CMS NSW Stage 5 Year 9 Mathematics Linear Relationships PDF

4 Linear relationships Maths in context: Animation, cartoon, and game developers The virtual world of animation has made computer and reduced. The finished shape has countless gaming and cartoon videos into global multimillion flat surfaces appearing o...

4 Linear relationships Maths in context: Animation, cartoon, and game developers The virtual world of animation has made computer and reduced. The finished shape has countless gaming and cartoon videos into global multimillion flat surfaces appearing on a computer screen as a dollar industries. Animation algorithms apply skills smoothly curved virtual 3D object. from algebra, coordinate geometry, linear relations, Linear algebra is used to control the position and trigonometry, and more advanced mathematics. movement when animating a virtual object. It can When designing a cartoon character, the animator be transformed using linear equations that alter the codes a subdivision process. Using the midpoint coordinates of its vertices. Up or down movement formula, a 2D polygon with straight sides is morphed is achieved by adding or subtracting, enlargement into a curve. By repeatedly subdividing lengths and or shrinking uses multiplication or division, and slicing off corners, a virtual curve illusion is created. rotation uses trigonometry. A similar subdivision process is coded to build The cartoon Bluey (pictured) is a successful pre- 3D shapes. The initial polyhedral object has flat school TV animated cartoon developed in Brisbane polygon faces that are then repeatedly subdivided and distributed internationally on Disney+. CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  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 In this chapter, a student: develops understanding and fluency in mathematics through exploring and connecting mathematical concepts, choosing and applying mathematical techniques to solve problems, and communicating their thinking and reasoning coherently and clearly (MAO-WM-01) determines the midpoint, gradient and length of an interval, and graphs linear relationships, with and without digital tools (MA5-LIN-C-01) graphs and interprets linear relationships using the gradient/slope- intercept form (MA5-LIN-C-02) describes and applies transformations, the midpoint, gradient/slope and distance formulas, and equations of lines to solve problems (MA5-LIN-P-01). © 2022 NSW Education Standards Authority Online resources A host of additional online resources are included as part of your Interactive Textbook, including HOTmaths content, video demonstrations of all worked examples, auto-marked quizzes and much more. CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  248 Chapter 4 Linear relationships 4A Introducing linear relationships CONSOLIDATING Learning intentions To review the features of the Cartesian plane and plot points using coordinates To know that a linear relationship gives a set of points that form a straight line To be able to complete a table of values and plot points to form a linear graph To be able to arrange the equation of a line in various ways, including general form To be able to decide if a point is on a line by substituting into the equation of the line To know which points are represented by the x-intercept and y-intercept To be able to identify the x-intercept and y-intercept from a table or graph Past, present and future learning This section consolidates Stage 4 concepts which are used in Stages 5 and 6 Some of these questions are more challenging than those in the Year 8 book These concepts will be revised and extended in Chapter 1 of our Year 10 book Expertise with these concepts is assumed knowledge for Stage 6 Advanced If two variables are related in some way, we can use mathematical rules to precisely describe this relationship. The most simple kind of mathematical relationship is one that can be illustrated with a straight line graph. These are called linear relationships. The volume of petrol in your car at a service bowser, for example, might initially be 10 L then be increasing by 1.2 L per second after that. This is an example of a linear relationship between volume and time because the volume is increasing at a constant rate of A car’s annual running costs can be $4000 p.a., 1.2 L/s. including fuel, repairs, registration and insurance. If a car cost $17 000, the total cost, C, for n years could be expressed as the linear relationship: C = 4000 n + 17 000. Lesson starter: Is it linear? Here are three rules linking x and y. y 1 2 y1 = _ 8 x +1 7 2 y2 = x 2 − 1 6 3 y3 = 3x − 4 5 First complete this simple table and graph. 4 x 1 2 3 3 y1 2 1 y2 x y3 O 1 2 3 −1 Which of the three rules do you think is linear? How do the table and graph help you decide it’s linear? CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4A Introducing linear relationships 249 KEY IDEAS Coordinate geometry provides a link between y geometry and algebra. 4 The Cartesian plane (or number plane) consists of 2 quadrant 3 nd 1st quadrant two axes which divide the number plane into four 2 (0, 2) (−4, 1) quadrants. 1 (3, 1) (−3, 0) (1, 0) The horizontal x-axis and vertical y-axis intersect at x O right angles at the origin O(0, 0). −4 −3 −2 −1−1 1 2 3 4 A point is precisely positioned on a Cartesian plane 3rd quadrant −2 4th quadrant using the coordinate pair (x, y) where x describes −3 (−2, −3) (0, −4) the horizontal position and y describes the vertical −4 (2, − 4) position of the point from the origin. A linear relationship is a set of ordered pairs (x, y) that when graphed give a straight line. Linear relationships have rules that may be of the form: y = mx + c. For example, y = 2x + 1. ax + by = d or ax + by + c = 0. For example, 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. x -intercept y x −2 −1 0 1 2 3 4 y-intercept y 8 6 4 2 0 −2 y -intercept x-intercept x O 2 BUILDING UNDERSTANDING 1 Refer to the Cartesian plane shown. y a Give the coordinates of all the points A − L. 4 b State which points lie on: 3 C B i the x-axis ii the y-axis. D 2 c State which points lie inside the: E 1 A i 2nd quadrant ii 4th quadrant. F L x O −4 −3 −2 −1 1 2 3 4 −1 2 For the rule y = 3x − 4, find the value of y for these G −2 J x-values. −3 I a x=2 b x = −1 H −4 K CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  250 Chapter 4 Linear relationships 3 For the rule y = − 2x + 1, find the value of y for these x-values. a x=0 b x = − 10 4 Give the coordinates of three pairs of points on the line y = x + 3. Example 1 Plotting points to graph straight lines Using − 3 ⩽ x ⩽ 3, construct a table of values and plot a graph for these linear relationships. a y=x+2 b y = − 2x + 2 SOLUTION EXPLANATION a Use − 3 ⩽ x ⩽ 3 as instructed and substitute x −3 −2 −1 0 1 2 3 each value of x into the rule y = x + 2. y −1 0 1 2 3 4 5 The coordinates of the points are read from the y table, i.e. (− 3, − 1), (− 2, 0), etc. 6 Plot each point and join to form a straight line. 4 Extend the line to show it continues in either 2 direction. x −4 −2−2O 2 4 −4 b x −3 −2 −1 0 1 2 3 Use − 3 ⩽ x ⩽ 3 as instructed and substitute y 8 6 4 2 0 −2 −4 each value of x into the rule y = − 2x + 2. For example: y x = − 3, y = − 2 × (− 3) + 2 8 = 6+2 6 = 8 4 Plot each point and join to form a straight line. 2 Extend the line beyond the plotted points. x −4 −2−2O 2 4 −4 Now you try Using − 3 ⩽ x ⩽ 3, construct a table of values and plot a graph for these linear relationships. a y=x−1 b y = − 3x + 1 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4A Introducing linear relationships 251 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. a x −2 −1 0 1 2 3 y 6 4 2 0 −2 −4 b y 6 x −4 O SOLUTION EXPLANATION a The x-intercept is at (1, 0). The x-intercept is at the point where y = 0 (on the x-axis). The y-intercept is at (0, 2). The y-intercept is at the point where x = 0 (on the y-axis). b The x-intercept is at (− 4, 0). The x-intercept is at the point where y = 0 (on the x-axis). The y-intercept is at (0, 6). The y-intercept is at the point where x = 0 (on the y-axis). Now you try Write down the coordinates of the x-intercept and y-intercept from this table and graph. a x −3 −2 −1 0 1 2 y 4 3 2 1 0 −1 b y x O 5 −3 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  252 Chapter 4 Linear relationships Example 3 Deciding 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 EXPLANATION a y = 2x + 10 Substitute x = − 2 Find the value of y by substituting x = − 2 into y = 2(− 2) + 10 the rule for y. = 6 ∴ the point (− 2, 4) is not on the line. The y-value is not 4, so (− 2, 4) is not on the line. b y = −x + 2 Substitute x = − 2 By substituting x = − 2 into the rule for the y = − (− 2) + 2 line, y is 4. = 4 ∴ the point (− 2, 4) is on the line. So (− 2, 4) is on the line. Now you try Decide if the point (− 1, 6) is on the line with the given rules. a y = − 2x + 4 b y = 3x + 5 Exercise 4A FLUENCY 1−3(1/2) 1−4(1/2) 1−4(1/2) Example 1 1 Using − 3 ⩽ x ⩽ 3, construct a table of values and plot a graph for these linear relations. a y=x−1 b y = 2x − 3 c y = −x + 4 d y = − 3x Example 2 2 Write down the coordinates of the x- and y-intercepts for the following tables and graphs. a x −3 −2 −1 0 1 2 3 y 4 3 2 1 0 −1 −2 b x −3 −2 −1 0 1 2 3 y −1 0 1 2 3 4 5 c x −1 0 1 2 3 4 y 10 8 6 4 2 0 d x −5 −4 −3 −2 −1 0 y 0 2 4 6 8 10 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4A Introducing linear relationships 253 e y f y 3 x O 7 x O 2 −3 g y h y x −2 O 5 x −5 −11 O Example 3 3 Decide if the point (1, 2) is on the line with the given rule. a y=x+1 b y = 2x − 1 c y = −x + 3 d y = − 2x + 4 e y = −x + 5 1x + _ f y = −_ 1 2 2 4 Decide if the point (− 3, 4) is on the line with the given rule. a y = 2x + 8 b y=x+7 c y = −x − 1 1 d y = _x + 6 e y = −_ 1x + 3 f y=_ 5x + _ 11 3 3 6 2 PROBLEM–SOLVING 5 5, 6 5(1/2), 6 5 Find a rule in the form y = mx + c (e.g. y = 2x − 1) that matches these tables of values. a x 0 1 2 3 4 y 2 3 4 5 6 x −1 0 1 2 3 b y −2 0 2 4 6 x −2 −1 0 1 2 3 c y −3 −1 1 3 5 7 x −3 −2 −1 0 1 2 d y 5 4 3 2 1 0 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  254 Chapter 4 Linear relationships 6 Match rules a, b, c and d to the graphs A, B, C and D. a y = 2x + 1 b y = −x − 1 c y = − 2x + 3 d x+y=2 A y B y C y D y 3 3 2 2 1 1 (1, 1) 1 x O x x x −2 −1−1 O 1 2 O 1 2 (2, −1) −1 O (−1, −1) −1 −2 REASONING 7 7, 8 8, 9 7 Decide if the following rules are equivalent. a y = 1 − x and y = − x + 1 b y = 1 − 3x and y = 3x − 1 c y = − 2x + 1 and y = − 1 − 2x d y = − 3x + 1 and y = 1 − 3x 8 Give reasons why the x-intercept on these graphs is at (_ 2 ) 3, 0. a b y y 4 4 3 3 2 2 1 1 x x −3 −2 −1−1O 1 2 3 4 −3 −2 −1−1O 1 2 3 4 −2 −2 −3 −3 9 Decide if the following equations are true or false. a _2x + 4 = x + 4 3x − 6 = x − 2 b _ 2 3 c 1 _ (x − 1) = _1x − _ 1 d 2 _ 2 x − 12 (x − 6) = _ 2 2 2 3 3 ENRICHMENT: Tough rule finding − − 10 10 Find the linear rule linking x and y in these tables. a b x −1 0 1 2 3 x −2 −1 0 1 2 y 5 7 9 11 13 y 22 21 20 19 18 c d x 0 2 4 6 8 x −5 −4 −3 −2 −1 y − 10 − 16 − 22 − 28 − 34 y 29 24 19 14 9 e x 1 3 5 7 9 f x − 14 − 13 − 12 − 11 − 10 y 1 2 3 4 5 y 1 5_ 5 1 4_ 4 1 3_ 2 2 2 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4B Graphing straight lines using intercepts 255 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 When linear rules are graphed, Businesses claim all the points lie in a straight line. a tax deduction for It is therefore possible to graph a the annual decrease in the value, V, of straight line using only two points. equipment over Two critical points that help draw n years. A straight line segment joining these graphs are the x-intercept the intercepts (0, V) and y-intercept introduced in the and (n, 0) shows that previous section. the rate of decrease V per year is _ n. Lesson starter: Two key points 1 x + 1 and complete this table and Consider the relation y = _ 2 y graph. 4 x −4 −3 −2 −1 0 1 2 3 y 2 1 What are the coordinates of the point where the line crosses x the y-axis? That is, state the coordinates of the y-intercept. −4 −3 −2 −1−1O 1 2 3 4 What are the coordinates of the point where the line crosses the x-axis? That is, state the coordinates of the x-intercept. −2 Discuss how you might find the coordinates of the x- and −3 y-intercepts without drawing a table and plotting points. −4 Explain your method. CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  256 Chapter 4 Linear relationships KEY IDEAS Two points are required to sketch a straight Example: 2x + 3y = 6 line graph. Often these points are the axes y y-intercept (x = 0) intercepts. 2(0) + 3y = 6 The y-intercept is the point where the line 3y = 6 2 y=2 intersects the y-axis (where x = 0). x-intercept (y = 0) Substitute x = 0 to find the y-intercept. x O 2x + 3(0) = 6 3 The x-intercept is the point where the line 2x = 6 intersects the x-axis (where y = 0). x=3 Substitute y = 0 to find the x-intercept. BUILDING UNDERSTANDING 1 Mark the following x- and y-intercepts on a set of axes and join in a straight line. a x-intercept: (− 2, 0), y-intercept: (0, 3) b x-intercept: (4, 0), y-intercept: (0, 6) 2 a Find the value of y in these equations. i 2y = 6 ii y = 3 × 0 + 4 iii − 2y = 12 b Find the value of x in these equations. i − 4x = − 40 ii 0 = 2x − 2 iii _1 x = 3 2 3 For these equations find the coordinates of the y-intercept by letting x = 0. a x+y=4 b x−y=5 c 2x + 3y = 9 4 For these equations find the coordinates of the x-intercept by letting y = 0. a 2x − y = − 4 b 4x − 3y = 12 c y = 3x − 6 Example 4 Sketching with intercepts Sketch the graph of the following, showing the x- and y-intercepts. a 2x + 3y = 6 b y = 2x − 6 SOLUTION EXPLANATION a 2x + 3y = 6 Only two points are required to generate a y-intercept (let x = 0): straight line. For the y-intercept, substitute 2(0) + 3y = 6 x = 0 into the rule and solve for y by dividing 3y = 6 each side by 3. y=2 ∴ the y-intercept is at (0, 2). State the y-intercept. CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4B Graphing straight lines using intercepts 257 x-intercept (let y = 0): Similarly to find the x-intercept, substitute 2x + 3(0) = 6 y = 0 into the rule and solve for x. 2x = 6 x=3 ∴ the x-intercept is at (3, 0). State the x-intercept. y Mark and label the intercepts on the axes and 2 sketch the graph by joining the two intercepts. 2x + 3y = 6 x O 3 b y = 2x − 6 y-intercept (let x = 0): y = 2(0) − 6 Substitute x = 0 for the y-intercept. y = −6 Simplify to find the y-coordinate. ∴ the y-intercept is at (0, − 6). x-intercept (let y = 0): Substitute y = 0 for the x-intercept. Solve the 0 = 2x − 6 remaining equation for x by adding 6 to both 6 = 2x sides and then divide both sides by 2. x=3 ∴ the x-intercept is at (3, 0). y Mark in the two intercepts and join to sketch x the graph. O 3 y = 2x − 6 −6 Now you try Sketch the graph of the following, showing the x- and y-intercepts. a 3x + 5y = 15 b y = 3x − 6 Exercise 4B FLUENCY 1−2(1/2) 1−3(1/2) 1−3(1/3) Example 4a 1 Sketch the graph of the following, showing the x- and y-intercepts. a x+y=2 b x − y = −2 c 2x + y = 4 d 3x − y = 9 e 4x − 2y = 8 f 3x + 2y = 6 g y − 3x = 12 h − 5y + 2x = − 10 i − x + 7y = 21 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  258 Chapter 4 Linear relationships Example 4b 2 Sketch the graph of the following, showing the x- and y-intercepts. a y = 3x + 3 b y = 2x + 2 c y=x−5 d y = −x − 6 e y = − 2x − 2 f y = − 2x + 4 3 Sketch the graph of each of the following mixed linear relationships. a x + 2y = 8 b 3x − 5y = 15 c y = 3x − 6 d 3y − 4x − 12 = 0 e 2x − y − 4 = 0 f 2x − y + 5 = 0 PROBLEM–SOLVING 4, 5 4, 5 5, 6 4 The distance d metres of a vehicle from an observation point after t seconds is given by the rule d = 8 − 2t. a Find the distance from the observation point initially (at t = 0). b Find after what time t the distance d is equal to 0 (substitute d = 0). c Sketch a graph of d versus t between the d and t intercepts. 5 The height, h metres, of a lift above the ground after t seconds is given by h = 100 − 8t. a How high is the lift initially (at t = 0)? b How long does it take for the lift to reach the ground (h = 0)? 6 Find the x- and y-axis intercept coordinates of the graphs with the given rules. Write answers using fractions. a 3x − 2y = 5 b x + 5y = − 7 c y − 2x = − 13 d y = − 2x − 1 e 2y = x − 3 f − 7y = 1 − 3x REASONING 7 7, 8 8, 9 7 Use your algebra and fraction skills to help sketch graphs for these relations by finding x- and y-intercepts. y y 2 − 4x a _x + _ = 1 b y=_ 8−x c _=_ 2 3 4 2 8 8 Explain why the graph of the equation ax + by = 0 must pass through the origin for any values of the constants a and b. 9 Write down the rule for the graph with these axes intercepts. Write the rule in the form ax + by = d. a (0, 4) and (4, 0) b (0, 2) and (2, 0) c (0, − 3) and (3, 0) d (0, 1) and (− 1, 0) e (0, k) and (k, 0) f (0, − k) and (− k, 0) ENRICHMENT: Intercept families − − 10 10 Find the coordinates of the x- and y-intercepts in terms of the constants a, b and c for these relationships. a ax + by = c b y=_ ax + c ax − by b c _ c =1 d ay − bx = c e ay = bx + c f a(x + y) = bc CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4C Lines with one intercept 259 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 using the origin and one other point Past, present and future learning This section addresses the Stage 5 Core Topic called Linear relationships 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 × number of acres; annual interest = interest rate p.a. × amount invested; weekly pay = pay rate in $/h × number of hours. Lesson starter: What rule satisfies all points? y Here is one vertical and one horizontal line. For the vertical line shown, write down the coordinates of all 4 the points shown as dots. 3 What is always true for each coordinate pair? 2 What simple equation describes every point on the line? 1 For the horizontal line shown, write down the coordinates of x O −4 −3 −2 −1−1 1 2 3 4 all the points shown as dots. What is always true for each coordinate pair? −2 What simple equation describes every point on the line? −3 Where do the two lines intersect? −4 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  260 Chapter 4 Linear relationships KEY IDEAS Vertical line: x = k y Parallel to the y-axis 3 Equation of the form x = k, where k is a constant x=k 2 x-intercept is at (k, 0) 1 x O k Horizontal line: y = c y Parallel to the x-axis Equation of the form y = c, where c is a constant y=c c y-intercept is at (0, c) x O 1 2 3 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 y y = 2x y = −2x 4 y = −1x 3 (1, 2) 2 2 y=x 1 (1, 1) x O −4 −3 −2 −1−1 1 2 3 4 −2 (1, − 1) 2 −3 (1, −2) −4 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4C Lines with one intercept 261 BUILDING UNDERSTANDING 1 State the coordinates of the x-intercept for these graphs. a y b y c y 3 3 3 2 2 2 y= 1x x = −3 2 1 1 1 x x x −1−1O 1 2 3 −3 −2 −1−1O 1 −2 −1−1O 1 2 3 −2 x=2 −2 −2 −3 −3 −3 2 State the coordinates of the y-intercept for these graphs. a y b y 3 3 y=3 2 2 1 1 x x −3 −2 −1 O −1 1 2 3 −3 −2 −1−1O 1 2 3 −2 −2 y = −2 −3 −3 c y 4 3 2 1 x −4 −3 −2 −1O 1 2 3 4 −1 −2 −3 y = − 4x −4 3 Find the value of y if x = 1 using these rules. a y = 5x b y = _1 x c y = − 4x d y = − 0.1x 3 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  262 Chapter 4 Linear relationships Example 5 Graphing vertical and horizontal lines Sketch the graph of the following horizontal and vertical lines. a y=3 b x = −4 SOLUTION EXPLANATION a y y-intercept is at (0, 3). Sketch a horizontal line through all points 3 where y = 3. y=3 2 1 x O 1 2 3 b y x-intercept is at (− 4, 0). Sketch a vertical line through all points where 2 x = − 4. x = −4 1 x −4 −3 −2 −1−1O Now you try Sketch the graph of the following horizontal and vertical lines. a y = −1 b x=3 Example 6 Sketching lines that pass through the origin Sketch the graph of y = 3x. SOLUTION EXPLANATION The x- and the y-intercept are both at (0, 0). The equation is of the form y = mx. Another point (let x = 1): As two points are required to generate the y = 3 × (1) straight line, find another point by substituting y = 3 x = 1. Another point is at (1, 3). Other x-values could also be used. y Plot and label both points and sketch the graph 3 (1, 3) by joining the points in a straight line. 2 y = 3x 1 x −3 −2 −1−1O 1 2 3 −2 −3 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4C Lines with one intercept 263 Now you try Sketch the graph of y = − 2x. Exercise 4C FLUENCY 1−3(1/2) 1−4(1/2) 1−4(1/2) Example 5 1 Sketch the graph of the following horizontal and vertical lines. a x=2 b x=5 c y=4 d y=1 e x = −3 f x = −2 g y = −1 h y = −3 Example 6 2 Sketch the graph of the following linear relationships that pass through the origin. a y = 2x b y = 5x c y = 4x d y=x e y = − 4x f y = − 3x g y = − 2x h y = −x 3 Sketch the graphs of these special lines all on the same set of axes and label with their equations. a x = −2 b y = −3 c y=2 d x=4 e y = 3x f y = −_ 1x g y = − 1.5x h x = 0.5 2 i x=0 j y=0 k y = 2x l y = 1.5x 4 Give the equation of each of the following graphs. a y b y c y 4 4 4 3 3 3 2 2 2 1 1 1 x x x −2 −1−1O 1 2 −2 −1−1O 1 2 −2 −1−1O 1 2 −2 −2 −2 d y e y f y 4 3 2 1.5 1 −6.7 x x x O 1 2 3 4 5 O O −1 −2 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  264 Chapter 4 Linear relationships PROBLEM–SOLVING 5, 6 5, 6, 7(1/2), 8 5(1/2), 7(1/2), 8, 9 5 Find the equation of the straight line that is: a parallel to the x-axis and passes through the point (1, 3) b parallel to the y-axis and passes through the point (5, 4) c parallel to the y-axis and passes through the point (− 2, 4) d parallel to the x-axis and passes through the point (0, 0). 6 If the surface of the sea is represented by the x-axis, state the equation of the following paths. a A plane flies horizontally at 250 m above sea level. One unit is 1 metre. b A submarine travels horizontally 45 m below sea level. One unit is 1 metre. 7 The graphs of these pairs of equations intersect at a point. Find the coordinates of the point. a x = 1, y = 2 b x = − 3, y = 5 c x = 0, y = − 4 d x = 4, y = 0 e y = − 6x, x = 0 f y = 3x, x = 1 g y = − 9x, x = 3 h y = 8x, y = 40 8 Find the area of the rectangle contained within the following four lines. a x = 1, x = − 2, y = − 3, y = 2 b x = 0, x = 17, y = − 5, y = − 1 9 The lines x = − 1, x = 3 and y = − 2 form three sides of a rectangle. Find the possible equation of the fourth line if: a the area of the rectangle is: i 12 square units ii 8 square units iii 22 square units b the perimeter of the rectangle is: i 14 units ii 26 units iii 31 units CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4C Lines with one intercept 265 REASONING 10 10−11(1/2) 10−12(1/2) 10 The rules of the following graphs are of the form y = mx. Use the points marked with a dot to find m and hence state the equation. a y b y 4 4 3 3 2 2 1 1 x x O −2 −1−1 1 2 3 4 −2 −1−1O 1 2 3 4 −2 −2 c y d y 8 3 6 2 4 1 2 x O −3 −2 −1−1 1 2 3 x −2 −1−2O 1 2 3 4 −2 −4 −3 11 Find the equation of the line that passes through the origin and the given point. a (1, 3) b (1, 4) c (1, − 5) d (1, − 2) 12 Sketch the graph of each of the following by first making y or x the subject. a y−8=0 b x+5=0 c x+ =0_1 d y − 0.6 = 0 2 e y + 3x = 0 f y − 5x = 0 g 2y − 8x = 0 h 5y + 7x = 0 ENRICHMENT: Trisection − − 13−15 13 A vertical line, horizontal line and another line that passes through the origin all intersect at (− 1, − 5). What are the equations of the three lines? 14 The lines y = c, x = k and y = mx all intersect at one point. a State the coordinates of the intersection point. b Find m in terms of c and k. 15 The area of a triangle formed by x = 4, y = − 2 and y = mx is 16 square units. Find the value of m given m > 0. CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  266 Chapter 4 Linear relationships 4D Gradient Learning intentions To understand what is meant by the gradient of a line To know that the gradient of a line can be positive, negative, zero or undefined To be able to find the gradient of a line from a graph or using two given points Past, present and future learning This section addresses the Stage 5 Core Topic called Linear relationships A These concepts will be revised, briefly, in Chapter 1 of our Year 10 book and also in Stage 6 Advanced The gradient of a line is a measure of its slope. It is a number that describes the steepness of a line and is calculated by considering how far a line rises or falls between two points rise within a given horizontal distance. The horizontal distance between two points is called the run and the vertical distance run is called the rise. Lesson starter: Which line is the steepest? y 5 D The three lines shown below right connect the points A−H. G 4 Calculate the rise and run (working from left to right) and 3 rise for these segments. 2 also the fraction _run 1 C F a AB b BC c BD d EF e GH x rise _ What do you notice about the fractions run for parts a, b −5 −4 −3 −2 −1 O −1 B 1 2 3 4 5 and c? H −2 How does the _ rise for EF compare with the _ rise for −3 run run parts a, b and c? Which of the two lines is the steeper? −4 rise for GH should be negative. Why is this the case? −5 E Your _ A run Discuss whether or not GH is steeper than AD. Use computer software (interactive geometry) to produce a set of axes and grid. Construct a line segment with endpoints on the grid. Show the coordinates of the endpoints. Calculate the rise (vertical distance between the endpoints) and the run (horizontal distance between the endpoints). Calculate the gradient as the rise divided by the run. Now drag the endpoints and explore the effect on the gradient. Can you drag the endpoints but retain the same gradient value? Explain why this is possible. Can you drag the endpoints so that the gradient is zero or undefined? Describe how this can be achieved. CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4D Gradient 267 KEY IDEAS rise Gradient (m) = _ run run (positive) Positive rise gradient (positive) rise Negative (negative) gradient run (positive) When we work from left to right, the run is always positive. The gradient can be positive, negative, zero or undefined. A vertical line has an undefined gradient. Negative y Undefined gradient gradient Zero gradient x Positive gradient Gradient is usually expressed as a simplified proper or improper fraction (not a mixed number). BUILDING UNDERSTANDING rise for these lines. Remember to give a negative answer if the line 1 Calculate the gradient using _ run is sloping downwards from left to right. a b (2, 3) (3, 6) (0, 2) (1, 2) c (−2, 2) d (4, 3) (7, 0) (−3, −1) e (−2, 3) f (−6, −1) (−2, −2) (−1, 0) CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  268 Chapter 4 Linear relationships 2 Use the words: positive, negative, zero or undefined to complete each sentence. a The gradient of a horizontal line is __________. b The gradient of the line joining (0, 3) with (5, 0) is __________. c The gradient of the line joining (− 6, 0) with (1, 1) is __________. d The gradient of a vertical line is __________. 3 Use the gradient m (= _run ) rise to find the missing value. a m = 4, rise = 8, run = ? b m = 6, rise = 3, run = ? c m = _3 , run = 4, rise = ? 2 d m = − _2 , run = 15, rise = ? 5 Example 7 Finding the gradient of a line For each graph, state whether the gradient is positive, negative, zero or undefined, then find the gradient where possible. a y b y 6 3 2 x x O 2 O 2 c y d y 4 x x O 2 O 2 SOLUTION EXPLANATION a The gradient is positive. By inspection, the gradient will be y positive since the graph rises from rise Gradient = _ left to right. Select any two points run 6 and create a right-angled triangle = 4 _ rise = 4 2 to determine the rise and run. 2 = 2 run = 2 x Substitute rise = 4 and run = 2. O 2 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4D Gradient 269 SOLUTION EXPLANATION b The gradient is negative. By inspection, the gradient will be y negative since y-values decrease rise from left to right. run = 2 Gradient = _ run 3 Rise = − 3 and run = 2. rise = −3 = 3 − _ 2 x = −_3 O 2 2 c The gradient is 0. The line is horizontal. d The gradient is undefined. The line is vertical. Now you try For each graph, state whether the gradient is positive, negative, zero or undefined, then find the gradient where possible. a y b y 2 x O x O 3 −4 c y d y x x O 5 −3 O −3 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  270 Chapter 4 Linear relationships Example 8 Finding the gradient between two points Find the gradient (m) of the line joining the given points. a A(3, 4) and B(5, 6) b A(− 3, 6) and B(1, − 3) SOLUTION EXPLANATION rise a m = _ Plot points to see a positive gradient and calculate rise and run. run y = −4 6 _ 5−3 (5, 6) = 2 _ 2 rise = 6 − 4 = 1 (3, 4) run = 5 − 3 x rise b m = _ y Negative gradient run = 9 or − _ − _ 9 or − 2.25 (−3, 6) run = 4 4 4 rise = −9 x (1, −3) Now you try Find the gradient (m) of the line joining the given points. a A(2, 6) and B(3, 8) b A(− 5, 1) and B(2, − 3) CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4D Gradient 271 Exercise 4D FLUENCY 1, 2(1/2) 1, 2(1/2) 1−2(1/2), 3 Example 7 1 For each graph state whether the gradient is positive, negative, zero or undefined, then find the gradient where possible. a y b y 4 4 3 3 2 2 1 1 x x −1−1O 1 2 3 4 5 −1−1O 1 2 3 4 5 −2 −2 c y d y 4 3 4 2 3 1 2 x 1 −1−1O 1 2 3 4 5 x −2 −6 −5 −4 −3 −2 −1 O e y f y 6 3 1 x x −2 −2 4 g y h y 3 x x −1 (−2, −1) −4 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  272 Chapter 4 Linear relationships Example 8 2 Find the gradient of the lines joining the following pairs of points. a A(2, 3) and B(3, 5) b C(2, 1) and D(5, 5) c E(1, 5) and F(2, 7) d G(− 2, 6) and H(0, 8) e A(− 4, 1) and B(4, − 1) f C(− 2, 4) and D(1, − 2) g E(− 3, 4) and F(2, − 1) h G(− 1, 5) and H(1, 6) i A(− 2, 1) and B(− 4, − 2) j C(3, − 4) and D(1, 1) k E(3, 2) and F(0, 1) l G(− 1, 1) and H(− 3, − 4) 3 Find the gradient of the lines A−F on this graph and grid. y 5 4 3 A B 2 1 F C x −5 −4 −3 −2 −1 O −1 1 2 3 4 5 −2 D E −3 −4 −5 PROBLEM–SOLVING 4, 5 4−6 5−7 4 Find the gradient corresponding to the following slopes. a A road falls 10 m for every 200 horizontal metres. b A cliff rises 35 metres for every 2 metres horizontally. c A plane descends 2 km for every 10 horizontal kilometres. d A submarine ascends 150 m for every 20 horizontal metres. 5 Find the missing number. a The gradient joining the points (0, 2) and (1, ? ) is 4. b The gradient joining the points (? , 5) and (1, 9) is 2. c The gradient joining the points (− 3, ? ) and (0, 1) is − 1. d The gradient joining the points (− 4, − 2) and (? , − 12) is − 4. 6 A train climbs a slope with gradient 0.05. How far horizontally has the train travelled after rising 15 metres? 7 Complete this table showing the gradient, x-intercept and y-intercept for straight lines. A B C D E F Gradient 3 −1 1 _ 2 −_ 0.4 − 1.25 2 3 x-intercept −3 6 1 y-intercept −4 1 _ 3 2 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  4D Gradient 273 REASONING 8 8, 9 8, 9 8 Give a reason why a line with gradient _ 3. 7 is steeper than a line with gradient _ 11 5 9 The two points A and B shown here have coordinates (x1, y1) and (x2, y2). y a Write a rule for the run using x1 and x2. B(x2, y2) b Write a rule for the rise using y1 and y2. c Write a rule for the gradient m using x1, x2, y1 and y2. d Use your rule to find the gradient between these pairs of points. i (1, 1) and (3, 4) A(x1, y1) ii (0, 2) and (4, 7) x O iii (− 1, 2) and (2, − 3) iv (− 4, − 6) and (− 1, − 2) e Does your rule work for points that include negative coordinates? Explain why. ENRICHMENT: Where does it hit? − − 10 2 which means that it falls 2 units for every 3 across. The y-intercept is 10 The line here has gradient − _ 3 at (0, 3). y 4 3 2 1 x −1−1O 1 2 3 4 5 6 a Use the gradient to find the y-coordinate on the line where: i x=6 ii x = 9 b What will be the x-intercept coordinates? c What would be the x-intercept coordinates if the gradient was changed to the following? i m = −_ 1 ii m = − _5 7 iii m = − _ 2 iv m = − _ 2 4 3 5 CambridgeMATHS NSW Stage 5 ISBN 978-1-009-40936-0 © Palmer et al. 2024 Cambridge University Press Year 9 Photocopying is restricted under law and this material must not be transferred to another party.  274 Chapter 4 Linear relationships 4E Gradient and direct proportion Learning intentions To understand that gradient is the rate of change of one variable with respect to another To understand what it means for two variables to be directly proportional To be able to form and work with equations involving direct variation Past, present and future learning This section addresses the Stage 5 Core Topics called Linear relationships A and Variation A These concepts will be revised, briefly, in Chapter 1 of our Year 10 book and also in Stage 6 Advanced The connection between gradient, rate problems and direct proportion can be illustrated by the use of linear rules and graphs. If two variables are directly related, then the rate of change of one variable with respect to the other is constant. This implies that the rule linking the two variables is linear and can be represented as a straight line graph passing through the origin. The amount of water spraying from a sprinkler, for example, is directly proportional to the time since the sprinkler was turned on. The gradient of the graph A bicycle’s speed, v m/min, is in direct proportion of water volume versus time will equal the rate at which to the number of pedal revolutions, P. For example, water is spraying from the sprinkler. v = 6P m/min for a wheel of circumference 2 m and gear ratio 3 : 1. If P = 90, v = 540 m/min, which is a speed of 32.4 km/h. Lesson starter: Average speed Distance (d km) Over 5 ho