mathsbooks.netmathsbooks.net/Maths Quest 11 Methods/By Chapter/Ch 01 Linear F… · 2 Mathematical...

1VCEVCEcocovverageerageAreas of studyUnit 1 • Functions and graphs

• Algebra

In thisIn this chachapterpter1A Solving linear equations1B Rearrangement and

substitution1C Gradient of a straight line1D Equations of the form

y = mx + c1E Sketching linear graphs

using intercepts1F Simultaneous equations1G Perpendicular lines1H Formula for finding the

equation of a straight line1I Distance between two

points1J Midpoint of a segment1K Linear modelling

Linear functions

2 M a t h e m a t i c a l M e t h o d s U n i t s 1 a n d 2

Solving linear equationsThroughout the Maths Methods course, the skill of being able to rearrange or solveequations is often called upon.

A linear equation is one that involves a pronumeral raised to the power of 1 only.Recall that x1 is the same as x, so a linear equation involving x would contain only x’s,

and not , x2, x3, x−1, x−2 and so on.For example, y = 7x − 3, ax + by = c and x + 1 = 9 are all linear equations.

Note that y = is not a linear equation, as the is really x−1. To isolate a particular

pronumeral — known as ‘making x (or whatever the pronumeral is) the subject’, wefocus on the pronumeral by ‘undoing’ other terms and operations. In doing so,remember to do the same to both sides of an equation, in the reverse order to thatoriginally used to make up the equation.

Though the focus of this chapter is linear equations, in this section some other typesof equation will be included for skills practice.

When there is only one pronumeral involved in an equation, we may attempt tosolve and find a numerical value by rearranging to make the pronumeral the subject.

x12---

83---

1x---

1x---

Solve the following linear equations.

a 7x − 4 = 17 b c

THINK WRITEa Write the equation. a 7x − 4 = 17

Add 4 to both sides. 7x = 21

Divide both sides by 7. x =

x = 3

b Write the equation. b + 5 = 1

Subtract 5 from both sides. = −4Multiply both sides by 4. 3x − 2 = −16Add 2 to both sides. 3x = −14


c Write the equation. c = 6

Divide both sides by 2. = 3

Add 1 to both sides. = 4

Multiply both sides by 5. 4x = 20


x = 5

3x 2–4

--------------- 5+ 1= 24x5

------ 1– 6=

12

3217

------

13x 2–

4---------------

2 3x 2–4

---------------

34

5 –143

------

1 2 4x5

------ 1–

24x5

------ 1–

34x5

------

4

5204

------

1WORKEDExample

C h a p t e r 1 L i n e a r f u n c t i o n s 3When an equation has pronumerals (for example, x’s) on both sides, at some stage

they must be gathered together on the same side of the equation.

Solve:a 4x − 3 = 3(6 − x) b c

THINK WRITE

a Write the equation. a 4x − 3 = 3(6 − x)Expand the right-hand side (RHS). 4x − 3 = 18 − 3xCollect x’s on one side, for example, the side which results in a positive x term, in this case, the left-hand side (LHS). (That is, add 3x to both sides.)

7x − 3 = 18

Add 3 to both sides. 7x = 21


x = 3

b Write the equation. b =

Find the lowest common denominator for all three terms. Here, we use 6.

Write all terms with the common denominator, adjusting numerators accordingly (so that numerator and denominator have been multiplied by the same amount).

=

Now that all terms have the same denominator, the numerators must be equal. (Multiply each term by 6.)

3(9x + 3) = 2(13x + 7)

Expand all brackets. 27x + 9 = 26x + 14Collect x’s on the LHS and numbers on the RHS. 27x − 26x = 14 − 9Simplify and solve. x = 5

c Write the equation. c =

Find the lowest common denominator for all three terms. Here, we use 20.

Write all terms with the common denominator, adjusting numerators accordingly (so that numerator and denominator have been multiplied by the same amount).

=

Now that all terms have the same denominator, the numerators must be equal. (Multiply each term by 20.)

5(3 − x) = 18(x + 7) + 20

Expand all brackets. 15 − 5x = 18x + 126 + 20Collect x’s on the RHS and numbers on the LHS. 15 − 126 − 20 = 18x + 5x

Simplify and solve.

−131 = 23x

− =

=

9x 3+2

---------------13x 7+

3------------------= 3 x–

4------------

9 x 7+( )10

-------------------- 1+=

123

4

5217

------

19x 3+

2--------------- 13x 7+

3------------------

2

33 9x 3+( )

6----------------------- 2 13x 7+( )

6--------------------------

4

56

7

13 x–

4----------- 9 x 7+( )

10-------------------- 1+

2

35 3 x–( )

20-------------------- 2 9 x 7+( )×

20----------------------------- 20

20------+

4

56

713123

--------- x

x13123

---------–

2WORKEDExample


Using a graphics calculatorSolving linear equations using solverPress , then continue to press until0:Solver is highlighted. Select it by pressing .

Press to move to the equation line.

Enter the equation to be solved in the form0 = . . .; in this case:

Press , the green key and [SOLVE].

Solving linear equations graphically1. Press and enter the equation. 2. Press . Select 6:ZStandard to

obtain a graph (of ‘standard’ dimensions).

3. If you cannot see where the graphcuts the x-axis, press ,select 3:Zoom Out and press .

5. Press to return to the graph, then [CALC] 2:zero.

MATH �

ENTER

�

2 x 6–( )7

-------------------- 8+

� ALPHA

Y= ZOOM

ZOOMENTER

GRAPH2nd

4. The axis divisions may not be suitable or may be cluttering the window. If so, change them by pressing and selecting new values for Xscl and Yscl.

WINDOW

6. To calculate where the line crosses the x-axis:(a) move the cursor to the left of where the graph cuts the axis and press ,(b) move the cursor to the right of where the graph cuts the axis and press ,(c) move the cursor closer to the crossing point and press again.

ENTER

ENTER

ENTER

C h a p t e r 1 L i n e a r f u n c t i o n s 5

Solving linear equations

1 Solve the following linear equations.

2 Solve:

3 Solve:

4 Solve:

5 Solve:

6 Solve:

7 Solve:

a 3x − 19 = −13 b −6x + 11 = 59 c 8x − 23 = −47d 15 − 2x = 1 e 4x + 25 = −7 f 63 − 7x = 21g −9x + 21 = 3 h 45x + 72 = −423 i 9x + 19 = −2j 15 − 6x = 2

a b c

d e f

g h

a b c

d e f

g h

a b c

d e f

a 2x − 9 = 3(2x − 11) b 5x + 6 = 2(3x + 4) c 7x − 1 = 17(3x − 13)d 5x + 9 = −4(x + 9) e x + 11 = 2(x + 12) f 5 − 2x = 3(3 − x)g 3x − 7 = 2(35 − 2x) h 16 − 4x = 7(1 − x)

a b c

d e f

a b c

d e f

rememberLinear equations can be solved by rearranging to make the pronumeral the subject.

remember

1AWORKEDExample

1a

EXCEL Spreadsheet

Equationsolver

GC program

Equationsolver

3x 1–4

--------------- 5= 4x 2+11

--------------- 2–= 2x 8+3

--------------- 6=

5x 20+7

------------------ 5–= 12 3x–3

------------------ 5= 10 x–4

-------------- 2–=

11 3x–7

------------------ 6= 6x 13+2

------------------ 4–=

WWORKEDORKEDEExamplexample

1b

Mathcad

Equationsolver

4x 6–3

--------------- 7– 3= 2x 7–5

--------------- 11+ 8= 12 3x–3

------------------ 5– 6=

12 9x–2

------------------ 7+ 5–= 7x 4+3

--------------- 8– 9–= 1 x–2

----------- 17+ 20=

x 16+5

--------------- 4– 0= 19 x–4–

-------------- 3+ 1–=


1c 2 3x5

------ 1– 10= 3

5x6

------ 4+ 27= 4

2x3

------ 5+ – 4=

7 8 x4---–

77= 5 8 2x7

------– – 20–= 6

20x9

--------- 15+ 150–=

WORKEDExample

2a


2b x 2+6

------------ x 5–3

-----------= x 11+3

--------------- 2 x 14+( )9

-----------------------= x 1–12

----------- 3x 17–8

------------------=

2x 4–5

--------------- 11 x–2

--------------= 4x 66+3

------------------ 13 3x–4

------------------= x 10+9

--------------- 2 7 3x–( )5

-----------------------=

WORKEDExample

2c 6x 7+5

--------------- 5x 1+4

--------------- 1+= 2x 29+3

------------------ x 44+8

---------------= 2+ 7x 9–2

--------------- 5x 3+4

--------------- 6+=

9x 28+5

------------------ 7x 6–2

--------------- 5–= 7x 9–9

--------------- 21 x–3

-------------- 18–= 17 x–2

-------------- 49 2x+5

------------------ 5+=


Rearrangement and substitutionWhen there is more than one pronumeral involved in an equation, we may rearrange tomake a particular pronumeral the subject using the same rules of equation solvingdescribed in the previous section.

Rearrange each of the following to make the pronumeral in red the subject.

a 6x + 8y − 48 = 0 y b s = ut + at2 u c T = k

THINK WRITE

a Write the equation. a 6x + 8y − 48 = 0Add 48 to both sides. 6x + 8y = 48Subtract 6x from both sides. 8y = 48 − 6x

Divide both sides by 8. y =

Cancel if possible. Here, divide the top and bottom of the fraction by 2.

y =

Other ways of representing the answer are shown opposite.

or y = − x

= 6 − x

= − x + 6

b Write the equation. b s = ut + at2

Subtract at2 from both sides. s − at2 = ut

Multiply both sides by 2. 2s − at2 = 2ut

Divide both sides by 2t. = u

Write the equation with the desired pronumeral on the left.

u =

c Write the equation. c T =

Divide both sides by 2π. =

Square both sides. =

Turn both sides upside down.Note: This can be done only if both sides are fractions.

=

Multiply both sides by m. = k

Write the equation with the desired pronumeral on the left. k =

12---

mk----2π

123

448 6x–

8------------------

524 3x–

4------------------

6 244

------ 34---

34---

34---

1 12---

212---

12---

3

42s at2–

2t-------------------

52s at2–

2t-------------------

1mk----2π

2 T2π------ m

k----

3T2π------

2 mk----

4 2πT------

2 km----

5 m2πT------

2

6 m2πT------

2

3WORKEDExample

C h a p t e r 1 L i n e a r f u n c t i o n s 7Once a pronumeral is isolated, we may substitute values of other pronumerals to

calculate various values of the isolated pronumeral. The following worked exampleillustrates some practical applications.

a The formula for converting temperature in degrees Fahrenheit (F) (which is the system

used in the USA) to degrees Celsius (C) is C = .

i Make F the subject.ii What is the temperature in Fahrenheit when the temperature measured in degrees

Celsius is 21°C?b The area (A) of a circle is given by A = πr2, where r is the radius. Find the value of r

when A = 20 cm2.

THINK WRITE

a i Write the equation. a i C =

Multiply both sides by 9. 9C = 5(F − 32)

Divide both sides by 5. =

Add 32 to both sides. = F

Write the equation with F first. Sometimes it may be appropriate to use a common denominator.

F =

or F =

a ii Replace C with 21. Note that 9C means 9 × C.

a ii F =

Evaluate F.F = + 32

F = 37.8 + 32F = 69.8

so 21°C = 69.8°Fb Write the equation. b A = πr2

Solve for r (make r the subject) as follows.

Divide both sides by π. =

Take the square root of both sides, and write r first. As r is the radius, we take the positive root only.

r =

Substitute A = 20 into the new formula. If A = 20,

r = Evaluate r. = 2.523 cm

5 F 32–( )9

------------------------

15 F 32–( )

9------------------------

2

39C5

------- F 32–

49C5

------- 32+

59C5

------- 32+

9C 160+5

----------------------

19 21×

5--------------- 32+

2

1895

---------

1

2

3Aπ--- r2

4 Aπ---

520π------

6

4WORKEDExample


Computer algebra systemsSoftware applications such as Mathcad,DERIVETM, MathView and the TI–89 and TI–92graphics calculators are able to isolate apronumeral in an equation. The screen at right isfrom a TI–92 graphics calculator.

Notice that the second ‘argument’ of the solvefunction is k, the pronumeral being solved for.

Note: In this example, m is short for metres, m/s is short for metres per second (velocity), and m/s2 is short for metres per second per second (acceleration).

The final velocity v m/s of an object that begins with velocity u m/s and accelerates at a m/s2 over a distance of s m is given by the equation .a Find the value of v when u = 16, a = 2 and s = 60.b Rearrange the given equation to make s the subject.c Find the distance travelled by an object which begins with a velocity of 10 m/s, and has

a final velocity of 4 m/s while accelerating at −1 m/s2.THINK WRITE

a Write the given equation. aSubstitute u = 15, a = 2 and s = 60.

Simplify and evaluate v.

Final velocity is ±4 m/s.

b Begin with the given equation. b v = Square both sides. v2 = u2 − 2as

Subtract u2 from both sides. v2 − u2 = −2as

Divide both sides by −2a. = s

Reverse so that s is given on the left. s =

c Match the pronumerals with the given information.

c u = 10v = 4a = −1

Write the formula that has s as the subject (see part b above).

s =

Substitute the values given in step 1. =

Simplify and evaluate. =

=

= −42Explain the answer in words. The object travels 42 m in the opposite

direction to its initial direction.

v u2 2as–=

1 v u2 2as–=

2 162 2 2 60××–=3 256 240–=

16= 4 ±=

1 u2 2as–

2

3

4v2 u2–

2a–----------------

5 v2 u2–2a–

----------------

1

2v2 u2–

2a–----------------

342 102–

2 1–×–-------------------

416 100–

2---------------------

84–2

---------

5

5WORKEDExample


Rearrangement and substitution

1 Each of the following is a real equation used in business, mathematics, physics or another area of science. Make the pronumeral shown in red the subject in each case.

2 Find the value of the subject (the first mentioned pronumeral), given the values of the other pronumerals.a W = Fd F = 10, d = 5.6b P = 2l + 2w l = 6, w = 9

c A = bh b = 10, h = 16

d s = 12, a = 4, b = 9, c = 11

e R1 = 50, R2 = 100

f k = 60, d = 15

g E = K + mgh K = 250, m = 2, g = 10, h = 5

h n = 3, λ = 2.8

i E = hf0 − W h = 6.62, f0 = 5000, W = 20 000

j ω = 2, r = 1.6, y = 1

a A = L + P P b A = lw l

c t d C = 2πr r

e R f V = πr3h r

g E = I4 R4 − I1 R1 + I2 R2 − I3 R3 I3 h R1 = R2(1 + αθ) α

i E = αθ + βθ2 β j r

k φ l V1 N2 = V2 N1 V2

m pV = nRT n n s = ut + at2 a

o Fd = mv2 − mu2 v p r

q U r γ

s S = 2w(l + h) + 2lh w t S = 2πr2 + 2πrH H

1B

rememberEquations may be rearranged by applying the same rules as those used to solve equations.

remember

Mathcad

Rearrangingequations


3

vdt---=

APRT100-----------= 1

3---

FkQqr2

----------=

Enφt

------=

12---

12--- 1

2--- F

µI2

2πr---------=

f 2

V f 1

V U+---------------= v γ rT=


4

Mathcad

Substitution

12---

A s s a–( ) s b–( ) s c–( )=

RT

R1R2

R1 R2+------------------=

Ikd2-----=

D n 12---–( )λ=

v ω r2 y2–=


3 Make the pronumeral in red the subject, and find its value using the giveninformation.

4 The perimeter, P, of a rectangle of length l and width w may be found using theequation P = 2(l + w).a Find the perimeter of a rectangle of length 16 cm and width 5 cm.b Rearrange the equation to make w the subject.c Find the width of a rectangle that has perimeter 560 mm and length 240 mm.

5 The net force, F, measured in Newton (N) acting on a mass m kilograms (kg) isfound using the equation F = ma, where a is the acceleration of the mass measuredin m/s2.a Find the net force required to accelerate a 2.5 kg rock at the rate of 4 m/s2.b Make a the subject of the equation.c Find the acceleration produced by a 700 N force acting on a 65 kg person.

6 The area of a trapezium (figure A) is given by

, where a and b are the lengths of the

parallel sides, and h is the height.

a Find the area of the trapezium shown in figure B.

b Using figure A, find an equation for determining side a interms of the area A and side b.

c Find a in figure C.

a A = l2 l A = 60

b V = πr3 r V = 1000

c F = mg − kv2 v F = 250, m = 60, g = 10, k = 0.1

d v = u + at a v = 25, u = 0, t = 6e S = πr(r + h) h S = 120, r = 2, π = 3.14

f T = 2π l T = 4, g = 9.8, π = 3.14

g d f = 2, l = 15

h H = U + PV V H = 26, U = 4.5, P = 2

i c K = 6.9, α = 0.05

j u Hi = 34, H0 = 4, v = 40


4

43---

lg---

fl2 d2–

4l---------------=

Kcα2

l α–-----------=

Hi

H0------- v

u---=


5

Aa b+

2------------

h=

a

h

b

AreaA

Figure A

9 m

16 m

21 mFigure B

62 cm50 cm

a

Area = 2000 cm2

Figure C

C h a p t e r 1 L i n e a r f u n c t i o n s 117 The size of a 2-year investment account with a particular bank is given by

, where A is the amount ($) in the account after two years, D is the

initial deposit ($) and r is the interest rate (%).a Find the amount in such an account after two years if the initial deposit was $1000

and the interest rate was 6%.b Make r the subject of the equation.c Find the rate required for an initial deposit of $1000 to grow to $2000 after 2

years.

8 The length of a circular curve is given by the formula

where L = length of curve, π = 3.14,

r = radius of curve and θ is the angle of the curve.

a Determine, to the nearestdegree, the angle required fora curve of radius 15 m andlength 40 m.

b Find the length of the curvededge of the fan at right.

9 The length, L, of metal striprequired to make the bracket at right is given by

.

a What length is required to make a bracket for whicha = 25 cm, b = 35 cm and w = 5 mm?

b Find the value of w (in cm) necessaryto make the bracket at right from a60 cm length of metal.

10 The object and image positions for a lens of focal length f are related by the formula

, where u is the distance of the object from the lens, and v is the distance

of the image from the lens.a Make f the subject of the equation.b Make u the subject of the equation.c How far from the lens is the image when an object is 30 cm in front of a lens of

focal length 25 cm?

A D 1r

100---------+

2=

L

r

rθ

Lπrθ180---------=

123 ο

8 cm

a

b

wL a bw2----+ +=

40 cm

17 cm

1u--- 1

v---+ 1

f---=


Qualifications: Nil (No formal qualifications)Employer: Self-employedCompany website: http://home.netc.net.au/~rick/

I grew up on the family vineyard and wasencouraged to develop an interest in wine-making at a very early age. Thus, I wasfamiliar with the workings of a vineyardfrom nearly day one. A usual day for me isto crush the previous day’s grape harvest,add yeast to establish the correctfermentation and then I am out to thevineyard for picking the next block ofgrapevines. Part of my job is to look aftercustomers by serving tastings of the wines.I then fill, cork and label more wine for thesales room. The wine fermentations arechecked and any wine that has completedits fermentation needs the skins to bepressed off.

Mathematics is very important inwinemaking as it is essential that quantitiesare calculated accurately. For example, tocalculate the amount of pure alcohol to addto a wine to fortify it to 18.5% by volume,the Pearson Square formula is used.

where X = litres of fortifying spirit requiredV = litres of wine to be fortifiedC = final alcohol strength of the wine

in % by volumeA = alcohol strength of the wine

before fortificationB = alcohol strength of fortifying

spiritIf we have 5000 litres of wine at 10%

alcohol by volume and our fortifying spirit is95% alcohol by volume then the litres ofspirit that are required is:

= 555.5.

That is, 555.5 litres of spirit is required.It is necessary to know the exact capacity

of the mixing tank where the fortification iscarried out. If the tank is cylinder shaped,then the formula is a straightforward volumeof a cylinder calculation. Then, once thenumber of cubic centimetres is known, it canbe converted into litres. A wooden dipstick isconstructed with either centimetre marks andthe number of litres per centimetrecalculated or marked in actual litres for thatparticular tank. The exact amount of spirit toadd to a wine can be calculated using maths.This can save a lot of money and reducewastage.

Questions1. What are the two ways in which a wooden

dipstick can be marked?2. Rick uses skills in substituting into an

equation to calculate the amount of purealcohol needed to fortify a wine. What otherareas of mathematics are useful to him?

3. Although Rick has no formal qualifications,are there any courses for vignerons? If so,what are they? Where else could you gainskills for working at a vineyard?

Testing the sugar content of the fermenting wine with an hydrometer

to measure the specific gravity uses a French graduation called

degrees Baum/e. 1 baum/e = 1.008 SG

XV C – A( )

B C–-----------------------=

5000 18.5 10 )–(95 18.5–

----------------------------------------

= 4250076.5

---------------

R I C K M O R R I S — Vi g n e r o n

Career profile


A cliff face with a steeper gradient provides a greater challenge for climbers.

Gradient of a straight lineThe gradient of a line describes its slope or steepness. You may recall from previousstudies the following types of gradient for straight lines.

The gradient may be calculated using the formula

gradient or . These terms are

illustrated at left.

Greate

r+ gr

adien

t

Positivegradients

y

x

Negativegradient

y

x

Zerogradient

y

x

Infinitegradient

y

x

y

x

(x1, y1)

(x2, y2)

Run

Rise

mriserun--------= m

y2 y1–

x2 x1–----------------=

Here are two examples of where gradient canaffect our everyday lives. Can you think ofothers?

Scientists calculate the required gradient of solar panels so that the maximum amount of energy is absorbed.


If the angle a line makes with the positive direction of the x-axis is known, thegradient may be found using trigonometry applied to the triangle shown below.

Calculate the gradient of this linear graph using the intercepts shown.

THINK WRITE

Identify the rise and run.rise = 14, run = 2

Calculate .

y

x–2

14

1

2 mriserun--------= m

142

------=

7=

6WORKEDExample

Calculate the gradient of the line passing through the points (3, −6) and (−1, 8).

THINK WRITE

Use the formula .

Match up the terms in the formula with the values given.(x1, y1) (x2, y2)(3, −6) (−1, 8)

Substitute the given values.

Simplify.

Cancel if possible.

1 my2 y1–

x2 x1–----------------= m

y2 y1–

x2 x1–----------------=

2

3 m8 6––−1 3–---------------=

4 14

4–------=

5 –72---=

7WORKEDExample

y

x

rise

run

tan = rise = mrun

θ

θ


a Find the gradient (accurate to 3 decimal places) of a line making an angle of 40o to the positive x-axis.

b Find the gradient of the line shown. Express your answer to 2 decimal places.

THINK WRITE

a Since the angle the line makes with the positive x-axis is given, the formula m = tan θ can be used.

b The angle given is not the one between the graph and the positive direction of the x-axis. Calculate the required angle θ.

a m = tan θ = tan 40° = 0.839

b θ = 180° − 60°= 120°

Use m = tan θ to calculate m to 2 decimal places.

m = tan θ = tan 120° = −1.73

y

x60°

1 y

x60° θ

2

8WORKEDExample

rememberThe gradient (m) of a straight line may be calculated using the followingformulas:

.

where (x1, y1) and (x2, y2) are points on the line.

m = tanθ where θ is the angle the line makes with the positive direction of the x-axis.

mriserun--------=

y

x

Rise

Run

my2 y1–

x2 x1–----------------=

remember


Gradient of a straight line

1 Calculate the gradient of each of the following linear graphs using the interceptsshown.

2 Without drawing a graph, calculate the gradient of the line passing through:

3 Calculate the gradient of the line joining each pair of points.

a b c

d e f

g h

a (2, 4) and (10, 20) b (4, 4) and (6, 14)c (10, 4) and (3, 32) d (5, 31) and (−7, 25)e (7, 2) and (12, −28) f (−3, 2) and (42, 17)g (0, −30) and (−8, −31) h (−11, −25) and (0, −3)i (217, 4) and (19, 4) j (3, −3) and (−45, 21)k (1, 32) and (67, −100) l (−2, −5) and (0, 0).

a b c

1CWWORKEDORKEDEExamplexample

6

6

–3

y

x

5

–1

y

x

1

–4

y

x

2

6

y

x

12

3

y

x

7

7

y

x

32

–10

y

x

100

45

y

x


7

Mathca

d

Gradientof astraightline

Cabri

Geometry


GCpro

gram


EXCEL

Spreadsheet


x

y

(1, 3)

(5, 5)

x

y(1, 6)

(6, 1)x

y

(3, –2)

(–5, –4)


4 Find the gradient of the line joining each pair of points.

d e f

g h

a b c

d e f

g h

x

y

(–2, –2)

(–7, –10)

x

y

(2, 2)

(–2, –2)

x

y

(–3, 8)

(–2, –4)

x

y

(4, 3)(–5, 3)

x

y

(6, 9)

(6, –3)


7

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y654321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y


5 Find the gradient of each line below.

6 Find the gradient (accurate to 3 decimal places) of a line making the angle given withthe positive x-axis.

7 Find the gradient of each line below. Give answers to 2 decimal places.

8 Which of these lines has:a a non-zero positive gradient?b a negative gradient?c a zero gradient?d an undefined gradient?

a b c

d e f

g h

a 50° b 72° c 10° d −30°e 150° f 0° g 45° h 89°

a b c d

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y654321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y

654321

–1–2–3–4–5–6

–5–6 –4 –3 –2 –1 1 2 3 4 5 60 x

y

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y

654321

–1–2–3–4–5–6

–5–6 –4 –3 –2 –1 1 2 3 4 5 60 x

y654321

–1–2–3–4–5–6

–5–6 –4 –3 –2 –1 1 2 3 4 5 60 x

y


8a


8b y

x43°

y

x69°

y

x28°

y

x15°

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y

A

C

B D


10 Arrange the following in order from smallest to largest gradient. The same scale hasbeen used to draw each graph.

11 Sketch two different graphs that have:a the same gradientb zero gradientsc different negative gradients.

12 Burghar plots the coordinates of a proposed driveway on a plan which is shownbelow. What is the gradient of the proposed driveway?

a Which of the following graphs has a gradient of −2?

b Which of the following lines has a gradient of 3?

c Which of the following lines has a gradient of ?

d Which of the following lines has a gradient of − ?

a b c d e

mmultiple choiceultiple choice

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

yA C

E

B D54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

yAC

E

B

D

12---

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

yA

CE

B

D

25---

54321

–1–2–3–4–5

–5 –4 –3 –2 –1 1 2 3 4 50 x

y

A

C

B

D

x

y

x

y

x

y

x

y

x

y

Driveway

Garage

2 m

17 m


13 An assembly line is pictured below. What is the gradient of the sloping section? (Giveyour answer as a fraction.)

14 A passenger jet takes off along the flight path shown below. What is the gradient ofthe path?

15 Find the value of a in each case so the gradient joining the points is equal to the valuegiven.a (3, 0) and (5, a) gradient: 2b (2, 1) and (8, a) gradient: 5c (0, 4) and (a, −11) gradient: 3d (a, 5) and (5, 1) gradient: −2

16 For safety considerations, wheelchair ramps are constructed under regulated specifi-cations. One regulation requires that the maximum gradient of a ramp exceeding1200 mm in length is to be .

a Does a ramp 25 cm high with a horizontal length of 210 cm meet therequirements?

b Does a ramp with gradient meetthe specifications?

c A 16 cm high ramp needs tobe built. Find the hori-zontal length of theramp required to meetthe specifications.

0.85 m

15 m

BOFFOMade inAustraliaBOFFOMade inAustralia

BOFFOMade inAustralia

150 m110 m

500 m

Runway

SkillSH

EET 1.1

114------

118------


Equations of the form y=mx+cA common form for linear equations is y = mx + c, sometimes referred to as ‘y =’ form.

The following exercise aims to demonstrate the effect of altering m and c. A graphicscalculator would be useful, but is not essential.

Equations of the form y= mx+ c

1 Use a graphics calculator or other method to sketch graphs of the following on thesame set of axes.

2 What is the effect on the graph of the number in front of the x (the ‘x-coefficient’ or‘gradient’)?



5 What is the effect on the graph of the number at the end of the equation (the‘y-intercept’)?

6 State the gradient for each of the following equations.

7 State the y-intercept for each of the equations in the previous question.

a y = x b y = 2x c y = 3x d y = −x e y = −2x

a y = x + 1 b y = x + 2 c y = x + 3 d y = x − 4

a y = 2x + 1 b y = 2x − 7 c y = −3x + 6 d y = 3x − 5

a y = 5x + 7 b y = 6x − 4 c y = −9x + 1 d y = 2x − 13e y = −8x − 5 f y = x + 2 g y = −x − 10 h y = 5x + 0i y = 3x j y = 0x + 17 k y = 2 l y = 0

y = mx + c

y

x

y

xGradient

Gradien

t m

y-intercept

y-intercept

x-intercept

These lines have identicalgradients (equal m values)

remember1. The general equation for a straight line of gradient m and y-intercept c is

y = mx + c.2. Lines with the same gradient (m) are parallel.

remember

1DMathcad

Lineargraphs

EXCEL Spreadsheet

Lineargraphs


8 Write the equation of a line having the following properties (where m = gradient andc = y-intercept).

9 Rearrange the following equations and hence state the gradient and y-intercept for each.

10 Rearrange the following equations and hence state the gradient and y-intercept for each.

11

Which of the following lines is parallel to (that is, has the same gradient as)y = −4x − 7?A 5 − 4y = 13 B x = −4y − 7 C 4x + y + 7 = 0 D y = 4x − 7 E y = 5x − 8

12

Which of the following lines has the same y-intercept as y = 18x − 2?A y = 2 − 18x B y − 18x − 2 = 0 C 3x + 7y = −2D 14 − 7y − 2x = 0 E 2x = −14 − 7y

13 Write three equations starting with y =, that each have a gradient of −7.

14 Write three equations starting with y =, that each have a y-intercept of −6.

15 Write three equations that have the same gradient as 3y + 5x = 17. Write them in thesame form as the equation in this question.

16 Write three equations that have the same y-intercept as 3y + 5x = 17. Write them inthe same form as the equation in this question.

17 State the equation for each of the following graphs.a b

a m = 2, c = 7 b m = −3, c = 1 c m = 5, c = −2d m = 0, c = 3 e m = 1, c = 0 f m = , c = −5g m = , c = h m = − , c = − i y-intercept 12, gradient −2j y-intercept −3, gradient

a y = 9 + 3x b y = −42 + 7x c y = 12 − 4xd y = −35 − 5x e y − 3x = 10 f y + 6x = 24g y + 16x = −15 h y + 9x + 1 = 0 i y − x + 23 = 0j y + 18 − 4x = 0

a 2y = 8x + 10 b 3y = 12x − 24 c −5y = −20x + 30d −y = 3x − 1 e 16 − 4y = 8x f 22 − 2y = −6xg 21x + 3y = −27 h −10x + 5y = 25 i 6y + 3x = −18j −11y − 2x = 66 k 8x + 3y − 2 = 0 l −3x − 4y + 13 = 0m 15 − 6y + x = 0 n 2y + 7 + 5x = 0

12---

23--- 1

3--- 3

4--- 1

2---

52---



GCpro

gram

Guesstheequation

654321

–1–2

–1 1 20 x

y

– 1–2

321

–1–2–3–4–5

–2 20 x

y

C h a p t e r 1 L i n e a r f u n c t i o n s 23c d

e f

18 a A set of axes is placed as shownover the Leaning Tower of Pisa.Find the equation of the linerepresenting the sloped left wallof the Tower.

b What angle does the Tower makewith the ground?

19 State the gradient and y-intercept(in that order) for the following.a y = ax + bb ax + by = cc ax + by + c = 0d 2y = 4kx − 6h

321

–1–2–3–4–5

–2 –1 1 20 x

y

54321

–1–2–3–4–5

–1 1 2 3 4 5 60 x

y

54321

–1–2–3–4–5

–1 1 2 3 40 x

y987654321

–1–2–3–4–5–6–7–8–9

–3 –2 –1 1 2 30 x

y

y

5.2 m

55.8 m

x WorkS

HEET 1.1


Sketching linear graphs using interceptsTo draw a graph, only two points are needed. A line may thenbe drawn through the two points, and will include all otherpoints that follow the given rule. So, rather than construct atable of values and plot several coordinates, it is quicker tomark only the points where a line cuts (or intersects) an axis.These points are called x- and y-intercepts. The x-interceptoccurs when y = 0, while the y-intercept occurs when x = 0.This information is the basis for the approach to sketchingillustrated in the following example.

y

x

Sketch the graph of y = −6x + 15, showing x- and y-intercepts.THINK WRITE

Find the y-intercept (when x = 0).Substitute x = 0 into the equation.

If x = 0 y = −6 × 0 + 15y = 15

Find the x-intercept (when y = 0).Substitute y = 0 into the equation.

If y = 0 0 = −6x + 156x = 15

x =

x = or 1

Mark the intercepts on a set of axes.Note that = (a little over 1 ).Join the intercepts with a straight line.

1

2

156------

53--- 2

3---

353--- 1.66̇ 1

2---

y

x5–3

154

9WORKEDExample

Sketch the graph of 3x − 2y = 12.

THINK WRITE

Find the y-intercept (when x = 0).Substitute x = 0 into the equation.

If x = 0 3 × 0 − 2y = 12−2y = 12

y =

y = −6Find the x-intercept (when y = 0).Substitute y = 0 into the equation.

If y = 0 3x − 2 × 0 = 123x = 12

x = 4

1

122–

------

2

10WORKEDExample


The graphs of some equations do not have two intercepts, as they pass through theorigin (0, 0). Such equations are of the form y = kx or ax + by = 0.

To sketch graphs of such equations, we use (0, 0) and any other point, for example,the point where x = 1. (We could choose any other non-zero value.)

THINK WRITEMark the intercepts on a set of axes.Join the intercepts with a straight line.

3 y

x

–6

4

4

Sketch the graph for the equation 4x − 3y = 0.THINK WRITE

Try substituting x = 0 to find the y-intercept.

If x = 0 4 × 0 − 3y = 0−3y = 0

y = 0Note that the graph passes through (0, 0). There is no point substituting y = 0, as we know we’ll get x = 0.Substitute another x-value. In this example we use x = 1.

If x = 1 4 × 1 − 3y = 04 −3y = 0

4 = 3y

y =

Plot the points (0, 0) and (1, ) on a set of axes. Note that is 1 , which is a little less than 1 .

1

2

3

43---

443---

43--- 1

3---

12---

y

x

4–3

(1, )

11WORKEDExample

rememberTo sketch a linear graph:1. Let x = 0 and find the y-intercept.2. Let y = 0 and find the x-intercept.3. If (0, 0) is an intercept, find another point on the line by substituting x = 1 (or

any other convenient non-zero value).4. Mark and join the intercepts with a straight line.

remember


Sketching linear graphs using intercepts

1 Sketch graphs of the following linear equations, showing x- and y-intercepts.

2 Sketch graphs for each of the following.

3 Sketch the graph for each equation.

4 Sketch the graph for each equation.

5Which of the following is in the form ax + by = c? (One or more answers.)

6The x- and y-intercepts for the equation 2y = −3x + 12 are (respectively):

7Which of the following has a y-intercept of −3?

To find x- and y-intercepts using a graphicscalculator, proceed as follows:

1. Enter the equation for y in the Y= menu.

2. Press and then or adjust settings to obtain a view of the graph thatincludes both intercepts.

3. To find the x-intercept, press [CALC] andselect 2:Zero. Scroll to the left of the x-interceptand press . Scroll to the right of thex-intercept and press . Press again.

4. To find the y-intercept, press [CALC] andselect 1:Value. Enter X = 0 and press .

a y = 6x + 18 b y = 3x − 21 c y = 5x + 12d y = −2x − 3 e y = 10 − 5x f y = 1 − xg y = −9x + 30 h y = 2(x − 8)

a 2x + 3y = 6 b 4x + 5y = 20 c −5x + 8y = 10d 6x − 3y = −18 e 7x − 5y = 35 f 8y − x = 4g x − y = 2 h −2x + 11 = 6y

a 6x + 7y + 42 = 0 b 5x − 2y + 20 = 0 c −3x + 4y − 16 = 0d y − 3x + 6 = 0 e 9x + 18 − 2y = 0

a x + y = 0 b x − y = 0 c 2x + y = 0

A 2x − 3y − 1 = 0 B 2x + 3y + 1 = 0 C 2x + 3y = x D 2x + 3y = 1 E y = x − 1

A 2 and 3 B −3 and 12 C −4 and 6 D −4 and −6 E 4 and 6

A y = −3x − 3 B y = −3x + 3 C x + 3y = 9 D x − 3y + 9 = 0 E 3x + y + 9 = 0

1EWORKEDExample

9Mathca

d

Gradientof astraightline WWORKEDORKED

EExamplexample

10

Cabri

Geometry



11mmultiple choiceultiple choice

23---



Graphics CalculatorGraphics Calculator tip!tip! Finding x- and y-intercepts

GRAPH ZOOM WINDOW

2nd

ENTERENTER ENTER

2ndENTER


Simultaneous equationsSimultaneous equations are groups of equationscontaining two or more pronumerals. In thissection, we look at pairs of linear equationsinvolving the pronumerals x and y. Each equation,as we have learnt in previous sections, may berepresented by a linear graph that is true for manyx- and y-values. If the graphs intersect (whenwouldn’t they?), the values of x and y at theintersection are those that make both equations true.

Graphical solutionThe following example shows how a graphics calculator may be used to solvesimultaneous equations graphically. Hand drawn sketch graphs may be used if graphicscalculators are not available. Refer to earlier sections if you need reminding how todraw linear sketch graphs, and make sure you use a consistent and accurate scale.

Both graphs havethe same x- andy-values here.

y

x

Use a graphics calculator to solve the following simultaneous equations and sketch the screen from which the solution was obtained.y = −3x + 54x − 7y + 8 = 0

THINK WRITE/DISPLAY

Convert the second equation into Y= form so it may be entered in the Y= menu of the graphics calculator.

4x – 7y + 8 = 04x + 8 = 7y

7y = 4x + 8

y = x +

Press . At Y1=, enter (-)3X+5.At Y2=, enter 4/7X+8/7.

Press , and adjust the WINDOW settings if required, to encompass the point of intersection of the two graphs.

Press [CALC]. Select 5:intersect and then press three times to find the coordinates of the intersection.

Sketch the screen.

1

47--- 8

7---

2 Y=

3 GRAPH

4 2ndENTER

5

12WORKEDExample


Algebraic solutionIt is possible to solve simultaneous equations without graphs, that is, algebraically. Themethods of substitution and elimination taught in earlier years may be used.

Use the substitution method to solve the following simultaneously.−2x + 7y = 8y = 3x − 1THINK WRITE

Write down and label the equations. −2x + 7y = 8 [1]y = 3x − 1 [2]

Substitute [2] into [1] and label the resulting equation.

Substitute [2] into [1].−2x + 7(3x − 1) = 8 [3]

Solve [3] for x and label the solution as equation [4].

−2x + 21x − 7 = 819x − 7 = 8

19x = 15

x = [4]

Use the solution to solve for y. Substitute [4] into [2].

y = 3 − 1

= − 1

= −

= or 1 [5]

State the complete answer. Solution: ,

Optional double-check: substitute equations [4] and [5] into [1] to check that these values for x and y make [1] true.

Check: In [1],LHS = −2x + 7y

= −2 + 7

= − +

=

= 8= RHS �

1

2

3

1519------

41519------

4519------

4519------ 19

19------

2619------ 7

19------

5 1519------

2619------

6

1519------

2619------

3019------ 182

19---------

15219---------

13WORKEDExample

Use the elimination method to solve these simultaneous equations.2x + 9y = −55x − 2y − 12 = 0THINK WRITE

Write down and label the equations. 2x + 9y = −5 [1]5x − 2y − 12 = 0 [2]

Rearrange [2] so it is in a similar form to [1]. Call this [3]. Write down [1] again.

5x − 2y = 12 [3]2x + 9y = −5 [1]

1

2

14WORKEDExample


THINK WRITEObtain 10x in both [1] and [3] as explained opposite.

2 × [3]: 10x − 4y = 24 [4]5 × [1]: 10x + 45y = −25 [5]

Eliminate x as shown. [5] − [4]: 49y = −49Solve for y.

y = − y = −1 [6]

Substitute [6] into [1] to find x. Substitute [6] into [1].2x + 9 (−1) = −5

2x − 9 = −52x = −5 + 92x = 4

x = 2 [7]State the solution. Solution: (2, −1)Again, [6] and [7] may be checked in [2] if desired.

Check: In [2],LHS = 5x − 2y − 12

= 5(2) − 2(−1) − 12= 10 + 2 − 12= 0= RHS �

3

45 49

49------

6

78

Two shoppers buy the following at a fruit shop, paying the amounts given. What was the cost of each apple and each banana?Shopper 1: 4 apples and 3 bananas for $2.59Shopper 2: 6 apples and 5 bananas for $4.11THINK WRITE

Decide on pronumeral names for the unknown quantities.

Let a = cost of an apple (in cents).Let b = cost of a banana (in cents).

Write equations involving these pronumerals. Work in terms of cents.

4a + 3b = 259 [1]6a + 5b = 411 [2]

Choose a pronumeral to eliminate. In this case b.

5 × [1]: 20a + 15b = 1295 [3]3 × [2]: 18a + 15b = 1233 [4]

Find [3] − [4] and solve for a. [3] − [4]: 2a = 62a = 31 [5]

Solve for b. Substitute [5] into [1].4 × 31 + 3b = 259

124 + 3b = 2593b = 135

b = 45 [6]State the answer using [5] and [6] as a guide.

The cost of an apple is 31 cents, and the cost of a banana is 45 cents.

1

2

3

4

5

6

15WORKEDExample

remember1. With a graphics calculator, express as Y1 = and Y2 = and find the intersection

using [CALC] and 5:intersect.2. If form is y = ax + b, y = cx + d consider using substitution.3. If form is ax + by = c, dx + ey = f consider using elimination.

2nd

remember


Simultaneous equations

1 Use a graphics calculator to solve the following simultaneous equations, and sketch the screen from which the solution was obtained.

2 Use the method of substitution to solve the following simultaneously.

3 Use the elimination method to solve these simultaneous equations.

4 At the conclusion of a tour of Wonky Willy’s confectionery factory, Nutrina buys 10choc balls and 8 fizz wizzers for $4.30, and her friend purchases 6 choc balls and 9fizz wizzers for $4.05. Determine the cost of each type of lolly.

5 The sum of two whole numbers, x and y, is 41. The difference between them is 3.Write two equations involving x and y and solve them to find the numbers.

6 A farmer counts emus and sheep in a paddock, and notes there are 57 animals and 196feet. Assuming no animal amputees, how many of each animal are there?

7 A sports store supplies 24 basketballs and 16 cricket balls to one school for $275.60,and delivers 12 basketballs and 32 cricket balls to another school for $211. If deliveryis free, how much did the supplier charge for each type of ball?

8 A businessperson hires a stretch limousine for 2 days and a sedan for 3 days while onan interstate trip. If the total car hire cost of the trip was $675, and the limousine costtriple the price of the sedan, find the cost per day of the limousine.

9A manufacturing plant produces fixed size square and circular metal panels. If themass of a square panel is 13 kg and that of a circular panel is 22 kg, how many ofeach panel are there in a truck loaded with 65 panels of total mass 1205 kg?The equations to solve are:

10Which of the following is a solution of 11x + 2y = −121 and 10x + 12y = −222?

a y = −2x, y = −4x − 6 b y = 4x, y = 3x − 5c y = 3x − 5, y = 20 d y = −2x − 4, y = −5x + 5e y = 3x + 5, y = 7x − 4 f y = −5x + 12, y = 2x − 1g −3x + y = −4, y = 6x + 5 h 2x − y − 11 = 0, y = −4x + 8i y = 10x + 1, 2x + y = −6 j 9x + y = 17, x + y = 14

a y = 3x + 1, y = 2x + 2 b y = −3x, y = 4x + 14c y = 5x + 5, y = −x − 19 d y = −4x − 3, y = 3x − 24e y = x + 2, 3x − 4y = −1 f y = 3x − 6, 2x + y = 9g y = −2x + 3, −5x + 2y = 1 h 6x − y = 8, y = x + 4i −4x − 3y = 2, y = −6x + 7 j y = 10 − x, 2x + 7y = 5

a 9x + 10y = 153, 3x − y = 12 b 2x + 4y = 8, 7x + y = −37c 7x − 11y = −13, x + y = 11 d 6x − 9y = 51, −6x + 11y = −49e 6x − 2y = 10, 2x + 5y = −8 f 2x + 7y = 16, 3x − 6y = 2g −3x + y = 8, 4x + 2y = 21 h 8x + 3y = −9, 4x − y = −3i 7y − x = 11, x + y = 10 j x − 11y = −15, y + 6x = 9

A 13s + 22c = 1205, s + c = 65 B 22s + 13c = 1205, s + c = 65C 13s + 22c = 65, s + c = 1205 D 22s + 13c = 65, s + c = 1205E 13s + 22c = 1205, s + c = 35

A (11, 2) B (−121, −222) C (10, 12) D (−9, −11) E (6, 10)

1FWWORKEDORKEDEExamplexample

12

Mathca

d

EXCEL

Spreadsheet

Simultaneous linear equations —graphical method

Mathca

d

Cabri

Geometry

Simultaneous linear equationsWWORKEDORKEDEExamplexample

13

GCprogram

Simultaneous linear equations

EXCEL

Spreadsheet


14


15




Using matrices to solvesimultaneous equations

The steps below show how to solve the following set of three simultaneousequations. They may be easily modified to handle any number of equations.

2x + 3y + 4z = 21 [1]x – 3y – z = 0 [2]6x + 7z = 44 [3]

In matrix notation, the above set may be written as follows:

or AX = B for short, where

A =

The solution to the set of equations may be found by calculatingA–1B.

A–1 is the ‘inverse’ of A, and may be found using the graphics calculator.The following steps show how to use matrices on the graphics calculator to solve

the given set of equations.

1. Press and select EDIT and 1:[A]. Overtype the dimensions with 3 × 3 if necessary, and press . Move the cursor to the various positions within the matrix and enter the required values (press after each one).

2. Now press and select EDIT and 2:[B]. Overtype the dimensions with 3 × 1, then press .

3. Fill in the required values.

4. Press [QUIT] to return to the home screen.

5. Press and select NAMES and 1:[A].

6. Press , and , and select NAME and 2:[B].

7. Press to find the solution matrix, which gives the solution for x, y and z from top to bottom, that is, in this example, x = 5, y = 1 and z = 2.

2 3 4

1 3 –1–

6 0 7

x

y

z

=

21

0

44

2 3 4

1 3 –1–

6 0 7

, X =

x

y

z

and B =

21

0

44

MATRIX

ENTER

ENTER

MATRIX

ENTER

2nd

MATRIX

x –1 X MATRIX

ENTER


Perpendicular linesThe reciprocal of a value is equal to one divided by the value.

Using algebra, the reciprocal of is

and the reciprocal of a is .

The reciprocal of a fraction is an ‘upside down version’ of the fraction. For example,

the reciprocal of is . The reciprocal of 7 (which may be thought of as ) is .

A negative reciprocal of a value is the opposite sign, and the reciprocal of the value.

The negative reciprocal of is − . The negative reciprocal of − is .

Perpendicular lines

The equations below are in pairs where each has the negative reciprocal gradient of the other.

1 Copy and complete the table below. You may use a graphics calculator to help you if one is available.

2 Sketch Y1 and Y2 for part a above using a graphics calculator (make the scale of theaxes equal by pressing and selecting 5:ZSquare) or using graph paper withidentical scales on each axis to produce accurate graphs. What do you notice about thegraphs? (Come back to this question later if you’re not sure.)

3 Repeat question 2 for graphs b to e. Do you notice anything special about each pair of graphs?

4 Find the gradient of a line perpendicular to another line which has a gradient of:

5 Find the gradient of a line which is perpendicular to the line with equation:

Equation 1(Y1)

Gradient of Equation 1

(m1)Equation 2

(Y2)

Gradient of Equation 2

(m2) m1 × m2

a y = 2x + 1 y = − x

b y = 3x − 4 y = − x + 2

c y = x + 6 y = −4x − 9

d y = x − 3 y = − x + 4

e y = – x y = x + 1

a 4 b −9 c d − e f 1

a y = −5x + 2 b y = x − 1 c y = x + 1d y = − x − 2 e 2x + y = 5 f 3x − 4y = 7

ab--- 1

ab---÷ 1

ba---× b

a---= =

1 a÷ 1a---=

23--- 3

2--- 7

1--- 1

7---

49--- 9

4--- 9

5--- 5

9---

1G

SkillSH

EET 1.2

12---

13---

14---

25---

52---

97---

79---

ZOOM

17--- 8

9--- 7

2---

23---

76---

rememberIf two graphs have negative reciprocal gradients, (m1 × m2 = −1), they are perpendicular; if they are perpendicular, then they have negative reciprocal gradients.

remember

GCpro

gram

Anglebetweentwo lines


Formula for finding the equation of a straight line

Consider a general linear graph containing the particularpoints (x1, y1), (x2, y2) and the general point (x, y) (whichcould be any point).

Using the first two of these points in the formula forgradient, we have

[1]

Using the first and last point in the same formula yields

[2]

Putting [2] = [1] gives which may be rearranged to

Since , equation [3] may be written

This last formula may be used to find the equation of a straight line when two pointsare given or when the gradient and only one point are given. When two points are

given, the gradient m may first be found using , and substituted into the

formula along with one of the points.

y

x

(x1, y1)

(x2, y2)

(x, y)

my2 y1–

x2 x1–----------------=

my y1–

x x1–--------------=

y y1–

x x1–--------------

y2 y1–

x2 x1–----------------=

[3]y y1–y2 y1–

x2 x1–----------------

x x1–( )=

my2 y1–

x2 x1–----------------=

[4]y y1– m x x1–( )=

my2 y1–

x2 x1–----------------=

y y1– m x x1–( )=

Find the equation of the line having gradient − , that passes through (7, 11).Express your answer in the form i ax + by + c = 0 and ii y = mx + c.

THINK WRITE

As one point and the gradient are known, use the formula y − y1 = m(x − x1).

y − y1 = m(x − x1)

List the given information. m = (x1, y1)

(7, 11)Substitute for all pronumerals except x and y.

y − 11 = (x − 7)

Simplify, expressing in the form.ax + by + c = 0

i 4y − 44 = 3(x − 7)4y − 44 = 3x − 21

3x − 4y + 23 = 0Express your answer in the form y = mx + c.

ii 3x + 23 = 4y

y = x +

34---

1

234---

334---

4

534--- 23

4------

16WORKEDExample


Find the equation of the straight line containing the points (2, −5) and (−3, 1).Express your answer in the form i ax + by + c = 0 and ii y = mx + c.

THINK WRITE

Write down the points so they match the pronumerals in the formula.

(x1, y1) (x2, y2)(2, −5) (−3, 1)

As two points are known, first use the

formula to find m.

m =

=

=

=

Write the formula y − y1 = m(x − x1). y − y1 = m(x − x1)

Substitute the calculated gradient

and the first point (x1, y1) =

(2, −5). Leave x and y as they are. y − −5 =

Simplify and express in the two forms required.

y + 5 = (x − 2)

−5y − 25 = 6(x − 2)−5y − 25 = 6x − 12

−5y = 6x − +13i So 6x + 5y + 13 = 0, or

ii y =

1

2

my2 y1–

x2 x1–----------------=

y2 y1–

x2 x1–----------------

1 5––−3 2–---------------

65–

------

–65---

3

4

m –65---=

–65--- x 2–( )

565–

------

65---x– 13

5------–

17WORKEDExample

remember1. If given a point and the gradient, use y − y1 = m(x − x1).

2. If given two points, use

or

first find m using , then use y − y1 = m(x − x1).

y y1–y2 y1–

x2 x1–----------------

x x1–( )=

my y1–

x x1–--------------=

remember


Formula for finding the equation of a straight line

1 Find the equation of a straight line passing through the point listed and having thegiven gradient. Express your answer in the form i ax + by + c = 0 and ii y = mx + c.

2 Determine the equation of the line containing each pair of points. Express youranswers in the form i ax + by + c = 0 and ii y = mx + c.

3A particular line has a gradient of and passes through the point (−1, 4). Theequation consistent with this information is:

4A line passes through the points (3, 4) and (−5, 6). An appropriate matching of pro-numerals and values could be:

5 Find the equation of the line passing through (3, −3) that makes an angle of 45° withthe positive x-axis.

6 Find the equation of the line containing (7, −2) that makes an angle of 71.565° withthe positive x-axis.

7To determine the formula of a straight line, you need to know:A one point only (that is, a particular x-value and y-value)B the gradient onlyC a point and the gradientD an x-value and the gradientE the x- and y-values for one point, and the x- or y-value for another point.

a (1, 2) gradient 3 b (5, 6) gradient 2c (4, 1) gradient 5 d (−1, 7) gradient 4e (3, −2) gradient −1 f (0, −5) gradient −3

g (−3, 2) gradient h (9, 0) gradient −

i (−6, −1) gradient j (12, 8) gradient −

k (−4, 4) gradient l (0, 3) gradient −

a (5, 2) (3, 1) b (1, 1) (5, 5)c (6, 3) (8, 2) d (2, −2) (0, 1)e (−5, 8) (−1, −4) f (9, −5) (7, −3)g (−1, −4) (2, 10) h (4, 0) (0, −3)i (0, −6) (−14, 2) j (−7, 8) (8, −7)

A y = 4 + (x + 1) B y + 4 = (x − 1) C y − 4 = (x − 1)

D y = −4 + (x + 1) E y = 4 − (x − 1)

A x1 = 3, y1 = −5, x2 = 4, y2 = 6 B x1 = 4, y1 = 3, x2 = 6, y2 = −5C x1 = 3, y1 = 4, x2 = −5, y2 = 6 D x1 = 3, y1 = 6, x2 = 4, y2 = −5E x1 = −5, y1 = 4, x2 = 6, y2 = 3

1HWWORKEDORKEDEExamplexample

16Mathcad

Equationof a

straightline

12--- 4

3---

45--- 1

6---

87--- 3

11------


17

EXCEL Spreadsheet

Equationof a

straightline

GC program

Equationof a

straightline

mmultiple choiceultiple choice34---

34--- 3

4--- 3

4---

43--- 4

3---




8 Find the equation of the line (in y = mx + c form) that is:a perpendicular to the line with equation y = 3x + 1, passing through (−3, 6).b parallel to the line with equation y = x − 9, passing through (4, −7).c parallel to the line with equation 3x + 6y = 8, passing through (2, 2).d perpendicular to the line with equation −6x + 7y − 2 = 0, passing through (4, 0).e having gradient 2, passing through the intersection of the lines with equations

y = 3x − 5 and y = −2x + 5.f having gradient − , passing through the intersection of the lines with equations

x + 4y = −14 and −5x + 2y = 4.

9 Find the equation of the line which passes through the point of intersection of thelines whose equations are 7x − 3y − 19 = 0 and 3x + 2y + 5 = 0, given that the requiredline is parallel to the line with equation −5x − 2y = 3.

10 Find the equation of a line containing the intersection of the lines with equationsy = −3x + 4 and 5x − 3y + 40 = 0 which:a has a gradient of b is perpendicular to the line with gradient c passes through the point (−1, 9)d is parallel with the line joining (−8, 5) and (0, 4).

11 A line passes through the points (−8, −5), (4, −3) and (a, 12). Find the value of a.

12 A factory produces 25 components at a cost of $830, and on another run produces 35components at a cost of $1050. Find an expression relating the cost (C) to the numberof components (n) produced, assuming a linear relationship.

13 The height of a particular young pine tree is found to increase in a linear manner eachweek in the first year after planting. Find an equation connecting height with time inmonths after planting, using the information supplied.

25---

34---

67---

23---

WorkS

HEET 1.2

52 cm

34 cm

After 2 months After 5 months


Distance between two pointsThe distance between two points on the Cartesian planemay be found using Pythagoras’ theorem applied to aright-angled triangle as shown below.

Using Pythagoras’ theorem c2 = a2 + b2 or

and replacing c with d, a with (x2 − x1) and

b with (y2 − y1) we have .

The TI-83 may be set up to calculate a value given by formula for different values ofcertain variables. The formula for the distance between two points is a good example.

We will use the following variables to representthe ‘usual’ values required in the distance formula

Usual variable x1 y1 x2 y2Calculator variable A B C D

The steps below show how to repeat calculationsusing this formula. Initially, consider finding thedistance between the points (1, 2) and (5, 9).1. Type the information in the figure into the home

screen, and press to calculate.2. Press [ENTRY] to recall the previous

calculation instructions.3. Use the arrow keys to move over old values of A, B, C and D, and type new values.

(Note: To insert a negative sign (–) or extra digits, press [INS] before typing thecharacter/s to be inserted.)

4. Press to calculate the expression for the new set of values.

y2

y1

d

x1 x2

y

x

(y2 – y1)

(x2, y2)

(x2–x1)(x1, y1)c a2 b2+=

d x2 x1–( )2 y2 y1–( )2+=

Find the distance between the points (−3, 7) and (5, −2) correct to 3 decimal places.THINK WRITE

Match up (−3, 7) and (5, −2) with (x1, y1) and (x2, y2).

(x1, y1) (x2, y2)(−3, 7) (5, −2)

Substitute into the formula for d and simplify.

1

2 d x2 x1–( )2 y2 y1–( )2+=

5 3––( )2 −2 7–( )2+=

8( )2 9–( )2+=

64 81+= 145= 12.042=

18WORKEDExample

Graphics CalculatorGraphics Calculator tip!tip! Repeated calculation of the distancebetween two points

d x2 x1 )2– y2 y1 )2:–(+(=

ENTER2nd

2nd

ENTER


Distance between two points

1 Find the distance between each of the following pairs of points.

2 Calculate the distance between the pairs of points below, accurate to 3 decimal places.

3 Find the perimeter of each shape below.a b

4 Find the perimeter of each shape, given the coordinates of the vertices.a b

5 Find the distance between the following pairs of points in terms of the given pro-numeral(s).

a (4, 5) and (1, 1) b (7, 14) and (15, 8)c (2, 4) and (2, 3) d (12, 8) and (10, 8)e (14, 9) and (2, 14) f (5, −13) and (−3, −7)g (−14, −9) and (−10, −6) h (0, 1) and (−15, 9)i (−4, −8) and (1, 4) j (12, 9) and (−4, −3)

a (−14, 10) and (−8, 14) b (6, −7) and (13, 6) c (−11, 1) and (2, 2)d (9, 0) and (5, −8) e (2, −7) and (−2, 12) f (9, 4) and (−10, 0)

a (a, 1), (2, 3) b (5, b), (0, 6) c (c, 2), (4, c) d (d, 2d) (1, 5)

rememberThe distance d between two points having coordinates (x1, y1) and (x2, y2) is given by

d x2 x1–( )2 y2 y1–( )2+=

remember

1I

Mathca

d

Distancebetweentwo points

WORKEDExample

18

EXCEL

Spreadsheet


GCpro

gram


Cabri

Geometry


–5 5

5

–5

x

y

–5 5

5

–5

x

y

y

x

(8, 2)(2, 2)

(6, 7)

y

(–2, –3)

(–3, 4)(5, 5)

C h a p t e r 1 L i n e a r f u n c t i o n s

39

6

A hiker is about to hike from A to B (shown on the map below). How far is it from Ato B ‘as the crow flies’, that is, in a straight line?

7

In the process of decorating a garment, Terrence places a sheet of material on a gridboard as shown. He needs to stitch between the two points marked. What distance is this?

(R3, B5)

(R1, T1)

RightLeft

Bottom

Top

= 10 cm

N

Grid spacing : 1 kmS

EW

B (E7, N4)

Lake Phillios

100 m

200 m

50 m

A (W12, S5)

300 m

200 m100 m

Ch 01 MM 1&2 YR 11 Friday, June 29, 2001 11:00 AM


8 An architect draws a plan of aproposed new house on a gridas shown. What length ofstormwater pipe is needed toconnect points A and B?

9 A military manoeuvreinvolves soldiers work-ing their way throughthe path shown on themap at right. What is thetotal distance along themarked path?

10 Using the coordinates shown on the aerial photo of the golf course, calculate (to thenearest metre):a the horizontal distance travelled by the golf ball for the shot down the fairwayb the horizontal distance that needs to be covered in the next shot to reach the point

labelled A on the green.

= 1 mScale:

Stormwater pipe

Stormwater pipe

Stormwater pipe

Inspection pity

x

Inspection pit

Silt trapA

BTo main road

(–8, –1)

(–2, –4)(5, –4)

(7, 3)

(–6, –7)

(–12, –6)

(–1, 2)

Wetlands

Tank

Tank

TankTrees

Trees

Bunkers

Bunker

Soldiers

Soldiers

SoldiersSoldiers

y

x

y (in metres)

xx (in metres)

A (320, 148)(225, 96)

(80, –64)


Approximating curve length using linear equations

The edge of a particular road corner is parabolic in shape, with the equation of the parabola given by y = x2 − 2x, for x between 0 and 3.

One way of estimating the length of the corner is to find the straight line distance between several pairs of points on the parabola. A printout from a program such as DERIVE TM for Windows may be used.

The total of the straight line segment distances using the points above is 1.414 21 + 1.414 21 + 3.162 27 = 5.990 69 or 6 if rounded to 1 decimal place.1 Complete the table below.

2 Calculate the sum of the straight line segment distances between the points inthe table. Which answer (6 or the one obtained from question 1) is the more accu-rate? Why?

3 Explain how a very accurate estimate of the corner length may be obtainedusing this method.

4 Find an approximation for the length of the curve y = x2 + 3x − 4 betweenx = −4 and x = 1.The arcLen function from the calc menu on

a TI–92 graphics calculator may be used to find a very accurate measure of the length of the curve between x = 0 and x = 3.

x 0 0.5 1 1.5 2 2.5 3

y

42

M a t h e m a t i c a l M e t h o d s U n i t s 1 a n d 2

Midpoint of a segment

The midpoint M(

x

m

,

y

m

) of a segmentjoining two general points A(

x

1

,

y

1

) andB(

x

2

,

y

2

) is shown on the Cartesian planebelow.

Since the horizontal and vertical sidesof the two triangles must be equal, wehave

So the point M has coordinates .

y

x

M(xm, ym)

y2 – ym

x2 – xm

ym – y1

xm – x1

A(x1, y1)

B(x2, y2)

xm − x1 = x2 − xm and ym − y1 = y2 − ym

Simplifying these,

2xm = x2 + x1 2ym = y2 + y1

xm = ym = x2 x1+

2----------------

y2 y1+

2----------------

x1 x2+

2----------------

y1 y2+

2----------------,

Find the midpoint of the segment joining (5, 9) and (−3, 11).

THINK WRITE

Match (5, 9) and (−3, 11) with (x1, y1) and (x2, y2).

(x1, y1) (x2, y2)(5, 9) (−3, 11)

Substitute values into the formula for M and simplify.

1

2 Mx1 x2+

2----------------

y1 y2+2

----------------, =

5 −3+

2--------------- 9 11+

2---------------,

=

22--- 20

2------,

=

1 10,( )=

19WORKEDExample

rememberThe midpoint M of a segment joining (x1, y1) and (x2, y2) is given by

Mx1 x2+

2----------------

y1 y2+2

----------------, =

remember




1 Find the midpoint of the segment joining each of the following pairs of points.

2 Find the midpoint of the segment joining each of the following pairs of points.

3 Find the coordinates of the midpoint of each of the following pairs of points, in terms of a pronumeral or pronumerals where appropriate.

4 The vertices of a triangle are (−3, 10), (7, 8) and (1, 2).a Sketch the triangle, and label the coordinates of each vertex.b Calculate the gradient of each side.c Calculate the coordinates of the midpoint of each side.d Find the gradients of the segments joining the midpoints.e Compare your answers to (b) and (d). What do you notice?

5 Find the value of a in each question below so that the point M is the midpoint of thesegment joining points A and B.

6 a Find a and b if the midpoint of the segment joining (a, b) and (−2, 5) is (3, −8).b Find a and b if the midpoint of the segment joining (a, b) and (−9, 3) is (−7, 1).c Find a and b if the midpoint of the segment joining (a, b) and (0, 5) is (0, 6).d Find a and b if the midpoint of the segment joining (a, b) and (−20, −11) is (14, 16).

7 A fun-run course is drawn (not to scale) atright. If drink stations D1, D2 and D3 are tobe placed at the middle of each straight sec-tion, give the map coordinates of each drinkstation.

8 Find the equation of a line that has a gradientof 5, and passes through the midpoint of thesegment joining (−1, −7) and (3, 3).

9 Find the equation of a line parallel to the line with equation 9x − 3y = 5 that passesthrough the midpoint of the segment connecting (0, −4) and (−2, 10).

a (1, 3) and (3, 5) b (6, 4) and (4, −2)c (2, 3) and (12, 1) d (6, 3) and (10, 15)e (4, 2) and (−4, 8) f (0, −5) and (−2, 9)g (8, 2) and (−18, −6) h (−3, −5) and (7, 11)i (−8, −3) and (8, 27) j (−4, 1) and (−2, 13)

a (7, −2) and (−4, 13) b (0, 22) and (−6, −29)c (−15, 8) and (−4, 11) d (−3, 40) and (0, −27)

a (2a, a) and (6a, 5a) b (7b, 2) and (b, 10)c (5, 3c) and (11, 3c) d (d, 2) and (3, d)e (2e, 6) and (8, 4e) f (3f, 5) and (g, −1)

a A(−2, a), B(−6, 5), M(−4, 5) b A(a, 0), B(7, 3), M(8, )c A(3, 3), B(4, a), M(3 , −6 ) d A(−4, 4), B(a, 0), M(−2, 2)

1JWWORKEDORKEDEExamplexample

19

Mathcad

Midpointof a

segment

EXCEL Spreadsheet

Midpointof a

segment

GC program

Midpointof a

segment

Cabri Geometry

Midpointof a

segment

32---

12--- 1

2---

y

x

(–4.5, 5)(1.5, 3.5)

(13, –8)(3, –7)

(1.5, –2)START/FINISH

Official tent

D1

D2

D3Coordinatesare in kilometres.

44


Linear modelling

Many real life applications such as fees charged for services, cost of manufacturing orrunning a business, patterns in nature, sporting records and so on follow linear relation-ships. These relationships may take the form of a linear equation, for example,

F

=

50

+

30

t

may be used by a tradesperson to calculate her fee (in dollars) for

t

hoursof work.

Here,

F

is the fee in dollars, and

t

the time in hours. The 50 represents an initialfee for simply turning up, while the 30

t

is the amount charged for the time spent onthe job.

For example, if

t

=

2 hours, 30

t

=

60, so that total charge for the work would be$(50

+

60)

=

$110.Equations like

F

=

50

+

30

t

are sometimes referred to as ‘linear models’, a commonform of which is

Total cost

=

Fixed cost

+

Cost per unit

×

Number of units.

This is, of course, similar to

y

=

mx

+

c

, or

y

=

c

+

mx

.

A generator company charges a $200 delivery fee, and a rental fee of $1500 per day.a Find an expression relating total charge to the number of days for which the generator

is hired.b Sketch a graph of the relationship.c What would be the charge for 4 weeks of rental?

THINK WRITE

a Define convenient pronumerals. a Let T = Total charge (in dollars) and n = number of days the generator is hired.

The fixed cost is $200, and the cost per unit is $1500.(c = 200, m = 1500)

T = 200 + 1500n

b Find the vertical intercept (when n = 0).

b If n = 0, T = 200

The total cost rises $1500 each day, so the graph must show this.

c After 4 weeks, n = 28. Substitute this in the equation from part a.

c If n = 28T = 200 + 1500 × 28 = 200 + 42 000 = 42 200

Write the answer in words. After 4 weeks, the total cost is $42 200.

1

2

1

2

1 2

3200

200

1700

T ($)

n (Days)

1

2

20WORKEDExample



‘Rent-a-Chef’ provides food cooked and served by a qualified chef at parties. The company charges $120 as a booking fee, and an additional $30 per hour. Another company, ‘Greased lightning’, provides fast food served by two students at a cost of $65 per hour, with no booking fee. Under what conditions would it be cheapest to hire ‘Greased lightning’?

THINK WRITE

Define convenient pronumerals. Let C = Cost (total) in dollars and t = time in hours.

Write an equation for the cost of hiring both organisations.

Rent-a-Chef C = 120 + 30t [1]Greased lightning C = 65t [2]

Use simultaneous equations to find when the cost is the same with each group.

Put [1] = [2]120 + 30t = 65t

120 = 35t

t =

= 3.4 hours

At 3.4 hours, the cost is the same. Since Greased lightning has the higher per hour cost, after 3.4 hours, they would be more expensive.

It is cheaper to hire Greased lightning for food preparation and service of less than 3.4 hours (3 hours and 24 minutes) duration.

Notes:1. 0.4 hours = 0.4 × 60 minutes = 24 minutes.2. An alternative approach would be to use a graphics calculator, and find the point at

which the two graphs crossed.

1

2

3

12035

---------

4

Rent-a-chef

Greased lightning

21WORKEDExample

rememberThough not all modelling questions involve costs, it is helpful to remember:

Total cost = Fixed cost + Cost per unit × Number of units.

y = c + mx

remember


Linear modelling

1 The cost of hiring a floodlit tenniscourt consists of a booking fee and anhourly rate.a Use the photo to write an equation

for the total hire cost in terms ofthe hourly rate.

b Sketch a graph of the relationship.c What would be the charge for

3 hours?

2 A singing telegram service charges a$60 appearance fee, and $8 perminute sung.a Write an equation for the total cost

of a singing telegram in terms ofthe number of minutes sung.

b Sketch a graph of the relationship.c What would be the charge for a

5 minute singing telegram?

3 Colleen delivers junk mail and is paid$32 to traverse a particular route, anda further 10 cents per leaflet delivered.a Write an equation for the total

payment she receives.b Sketch a graph of the relationship expressed in a.c What would be Colleen’s pay if she delivers 1650 leaflets along the route?

4 A pay-TV salesperson receives $300 per week plus $20 for every household he signsup to have pay-TV connected. How much does the salesperson receive for a week inwhich he signs up 33 households?

5 A computer firm, SuperComputers Inc., offers a back-up plan covering the ongoing ser-vice and troubleshooting of its systems after sale. The cost of signing up for the serviceplan is $215, and there is an hourly rate of $65 for the serviceperson’s time. Purchasersnot signing up for the plan are charged a flat rate of $150 per hour for service. Wouldit be advisable to sign up for the service plan if you expected to need 3 hours of serviceassistance during the life of a computer purchased from SuperComputers Inc?

6 A telephone company, Opus, offers calls to Biddelonia for a connection fee of $14,and thereafter $1 per minute. Its rival, Belecom, offers calls for $2 per minute (noconnection fee) to the same country.a Compare the cost of a 10 minute call to Biddelonia using each company.b At what point would it be cheaper to use Opus?

7 It costs you $6 to get into a taxi (the ‘flagfall’), and $1.50 per kilometre if you use‘PinkCabs’, while NoTop taxis charge $8 flagfall, and $1.20 per kilometre.a How much would it cost with each company to travel 15 km in one of its cabs?b When would it cost the same to use both companies?

1K

Mathca

d

Simultaneouslinearequations

EXCEL

Spreadsheet

Simultaneouslinearequations

Hire ChargesBooking fee $25Hourly rate $5/hr


20


21

C h a p t e r 1 L i n e a r f u n c t i o n s 478 Two amusement parks show the following information for school-age tourists in a

promotional brochure.

After how many rides does an excursion to Fun World become the cheaper option forthe same number of rides?

9 Medirank, a health insurance company, charges $860 per year (for a single person),and requires customers to pay the first $100 of any hospital visit. HAB, on the otherhand, charges an annual fee of $560 and requires its members to pay the first $150 ofany hospital visit. Determine the number of hospital visits in a year for which the costof health services is the same whichever company insures you.

10 Nifty is a car hire firm that charges insurance of $135, and $50 per day car hire. Acompetitor, Savus, simply charges $65 per day and offers ‘free’ insurance. You areplanning a holiday, and would prefer to use Savus. Under what conditions (dayshired) could you justify this choice?

Water World$8.00 entry$2.50 per ride

$12 entry$1.50 per ride

48


Rearrangement and substitution

• Do the same to both sides and remember inverse operations including

+

and

−

,

×

and

÷

, and

2

. • ‘Make

x

the subject’ means manipulate into the form ‘

x

= . . .’.• ‘Substitute’ means to replace a pronumeral with a value.

Solving linear equations

• Do the same to both sides and remember inverse operations

+

and

−

,

×

and

÷

, and

2

. • Aim to get a single pronumeral by itself.

Gradient of a straight line

•

m

=

tan

θ

Equations of the form

y

=

mx

+

c

• The general equation for a straight line of gradient

m

and

y

intercept

c

is

y

=

mx

+

c

.• Lines with the same gradient (

m

) are parallel.

Sketching linear graphs using intercepts

• To find the

y

-intercept, let

x

=

0, and find

y

.• To find the

x

-intercept, let

y

=

0, and find

x

.• If

y

=

0 when

x

=

0, substitute another

x

-value (for example,

x

=

1) to find another point on the line.

• Join two points and/or intercepts with a straight line.

Simultaneous equations

• With a graphics calculator, express as

Y

1

=

, and

Y

2

=

and find intersection using

[CALC]

and

5:intersect

.•

y

=

ax

+

b

,

y

=

cx

+

d

consider using substitution•

ax

+

by

=

c

,

dx

+

ey

=

f

consider using elimination

Perpendicular lines

•

m

1

×

m

2

= −

1

Formulas for finding the equation of a straight line

• .

Distance between two points

•


•

Linear modelling

• Total cost

=

Fixed cost

+

Cost per unit

×

Number of units•

y

=

c

+

mx

summary

my2 y1–

x2 x1–----------------=

2nd

y mx c+= y y1–y2 y1–

x2 – x1-----------------

x x1–( )= y y1– m x x1–( )=

d x2 x1–( )2 y2 y1–( )2+=

Mx1 x2+

2----------------

y1 y2+

2----------------,

=



Multiple-choice

1 The first step in solving would be to:

2 x = −5 is a solution to the equation:

3 When c2 = a2 + b2 is rearranged to make a the subject, the equation becomes:

4 Which values, when substituted into give a value for K of 4?

5 Using the equation P = m1 v1 + m2 v2, if P = 10, m1 = 2, m2 = 6 and v1 = 4, v2 would equal:

6 The line shown has a gradient of:A −6B −3C −2D 2E

7 The gradient of the line shown below is 3. The value of a must be:A −2B −1C 5D 7E 11

A add 23 to both sides B divide both sides by 3 C divide both sides by 7D multiply both sides by 3 E multiply both sides by 7

A B C

D E

A B C

D E

A m = 2, v = 4 B m = 4, v = 2 C m = 8, v = D m = 8, v = 1 E m = 1, v = 16

A B C 1

D 2 E 3

CHAPTERreview

1A7x 23–

3------------------ 99=

1A3x 7+ 8–= 2x 17– 5–= x 25+

6--------------- 5=

2 x 3+( ) 10= 5x–9

--------- 45=

1Bc a2 b2+= a2 b2 c2+= a2 c2 b2–=

a c2 b2–= a b c+=

1BK 12---mv2=

12---

1B13--- 1

2---

1Cy

x3

6

63---

1Cy

x

(2, –5)

(6, a)


8 The gradient of the line joining (−1, 0) and (4, −10) is:

9 Which of the graphs below has a gradient of ?

10 The gradient of the line with equation y = x − 1 is:

11 The y-intercept of the line with equation y = 12x + is:

12 The gradient and y-intercept (in that order) of the line with equation 2x − 3y = 7 are:

13 Which of the following could be the graph of y = 2x + c?

A −4 B −2 C 2 D 4 E 5

A B C

D E

A −1 B C D 6 E 7

A B C 2 D 3 E 12

A 2 and −3 B 2 and 7 C − and 7 D and − E −3 and

A B C

D E

1C

1C67---

y

x

7

–6

y

x

6

–7

y

x

6

7

y

x–7

–6

y

x

–7

6

1D67---

67--- 7

6---

1D23---

23--- 3

2---

1D 23--- 2

3--- 7

3--- 7

2---

1Ey

x

–c

y

x

c

y

xc

y

x–c

y

x

c

C h a p t e r 1 L i n e a r f u n c t i o n s 5114 The equation of this line is:

A 2x − 5y = 1B 2x − y = 4C 15x + 6y = −30D 10y − x = −2

E

15 To solve the equations 2x + y = 5 and 3x − 6y = 12 graphically on a graphics calculator, you would enter the following in the Y= menu:

A and

B and

C and

D and

E and

16 Which of the following would be an effective way to solve the following equations simultaneously?y = 2x − 13 [1]y = 7x + 2 [2]

A Multiply [1] by 2 and [2] by 13 and add the newly formed equations.B Mutliply [1] by 7, and put it equal to [2].C Multiply [2] by 2 and put it equal to [1].D Multiply [1] by 2 and [2] by 7 and subtract the newly formed equations.E Put [1] equal to [2].

17 The gradient of a line perpendicular to is:

18 The equation of the line containing (1, −2) and (2, −3) could be expressed as:

19 A line with equation y − 7 = 5(x − 1) has:A gradient 5 and contains the point (7, 1)B gradient −7 and contains the point (−1, −7)C gradient 5 and contains the point (1, 7)D gradient −5 and contains the point (1, −7)E gradient −1 and contains the point (5, 7)

20 The distance between (4, 3) and (−2, 1) is equal to:

A − B C D E −

A y − 2 = x − 1 B y + 2 = 1 − x C y + 3 = x − 1

D x − 2y = −3 E 3x − 5y = 1

A B C

D E

1Ey

x

–5

2

x2--- y

5---– 1=

1FY1=2X+Y Y2=3X−6Y

Y1=5 Y2=12

Y1=5−2X Y2=12+6Y

Y1=2X+5 Y2=3X+12

Y1=−2X+5 Y2=1/2X−2

1F

1Gy−7x 5+

16-------------------=

17--- 1

7--- 16

7------ 5

16------ 7

16------

1H

1H

1I−2 4–( )2 1 3–( )2+ 4 3–( )2 −2 1–( )2+ 42 32+( ) −22 12+( )–

4 2+( )2 3 1–( )2+ −2 4–( )2 1 3–( )2–


21 The midpoint of the segment joining (11, −3) and (−5, 17) is:

22 Bote lives 5 kilometres from the nearest post office. At noon one day he begins cycling (from home) at 20 kilometres per hour in a constant direction away from the post office. t hours after he begins cycling, the distance D km Bote is from the post office is given by:

Short answer

1 Solve the equation .

2 Find the value of x where .

3 The following formula may be used to study planetary motion.

Make T the subject of the equation.

4 Calculate the gradient of each of the following lines.a b

c d

5 Find the gradient of the line joining (−7, 15) and (2, −6).

6 Find the gradient of the line shown.

A (−5 , −1 ) B (3, 7) C (6, 14) D (−2 , 8 ) E (4, 6)

A D = 5t B D = 20t C D = 5t + 20 D D = 20t + 5 E D = 20t − 5

1J 12--- 1

2--- 1

2--- 1

2---

1K

1A3 5x 4–( )

7----------------------- 6 4x 3+( )

5-----------------------=

1A 37x 8+

10---------------

4x 9–=

1BGmM

R 2-------------- m4π 2R

T 2------------------=

1Cy

x

6

–8x

y654321

–1–2–3

–5 –4 –3 –2 –1 1 2 3 4 5 60

x

y1 gridsquare = 1 unit

y

x

(–4, –10)

(–12, –3)

1C1C y

x77°

C h a p t e r 1 L i n e a r f u n c t i o n s 537 State the gradient of the line at right.

8 State the gradient and y-intercept (in that order) for each of the following.

9 Find the equation for a linear graph having gradient and y-intercept −3.

10 Sketch graphs of the following showing intercepts.

11 Solve graphically (using a graphics calculator if available): y = 3x + 10 and y = −2x − 15.

a Sketch the solution on a set of axes.b State the solution (point of intersection).

12 Solve y = −3x, y = 6x − 15 using substitution.

13 Use the method of elimination to solve 4x − 7y = 21, −2x + y = 6.

14 A piggybank contains 67 coins. If there are only one and two dollar coins in the piggybank, and there are 25 more one dollar coins than two dollar coins, how many of each type are there?

15 Find the gradient of a line perpendicular to 3x − 9y = 7.

16 Find the equation of the line containing (−4, 8) and (3, 1).

17 Find the equation of the line having gradient − that passes through (1, 4).

18 Find the equation of the line perpendicular to y = x − 5, passing through (−8, 6).

19 The distance between (2, −7) and (a, −2) is units. Find the value of a if it is positive.

20 The mid-point of the line joining (k, 2h) and (9k, 6h + 2) is (20, −11). Find k and h.

21 The washing machine repair company ‘Washed out’ charges $75 to come to your house, as well as an hourly charge of $65, calculated to the nearest half hour.

a Write an equation that may be used to calculate the cost of any service call if the time taken by the repairer is known.

b Sketch a graph of the relationship between repair cost and time taken to do a repair.

c How much would it cost to have a repair done that takes 3 hours?

a y = 3x − 7 b 5x + 3y = 30 c 2x − 4y − 8 = 0

a y = −3x + 24 b −x + 8y = 40 c 9x − 7y − 63 = 0 d y + 6x = 0

y

x5

1C

1D

1D25---

1E

1F

1F1F1F

1G1H1H

67---

1H14---

1I41

1J1K

12---


Analysis1 The following is a rough sketch of three points on a section

of sheet metal that are to be drilled by a programmed robotic drilling arm. Any deviation from a straight path, no matter how slight, means the arm must be programmed for more than one direction. The coordinates marked are correct. Will the robotic arm be able to move in one direction only to drill all three holes?

2 Points A, B and C have the coordinates (1, 6), (0, 0) and (−2, 2). Find 3 possible coordinates for a point D so that the four points form a parallelogram. (Exclude the case where all points lie in a straight line.)

3 Consider the points (−4, −2), (6, 2), (4, −1) and (0, −7).a Find the coordinates of the midpoints of each side of the quadrilateral formed by the

points.b Show that the shape formed by the midpoints is a parallelogram.c Repeat parts a and b for a different set of starting points.d What can you conjecture based on your answers?

4 The cost of a parachuting course consists of a charge of $250 which covers equipment hire and tuition, and a further expense of $55 per jump.a Express the total cost, C, as a

function of j, the number of jumps.

b How many jumps could a person doing the course afford if she was prepared to spend up to $1000?

c Why would anyone want to jump from an aircraft anyway?

y

x

6

8 14 20

12

16

A

B

C

testtest

CHAPTERyyourselfourself

testyyourselfourself

1

mathsbooks.netmathsbooks.net/Maths Quest 11 Methods/By Chapter/Ch 01 Linear F… · 2 Mathematical...

Documents

Transcript of mathsbooks.netmathsbooks.net/Maths Quest 11 Methods/By Chapter/Ch 01 Linear F… · 2 Mathematical...