using digital  form.

4F gradients intercept form bulding understaninding.

• State a rule (in gradient–intercept form) for a straight line with the given properties

  • Substitute x with -3 values to with variable linear numbers and see for the following gradient with the variables numbers for rules.
    Rearrange and to make and write the subject

4F EXAMPLE 10 4F:

• State the gradient a try new method of trying to state coordinates with y intercept from both lines for numbers/variables and gradient methods

4F rearrange and to make linear into graphs form for new methods .

4FE EXCERSISE FLUENCY PROBLEMS: 1-3 all 1/3rd questions

4F find the and and sketch with equation first numbers EXAMPLE.: 12A and 1B to be solved.

4F find gradients for following graphs and variables for all equations to find graph

EXCERCISE: 12 AND 12B problems

End of 4 f gradients solving methods

4G Finding the of line with and formulas being applied and used

• Using the gradient–intercept form, the rule (or equation) of a line can be found by calculating the value. Using the rule and value for the constant value can work.

4G Finding the equation of a line using Building Understanding:

what to do for formula to be the result.

4G EXAMPLE 13

what is the to do for this gradient question finding and being solved:

4G EXCERSISE 4G:

Solving equations as a key part

Find equation with given formulas values and and or points questions 4,5,and points.

Reasoning 9 and 11 reason of why formulas used.

Enrichment: missing gradient finding the equation to solve and variables with points.

4H Midpoint and Length of a Line Segment

Midpoints and the length of that seg are needed

The lesson starter will relate to choosing a formula/equation

4H Past Recent Future Learning:

What this learning will address and build on. to build on to recent topics.

4H: Key terms for the chapter:

formula information for how the key point relate to certain situations with length midpoint and formulas/what’s needed all equations and segments

Example 4H Building Understanding

finding values for questions needed.

What to do in that situation

Example 4H 15

The midpoint formula being used / shown and what to do for those types of gradient segment problems.

Example 4H 16

Finding length as a new formula to learn. / solving equations using variables.

4H EXCERSISE 4H

All questions having to deal with finding numbers for formula as example 15,16 all variable questions

EXCERSISE 4H: Problems

3 and 4 to work

**EXCERSISE 4H Reasoning questions:

7 to 8 reason why equations and things occur within the equation questions**
Enrichment gradient with length segment with given equations / points numbers gradient and ratio: numbers nine to the learning.
4l perpendicular Key: and formula how that all and

L41 Key ideas: what all is need to make gradient with slope information. .

What and how it to find line information with gradient or the negative
Reciprocal and point with variable points.

L41 EXCERSISE QUESTION

find the equation line and point to be related and the answer
H. all problem points with variable info for equations and answers

Reasoning for M1 and to find gradient information reasons and work of the following in equations variable point and info points

enrichment for bisectors what formulas key. key to success.

Alright, let's break down these math topics with some weird analogies:

4H: Midpoint and Length of a Line Segment

• Midpoint: Imagine you have a Twinkie. You want to cut it perfectly in half so you and your friend get equal parts. The midpoint is where you'd make that cut. In math terms, it's the exact middle point between two coordinates.

• Length: Now, imagine you're measuring how long that Twinkie is. That's the length of the line segment. In math, you use a formula (like the Pythagorean theorem in disguise!) to find the exact distance between two points.

4F: Gradient – Intercept Form

• Gradient: Think of a hill. The gradient is how steep the hill is. A high gradient means a very steep hill, and a low gradient means a gentle slope. If the hill goes downwards, it's a negative gradient.

• Intercept: The y-intercept is where your hill climber starts on the y-axis (the vertical one). It's the climber's initial height before they start walking up the hill.

• Gradient-Intercept Form (y = mx + c): This is the hill climber's equation. 'm' is the steepness of the hill (gradient), and 'c' is where the climber starts (y-intercept).

4E: Gradient and Direct Proportion

• Direct Proportion: Think of baking a cake. The more flour you use, the bigger the cake you can make. If you double the flour, you double the cake size. That's direct proportion. As one thing increases, the other increases at the same rate.

• Gradient as Rate of Change: The gradient here is like the recipe. It tells you how much the cake size changes for every scoop of flour you add. A steeper gradient means your cake grows faster with each scoop.

4K: Graphical Solutions to Simultaneous Equations

• Simultaneous Equations: Imagine you and a friend are trying to figure out the price of apples and bananas. You know that 2 apples + 3 bananas cost 5$, and 1 apple + 1 banana cost 2$. These are simultaneous equations – you need to solve them together to find the individual prices.

• Graphical Solution: Draw a line for each equation on a graph. The point where the lines cross is the solution! It tells you the price of one apple and one banana that satisfies both equations.

4I: Perpendicular and Parallel Lines

• Parallel Lines: Think of train tracks. They run side by side and never meet. Parallel lines have the same steepness (gradient) so they will never Intersect.

• Perpendicular Lines: Imagine a street crossing another street at a perfect right angle. Those streets are perpendicular. Perpendicular lines have gradients that are negative reciprocals of each other.

  Think of negative reciprocal as flipping a fraction and changing its sign. If a line has a gradient of 2/3, a line perpendicular to it has a gradient of -3/2.

4G: Finding the Equation of a Line Using y = mx + c

• Finding the Equation of a Line:* Think of code and creating a line. You already have the slope and a point the line makes contact with.

• Figuring Out y = mx + c:* You take the line's code and plug in what you know (slope, point) to fill in the missing and discover what formula results, with it results you unlock the line's full code!