SA Yr 11 geom & trig (gm)

Robert HaeseSandra HaeseMark BruceKim HarrisCarol MouleMarion Gaertner-JonesDerk KappelleDean Bennett

MATHEMATICS

fifth edition

geometry andtrigonometry

geometry andtrigonometry

Year 11

Haese & Harris Publications

SA_11 geomcyan black

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Color profile: DisabledComposite Default screen

MATHEMATICS FOR YEAR 11 (Fifth Edition)

GEOMETRY AND TRIGONOMETRY

This book is copyright

Copying for educational purposes

Disclaimer

Robert Haese B.Sc.Sandra Haese B.Sc.Mark Bruce B.Ed.Kim Harris B.Sc., Dip.Ed.Derk Kappelle Adv.Dip.T M.Ed.Carol Moule B.Sc, M.Ed, Dip.Ed, Dip.T, Grad.Dip.Ed.AdminMarion Gaertner-Jones B.A.(Hons.), Dip.Ed.M.A.C.E., M.A.C.E.ADean Bennett B.Sc.(Hons.), Dip.Ed

Haese & Harris Publications3 Frank Collopy Court, Adelaide Airport SA 5950Telephone: (08) 8355 9444, Fax: (08) 8355 9471email:

National Library of Australia Card Number & ISBN 1 876543 31 0

© Haese & Harris Publications

Published by Raksar Nominees Pty Ltd, 3 Frank Collopy Court, Adelaide Airport SA 5950

First Edition 1980 1981, 1983Second Edition 1984 1985, 1986, 1987, 1988, 1989, 1990, 1991Third Edition 1992 1993, 1994, 1995, 1996, 1997Fourth Edition 1998Fifth Edition 2001

Cartoon artwork by John Martin. Artwork by Piotr Poturaj, Joanna Poturaj and David PurtonCover design by Piotr Poturaj. Cover photography Piotr PoturajComputer software by David Purton and Mark Foreman

Typeset in Australia by Susan Haese (Raksar Nominees). Typeset in Times Roman 10 /11

. Except as permitted by the Copyright Act (any fair dealing for thepurposes of private study, research, criticism or review), no part of this publication may bereproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic,mechanical, photocopying, recording or otherwise, without the prior permission of the publisher.Enquires to be made to Haese & Harris Publications.

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While every attempt has been made to trace and acknowledge copyright, the author and publishersapologise for any accidental infringement where copyright has proved untraceable. They wouldbe pleased to come to a suitable agreement with the rightful owner.

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FOREWORD

Mathematics for Year 11 Geometry and Trigonometry 5 edition has been written to embrace

the concepts outlined in the Stage 1 Mathematics Curriculum Statement. It is not our

intention to define a course.

This package is the first step in a new approach to mathematics education. You are provided

with a text book and a CD-Rom which displays the contents of the book plus many exciting

new interactive features which will assist teachers and students.

The book is language rich and technology rich. Whilst some of the exercises are simply

designed to build skills, every effort has been made to contextualise problems, so that

students can see everyday uses and practical applications of the mathematics they are

studying.

The book contains many problems, from the basic to the advanced, to cater for a wide range

of student abilities and interests. Much emphasis has been placed on the gradual development

of concepts with appropriate worked examples. However, we have also provided extension

material for those who wish to go beyond Stage 1 and look towards further studies or

applications of mathematics for their career choices. It is not our intention that each chapter

be worked through in full. Time constraints will not allow for this. Consequently, teachers

must select exercises carefully, according to the abilities and prior knowledge of their

students, in order to make the most efficient use of time and give as thorough coverage of

work as possible.

The extensive use of graphics calculators and computer packages throughout the book

enables students to realise the importance, application and appropriate use of technology. No

single aspect of technology has been favoured. It is as important that students work with a

pen and paper as it is that they use their calculator or graphics calculator, or use a spreadsheet

or graphing package on computer.

The interactive features of the CD-Rom allow immediate access to our own specially

designed geometry packages, graphing packages and more. Teachers are provided with a

quick and easy method of demonstrating concepts, or students can discover for themselves,

and revisit when necessary.

Teachers should note that instructions appropriate to each graphics calculator problem are

available on the CD-Rom and can be printed for students. These instructions are written for

Sharp, Texas Instruments, Casio and Hewlett-Packard calculators.

In this changing world of mathematics education, we believe that the contextual approach

shown in this book, with the associated use of technology, will enhance the students’

understanding, knowledge and appreciation of mathematics.

RCH SHH

KPH MFB

DWK CEM

MG-J DRB

th


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Coordinategeometry

Coordinategeometry

2Chapter

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

locate points on a plane usingcoordinatesdescribe the points on a straight lineusing an equationmeasure and interpret the slope of alinefind the distance between pointsfind the midpoint of the line joiningtwo pointsfind the equations of parallel andperpendicular linesdivide a line segment in a given ratiofind where two lines meet

find how far a point is from a linefind the equation of a circle from itscentre and radius and vice versadetermine the intersection of twocirclesdetermine the intersection of a lineand a circledefine and discuss the behaviour ofrectangular hyperbolaefind the areas of polygons given theircoordinatesdescribe the region defined by alinear inequality or a set of them

CD Link

By the end of this chapter you should be able to

Outcomes · · · · · · · · · · · · · · · · · · · · · · · · · ·

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·


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OPENING PROBLEM

0

HISTORICAL NOTE VORONOI DIAGRAMS

68 COORDINATE GEOMETRY (Chapter 2) (T11)

INTRODUCTION

In previous courses we have already been introduced to many of the concepts and ideas of

this chapter. We are familiar with:

² the use of Cartesian coordinates to describe the position of points

² the method of finding distances between two points

² the slope of the line segment between two points

² the midpoint of the line segment joining two points

² the equation of a line

² how to find where two lines meet by using simultaneous solution

² the properties of parallel and perpendicular lines.

We will revisit these concepts and extend coordinate geometry into other useful areas.

P Veter Dirichlet oronoi( ) ( )

a

1850 1908and later considered the followingproblem:

“For collection of points in the two dimensional plane, how can we partitionthe plane so that the partition lines are equidistant to two or more points?”

Three small country towns de-

cide to enter a combined team

in the local football association.

However, they cannot agree which

town oval to use as each wants to use its own

venue. This is a problem as there could be

long distances for some players to travel to

practice.

At a meeting they decide to place a new oval

in such a position that it is equidistant from

all three towns. The coordinates of each town

are known from the local council map of the

area, and are marked on the grid above.

Where exactly should the new venue be located? For you to think about:

What vital information given above is necessary to solve the problem?

Can you see how to use coordinate geometry (midpoints, slopes, equations of lines,

etc) to solve the problem?

Can the problem be solved without using coordinate geometry?

Jason Jolly says the answer is obvious. He says “Just average the -coordinates and

then the -coordinates to give the correct answer”. Is he correct?

Can you see any problems with a mathematical solution to this problem?

�

�

�

�

�

xy

(111"9' 152"5) (193"1' 136"9)

(127"2' 87"7)

N


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COORDINATE GEOMETRY (Chapter 2) (T11) 69

Notice that the plane has been partitioned into three

regions and that each partition line is equidistant

to any two of the points.

Points A, B and C are called sites.

This is a Voronoi edge, a boundary line between

Voronoi regions. O is called a Voronoi vertex.

THE NUMBER PLANE

The position or location of any point in the number plane can be specified in terms of an

ordered pair of numbers (x, y), where:

x is the horizontal step from a fixed point O, and y is the vertical step from O.

Once an origin O, has been given, two perpendicular axes are drawn.

The x-axis is horizontal and the y-axis is vertical.

The number plane is also known as either:

² the 2-dimensional plane, or

² the Cartesian plane, (named after Rene Descartes).

Note:

(a, b) is called an ordered pair, where a and b are

often referred to as the coordinates of the point.

a is called the x-coordinate and

b is called the y-coordinate.

As oronoi published solutions to the problem, the diagrams were named after him.

oronoi diagrams have applications in zoology archaeology communications, crystallog-

raphy and motion planning.

For example, oronoi diagrams are useful in the solution to the problem of where to site

mobile telephone towers. Given say three towers, how can they be sited so that any par-

ticular mobile call is carried by the closest tower?

arious methods (algorithms) are used to construct oronoi diagrams. However Steve

Fortune and his method greatly reduced the time to draw these dia-

grams. Below is oronoi diagram for three distinct, non collinear points A, and C.

V

V , ,

V

V V ,

( )

a V B

1985 plane-sweep

V a ,oronoi diagrams are feature of which is worth your investigation.P eeanuts SoftwarVoronoi diagrams are found at http://dimacs.rutgers.edu/drei/96/classroom/

P( , )a b

b

a

y-axis

x-axis

A

B

C

ASSUMED KNOWLEDGE (REVIEW)A

This region is called the for siteC. Every point in this region is closer to thanany other site.

V roronoi egion

C

DEMO


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THE DISTANCE FORMULA

If A(x1, y1) and B(x2, y2) are two points in a plane, then the distance d,

between these points is given by AB =p(x2¡x1)2 + (y2 ¡y1)2

or AB =q(x-step)2 + (y-step)2:

THE MIDPOINT FORMULA

If M is halfway between points A and B then

M is the midpoint of AB.

If A(x1, y1) and B(x2, y2) are two points then the midpoint M of

AB has coordinates

µx1 + x2

2,y1 + y2

2

¶:

Example 2

Find the coordinates of the midpoint of AB for A(¡1, 3) and B(4, 7).

x-coordinate of midpoint

=¡1 + 4

2

= 32

= 112

y-coordinate of midpoint

=3 + 7

2

= 5

) the midpoint of AB is (112 , 5)

SLOPE (OR GRADIENT) OF A LINE

When looking at line segments drawn on a set of axes, it is clear that different line segments

are inclined to the horizontal at different angles, i.e., some appear to be steeper than others.

Find the distance between A(¡2, 1) and B(3, 4).

A(¡2, 1) B(3, 4)

" " " "x1 y1 x2 y2

AB =p

(3 ¡ ¡2)2 + (4 ¡ 1)2

=p

52 + 32

=p

25 + 9

=p

34 units

Example 1

A

B

M

This distanceformula saves ushaving to graphthe points eachtime we want tofind a distance.


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The slope or gradient of a line is a measure of its steepness.

If A is (x1, y1) and B is (x2, y2) then the slope of AB isy2¡ y1x2¡x1

:

Note: ² horizontal lines have a slope of 0 (zero)

² vertical lines have an undefined slope

² forward sloping lines have positive slopes

² backward sloping lines have negative slopes

² parallel lines have equal slopes

² the slopes of perpendicular lines are negative reciprocals

i.e., if the slopes are m1 and m2 then m2 =¡1

m1or m1m2 = ¡1:

EQUATION OF A LINE

The equation of a line is an equation which connects the x and y values for every

point on the line.

Equations of lines have various forms:

² All vertical lines have equations of the form x = a, (a is a constant).

² All horizontal lines have equations of the form y = c, (c is a constant).

² If a straight line has slope m and passes through (a, b) then it has equation

y ¡ bx¡ a = m

which can be rearranged into y = mx+ c fslope-intercept formgor Ax+By = C fgeneral formg

Find the slope of the line through (3, ¡2) and (6, 4).

(3, ¡2) (6, 4)

" " " "x1 y1 x2 y2

slope =y2 ¡ y1x2 ¡ x1

=4 ¡ ¡2

6 ¡ 3

= 63

= 2

Example 3


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INTERCEPTS

Axis intercepts are the x- and y-values where a graph cuts

the coordinate axes.

The x-intercept is found by letting y = 0.

The y-intercept is found by letting x = 0.

Find, in slope-intercept form, the equation of the line

through (¡1, 3) with a slope of 5.

The equation of the line isy ¡ 3

x¡ ¡1= 5

i.e.,y ¡ 3

x+ 1= 5

) y ¡ 3 = 5(x+ 1)

) y ¡ 3 = 5x+ 5

) y = 5x+ 8

Example 4

Graph the line with equation 2x¡ 3y = 12 using axis intercepts.

For 2x¡ 3y = 12,

when x = 0, ¡3y = 12

) y = ¡4

when y = 0, 2x = 12

) x = 6

Example 6

Find, in general form, the equation of the line through , and , .(1 5) (5 2)¡ ¡

So, the equation of the line isy ¡ ¡2

x¡ 5=

3

4The slope

¡ ¡¡ 52

5 ¡ 1

=

=

3

4 i.e.,y + 2

x¡ 5=

3

4

) 4(y + 2) = 3(x¡ 5)

) 4y + 8 = 3x¡ 15

) 3x¡ 4y = 23

Example 5

To find the equationof a line we need toknow its slope and a

point on it.

x

y

y-intercept

x-intercept


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A point lies on a line if its coordinates satisfy the equation of the line.

There are three possible situations which may occur. These are:

DOES A POINT LIE ON A LINE?

Does (3, ¡2) lie on the line with equation 5x¡ 2y = 20?

Substituting (3, ¡2) into 5x¡ 2y = 20 gives

5(3) ¡ 2(¡2) = 20i.e., 19 = 20 which is false

) (3, ¡2) does not lie on the line.

Example 7

WHERE GRAPHS MEET

Case 1: Case 2: Case 3:

The lines meet in a

single

.

point of

intersection

The lines are

and . So,

there is no point of

intersection.

parallel

never meet

The lines are

(the same line) and so

there are infinitely many

points of intersection.

coincident

Use graphical methods to find where the lines x+ y = 6 and 2x¡ y = 6 meet.

For x+ y = 6

when x = 0, y = 6when y = 0, x = 6 x 0 6

y 6 0

For 2x¡ y = 6

when x = 0, ¡y = 6, ) y = ¡6when y = 0, 2x = 6, ) x = 3

x 0 3

y ¡6 0

The graphs meet at (4, 2).

Check: 4 + 2 = 6 X and 2 £ 4 ¡ 2 = 6 X

��

��

�

��

�

�

�

� ��

y

x

2!-@=6

!+@=6

(4, 2)

Example 8


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INVESTIGATION 1 FINDING WHERE LINES MEET USING TECHNOLOGY

Step 1: We rearrange each equation into the form y = mx+ c, i.e.,

4x+ 3y = 10

) 3y = ¡4x+ 10

) y = ¡43x+ 10

3

and x¡ 2y = ¡3

) ¡2y = ¡x¡ 3

) y =x

2+

3

2

Graphing packages graphics calculatorsand can be used to plot

straight line graphs and hence find the point of intersection of the

straight lines. This can be useful if the solutions are not integer values,

although an algebraic method can also be used. However, most graphing

packages and graphics calculators require the equation to be entered in

the form .y mx c� � � �= +

Consequently, if an equation is given in , it must be rearranged into

.

For example, if we wish to use technology to find the point of intersection of

and :

general form

slope-intercept form

4 +3 =10

2 = 3

x y

x y

� � � �� ¡ ¡

Step 2: If you are using the graphing package, click on the icon to

open the package and enter the two equations.

If you are using a graphics calculator, enter the functions

Y1 = ¡4X=3 + 10=3 and Y2 = X=2 + 3=2.

Step 3:

Step 4:

1 Use technology to find the point of intersection of:

a y = x+ 45x¡ 3y = 0

b x+ 2y = 8y = 7 ¡ 2x

c x¡ y = 52x+ 3y = 4

d 2x+ y = 73x¡ 2y = 1

e y = 3x¡ 13x¡ y = 6

f y = ¡2x

3+ 2

2x+ 3y = 6

2 Comment on the use of technology to find the point(s) of intersection in 1 e and 1 f.

What to do:

GRAPHINGPACKAGE

TI

C

HP

S

Draw the of the functions on the same set of axes. (You

may have to change the viewing if using a graphics

calculator.)

graphs

window

Use the built in functions to calculate

the point of .

Thus, the point of intersection is ( , ).

intersection

1 2


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1 Use the distance formula to find the distance between the following pairs of points:

a A(1, 3) and B(4, 5) b O(0, 0) and C(3, ¡5)

c P(5, 2) and Q(1, 4) d S(0, ¡3) and T(¡1, 0)

e A(¡3, ¡2) and B(1, 4) f M(4, 3) and N(¡1, 2)

2 Find the midpoint of AB for:

a A(3, 6) and B(1, 0) b A(5, 2) and B(¡1, ¡4)

c A(7, 0) and B(0, 3) d A(5, ¡2) and B(¡1, ¡3)

3 By finding an appropriate y-step and x-step, determine the slope of each of the following

lines:

a b c

d e f

4 Find the slope of the line passing through:

a (2, 3) and (4, 7) b (3, 2) and (5, 8)

c (¡1, 2) and (¡1, 5) d (4, ¡3) and (¡1, ¡3)

e (0, 0) and (¡1, 4) f (3, ¡1) and (¡1, ¡2)

5 Classify the following lines pairs as parallel, perpendicular or neither.

Give a reason for your answer.

a b c

d e f

EXERCISE 2A (review)


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6 State the slope of the line which is perpendicular to the line with slope:

a 34 b 11

3 c 4 d ¡13 e ¡5 f 0

7 Find the equation of the line in slope-intercept form, through:

a (4, 1) with slope 2 b (1, 2) with slope ¡2 c (5, 0) with slope 3

d (¡1, 7) with slope ¡3 e (1, 5) with slope ¡4 f (2, 7) with slope 1

8 Find the equation of the line, in general form, through:

a (2, 1) with slope 32 b (1, 4) with slope ¡3

2 c (4, 0) with slope 13

d (0, 6) with slope ¡4 e (¡1, ¡3) with slope 3 f (4, ¡2) with slope ¡49

9 Find the equations of the lines through:

a (0, 1) and (3, 2) b (1, 4) and (0, ¡1)

c (2, ¡1) and (¡1, ¡4) d (0, ¡2) and (5, 2)

e (3, 2) and (¡1, 0) f (¡1, ¡1) and (2, ¡3)

10 Find the equations of the lines through:

a (3, ¡2) and (5, ¡2) b (6, 7) and (6, ¡11) c (¡3, 1) and (¡3, ¡3)

11 Copy and complete:

Equation of line Slope x-intercept y-intercept

a 2x¡ 3y = 6

b 4x+ 5y = 20

c y = ¡2x+ 5

d x = 8

e y = 5

f x+ y = 11

g 4x+ y = 8

h x¡ 3y = 12

12 a Does (3, 4) lie on the line with equation 3x¡ 2y = 1?

b Does (¡2, 5) lie on the line with equation 5x+ 3y = ¡5?

c Does (6, ¡12 ) lie on the line 3x¡ 8y = 22?

13 Use graphical methods to find where the following lines meet:

a x+ 2y = 8 b y = ¡3x¡ 3 c 3x+ y = ¡3

y = 2x¡ 6 3x¡ 2y = ¡12 2x¡ 3y = ¡24

d 2x¡ 3y = 8 e x+ 3y = 10 f 5x+ 3y = 10

3x+ 2y = 12 2x+ 6y = 11 10x+ 6y = 20

If a line has equationthen the

slope of the line is ‘m’.

y mx c��


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DISCUSSION


² Street directories use grids such as the

one illustrated. The directory may

say that a street or place is located at

B4. Why is this reference system used

rather than number coordinates?

² Why are coordinates also called ‘ordered pairs’ and

what is the significance of ‘order’ in ordered pairs?

² Why is slope defined asy-step

x-steprather than

x-step

y-step?

A

1

2

3

4

B C D E F

COORDINATE SYSTEMSBIf we consider a point with an -coordinate

of , the point could lie anywhere on the

vertical line with equation .

If we also know that the -coordinate is ,

this fixes the point at ( , ).

Could we use the distance from the origin

to locate the position of a point?

If we know that a point is units from the

origin O, then the point must lie on a cir-

cle, centre O and radius units.

What other information would we need to

know to fix a point?

Knowing that the -coordinate is is not

sufficient as the point could be at P or P .

So, a coordinate system ( , ) is not good

enough to describe one and only one point.

Consider a point of reference O and a hori-

zontal axis in the positive direction.

O is called the pole and the horizontal line

is called the polar axis.

( , ) are the polar coordinates of

point P where is the distance of P

from O ( ) and is the angle OP

makes in an anticlockwise direction

with the polar axis.

So, A( , ) is located on a circle, centre

O, with radius units and the angle from

the polar axis is .

x

x

y

x

r x

r µr

r µ

3= 3

23 2

5

5

3

0

4 60460

1 2

POLAR COORDINATES

>

o

o

A

4

60°O

P( , )r �

r

�O

polepolar axis

P1

(3, 2)

P2

x

x

y

y

x��

x��

2


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DISCUSSION


² Where could polar coordinates be useful?

² Can every point of the plane be described using polar coordinates?

²

² In the Cartesian coordinate system distances are easy to find using the

distance formula, d =p

(x2 ¡ x1)2 + (y2 ¡ y1)2:

Consider: angle AOB = µ1 ¡ µ2

How can distances be found in a polar coordinate system?

1 On the same set of axes, plot the points with polar coordinates:

a (4, 90o) b (3, 180o) c (2, 45o) d (5, 120o) e (3, 0o)

2 Determine the equivalent polar coordinates for a point with Cartesian coordinates:

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

3 Determine the equivalent Cartesian coordinates for a point with polar coordinates:

a (3, 0o) b (7, 90o) c (4, 180o) d (2, 70o) e (3:2, 123o)

4 Find the distance between the points with polar coordinates:

a (3, 20o) and (4, 70o) b (5, 72o) and (2, 138o)

The equation of a line can be determined if we know, or have sufficient information to

determine, the slope and a point on the line.

The line segment connecting (0, ¡1) to (¡2, 9) has

equation y = ¡5x ¡ 1. So to describe the line

segment exactly we say that y = ¡5x¡ 1 where

¡2 6 x 6 0.

We say that ¡2 6 x 6 0 is the restricted domain

in this example.

A( , )r1 1�

B( , )r2 2�

polar axisO�1�1 �2

r1

r2

EXERCISE 2B

EQUATIONS OF LINESC

RESTRICTED DOMAINS

��

��

x

y

Could the direction of line segments be easily found in the polar coor-dinate system?


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1 Find the equation of the:

a horizontal line through (3, ¡4)

b vertical line with x-intercept 5

c vertical line through (¡1, ¡3)

d horizontal line with y-intercept 2

e x-axis

f y-axis

A straight road is to pass through points A(5, 3) and B(1, 8).

a Find where this road meets the road given by:

i x = 3 ii y = 4

b If we wish to refer to points on road AB, but

between A and B, how can we indicate this?

c Does C(23, ¡20) lie on the road?

First we must find the equation of the line representing the road.

Its slope is m =3 ¡ 8

5 ¡ 1= ¡5

4

So, its equation isy ¡ 3

x¡ 5= ¡5

4

i.e., 4(y ¡ 3) = ¡5(x¡ 5)

i.e., 4y ¡ 12 = ¡5x+ 25

i.e., 5x+ 4y = 37

a i when x = 3, 5(3) + 4y = 37) 15 + 4y = 37

) 4y = 22) y = 512

i.e., meets at (3, 512 )

ii when y = 4, 5x+ 4(4) = 37) 5x+ 16 = 37

) 5x = 21) x = 4:2

) meets at (4:2, 4)

b We state the possible x-value restriction i.e., 1 6 x 6 5.

c If C(23, ¡20) lies on the line, its coordinates must satisfy the line’s equation.

Now LHS = 5(23) + 4(¡20)= 115 ¡ 80= 356= 37

) C does not lie on the road.

Example 9

EXERCISE 2C.1

A(5' 3)

B(1' 8)

x � 3

y � 4


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2 Find the equation of the line:

a through A(¡1, 4) and with slope 34

b through P(2, ¡5) and Q(7, 0)

c which is parallel to the line with equation y = 3x¡ 2, but passes through (0, 0)

d which is parallel to the line with equation 2x+ 3y = 8, but passes through

(¡1, 7)

e which is perpendicular to the line with equation y = ¡2x+5 and passes through

(3, ¡1)

f which is perpendicular to the line with equation 3x¡y = 11and passes through (¡2, 5).

3 A is the town hall on Scott Street and D

is a Post Office on Keach Avenue. Diag-

onal Road intersects Scott Street at B and

Keach Avenue at C.

a Find the equation of Keach Avenue.

b Find the equation of Peacock Street.

c Find the equation of Diagonal Road

(Be careful!).

d Plunkit Street lies on map reference line x = 8.

Where does Plunkit Street intersect Keach Avenue?

Find the equation of the tangent to the circle with centre (2, 3) at the point (¡1, 5).

slope of CP is3 ¡ 5

2 ¡ (¡1)=

¡2

3= ¡2

3

) the slope of the tangent at P is 32 ,

and since the tangent is through (¡1, 5)

the equation is

y ¡ 5

x¡ ¡1=

3

2

) 2(y ¡ 5) = 3(x+ 1)

) 2y ¡ 10 = 3x+ 3

i.e., 3x¡ 2y = ¡13

4 Find the equation of the tangent to the circle with centre:

a (0, 2) at (¡1, 5) b (0, 0) at (

(

3

5

,

,

¡¡2

2

)

)c d( (3 2, ,¡ ¡1 2) )at at(¡1, 1)

The tangent isperpendicular tothe radius at thepoint of contact.

A(3,17)

B(7,20)

D(13,12)C(5,11)E

Dia

gonal

Rd

Scott

StP

eacock

St

Keach Av

C(2, 3)�

P( 1, 5)�

Example 10


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C:\...\SA11G_02\080SG102.CDRMon Nov 19 13:56:42 2001


Find the equation of the line:

a with slope 34 , passing through (5, ¡2)

b with slope ¡34 , passing through (1, 7)

a The equation is 3x¡ 4y = 3( )5 ¡ 4(¡2)i.e., 3x¡ 4y = 23

b The equation is 3x+ 4y = 3( )1 + 4( )7i.e., 3x+ 4y = 31

Example 11


If a line has slope 34 , it must have form y = 3

4x+ c

i.e., 4y = 3x+ 4c or 3x¡ 4y = a constant.

If a line has slope ¡34 , using the same working we would obtain 3x+ 4y = a constant.

This suggests that for slopeA

Bthe form of the line is Ax¡By = ::::::

for slope ¡AB

the form of the line is Ax+By = ::::::

The constant term on the RHS is obtained by substituting a point which lies on the line into

this form.


a through (4, 1) with slope 12 b through (¡2, 5) with slope 2

3

c through (5, 0) with slope 34 d through (3, ¡2) with slope 3

e through (1, 4) with slope ¡13 f through (2, ¡3) with slope ¡3

4

g through (3, ¡2) with slope ¡2 h through (0, 4) with slope ¡3

2 We can use the reverse process to question 1 to write down the slope of a line given in

general form. Find the slope of the line with equation:

a 2x+ 3y = 8 b 3x¡ 7y = 11 c 6x¡ 11y = 4

d 5x+ 6y = ¡1 e 3x+ 6y = ¡1 f 15x¡ 5y = 17

3 Explain why:

a a line parallel to 3x+ 5y = :::::: has form 3x+ 5y = ::::::

b a line perpendicular to 3x+ 5y = :::::: has form 5x¡ 3y = ::::::

4 Find the equation of the line which is:

a parallel to the line 3x+ 4y = 6 and passes through (2, 1)

Using this method youwill find with practice

that you can writedown the equation.

FINDING THE EQUATION OF A LINE QUICKLY

EXERCISE 2C.2


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b perpendicular to the line 5x+ 2y = 10 and passes through (¡1, ¡1)

c perpendicular to the line x¡ 3y + 6 = 0 and passes through (¡4, 0)

d parallel to the line x¡ 3y = 11 and passes through (0, 0).

5 2x¡ 3y = 6 and 6x+ ky = 4 are two straight lines.

a Write down the slope of each line. b Find k if the lines are parallel.

c Find k if the lines are perpendicular.

A very common question is “What is the distance from point A to point B?” When travelling,

time required and costs are directly affected by distance. Likewise, the costs of supplying

services such as power lines, water mains, gas and sewerage pipes etc. depend on distance.

In the Cartesian coordinate system,

the distance between (x1, y1) and (x2, y2) is given by d =p(x2¡x1)2+ (y2¡y1)2:

The distance formula is a useful tool in coordinate geometry problem solving.

1 Show using distances only that:

a the triangle with vertices A(3, 8), B(¡11, 3) and C(¡8, ¡2) is isosceles

b the triangle with vertices P(¡1, 0), Q(0,p

3) and R(1, 0) is equilateral

DISTANCED

EXERCISE 2D

We let a general point on the line x = 11 have coordinates (11, a) say.

Now BS = AS

)p

(11 ¡ 1)2 + (a¡ 6)2 =p

(11 ¡ 5)2 + (a¡ 2)2

) 102 + (a¡ 6)2 = 62 + (a¡ 2)2 fsquaring both sidesg) 100 + a2 ¡ 12a+ 36 = 36 + a2 ¡ 4a+ 4

) ¡12a+ 4a = 4 ¡ 100

) ¡8a = ¡96

) a = 12

So, the railway siding should be located at (11, 12).

Example 12

x�11

B(1, 6)

A(5, 2)

railwayline

SMining towns are situated at B( andA( ). Where should railway siding Sbe located so that ore trucks from either

or would travel equal distances to arailway line with equation ?

1 65 2

, ), a

A Bx = 11


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c the triangle with vertices K(6, ¡5), L(¡2, ¡3) and M(¡1, 1) is right angled

d the points A(1, ¡1), B(3, 5) and C(¡4, ¡16) are collinear

e the triangle with vertices A(7, 5), B(2, 3) and C(6, ¡7) is right angled.

2 A(5, 5) and B(7, 10) are houses and y = 8 is a gas

pipeline. Where should the one outlet from the

pipeline be placed so that it is the same distance

from both houses so they pay equal service costs?

3 CD is a water pipeline. A and B are two towns.

A pumping station is to be located on the pipeline

to pump water to A and B. Each town is to pay

for their own service pipes and insist on equality

of costs.

a Where should C be located for equality of

costs to occur?

b What is the total length of service pipe re-

quired?

c If the towns agree to pay equal amounts, would it be cheaper to install the service

pipeline from D to B to A?

A tunnel through the mountains connects town

Q(2, 4) to the port at P.

P is on grid reference x = 6 and the distance

between the town and the port is 5 km.

Assuming the diagram is reasonably accurate,

what is the horizontal grid reference of the

port?

Let P be (6, a) say.

Now PQ = 5

)p

(6 ¡ 2)2 + (a¡ 4)2 = 5

)p

16 + (a¡ 4)2 = 5

) 16 + (a¡ 4)2 = 25

) (a¡ 4)2 = 9

) a¡ 4 = §3

) a = 4 § 3 = 7 or 1

But P is further North than Q ) a > 4

So P is at (6, 7) and the horizontal grid reference is y = 7.

y��

B��

A��

Example 13

Scale: each grid unit is 1 km.

Q(2, 4)

P

x��x��

C Dy��

A(2, 3)B(5, 4)

Scale: 1 unit 1 km�


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4 Find a if:

a the distance from P(3, a) to Q(¡3, 6) is 10 units

b the distance from A(a, 2) to B(¡1, 5) isp

34 units

c Q(a, 2a) isp

5 units from R(¡1, 1).

5

Scale: 1 unit ´ 1 km.

6 Find the coordinates of any point:

a on the x-axis which is equidistant from A(¡3, 2) and B(5, ¡1)

b on the y-axis which is equidistant from C(5, 0) and D(¡1, ¡3)

c on the line with equation y = 3x and 5 units from (2, 1).

(Hint: Let (a, 3a) lie on the line y = 3x:)

J(4, 1)

y � 8 Clifton

Highway

Find any points on the x-axis which are equidistant from A(1, ¡2)

and B(4, 4). Illustrate.

Any point on the x-axis can be represented by P(a, 0).

So, given that AP = BP,p(a¡ 1)2 + (0 ¡ ¡2)2 =

p(a¡ 4)2 + (0 ¡ 4)2

)pa2 ¡ 2a+ 1 + 4 =

pa2 ¡ 8a+ 16 + 16

) a2 ¡ 2a+ 5 = a2 ¡ 8a+ 32 f gsquaring both sides

) 6a = 27

) a = 276 = 9

2

) (92 , 0) is equidistant from A and B.

y

x�4

4

4

�4

B(4, 4)

A(1, 2)�

P( , 0)a

Example 14

Jason’ girlfriend lives in house on CliftonHighway which has equation The dis-tance ‘as the crow flies’ from Jason’ houseto his girlfriend’ house is km. If Jasonlives at ), what are the coordinates of hisgirlfriend’ house?

s a.

ss

( ,s

y 8

11 73:4 1

=


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7 For each of the following:

i draw a sketch of the path mapped out by P(x, y)

ii find the equation connecting x and y.

a P(x, y) is always 3 units from the point A(¡1, 3)

b P(x, y) is equidistant from R(2, 0) and S(¡2, 0)

c the distance of P(x, y) to (3, 0) is twice the distance of P from (0, 3).

8 Triangle ABC sits on the x-axis so that vertices

A and B are equidistant from O.

a Find the length of AC.

b Find the length of BC.

c If AC = BC, deduce that ab = 0.

d Copy and complete the following

statement based on the result of c.

“The perpendicular bisector of the base of

an isosceles triangle ....”

Recall that the midpoint of a line segment connecting (x1, y1) to (x2, y2) isµx1 + x2

2,y1+ y2

2

¶average of average of

x-coordinates y-coordinates

If P(x, y) is equidistant from A(¡1, 7) and B(2, ¡3):

a draw a sketch of possible positions of P

b find the equation connecting x and y:

a

b As AP = BPp(x¡ ¡1)2 + (y ¡ 7)2 =

p(x¡ 2)2 + (y ¡ ¡3)2

) (x+ 1)2 + (y ¡ 7)2 = (x¡ 2)2 + (y + 3)2

) x2 + 2x+ 1 + y2 ¡ 14y + 49 = x2 ¡ 4x+ 4 + y2 + 6y + 9i.e., 6x¡ 20y = ¡37

A( 1, 7)�

B(2, 3)�

P could be anywhere on

this line

Example 15

x

y

B( , 0)a

C( , )b c

A( , 0)�a

MIDPOINTS AND

PERPENDICULAR BISECTORS

E


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Note: Diagrams are often drawn to display facts or to help us solve a problem. Often no

attempt is made to place points in correct positions relative to each other.

1 Given find the coordinates of A, B and C.

2 Find the coordinates of C, D and E.

3 Recall that “the diagonals of a parallelogram bisect each other.”

a Using midpoints check that ABCD is a parallelogram for A(¡1, ¡2), B(0, 1),

C(¡3, 2) and D(¡4, ¡1).

b Check that ABCD is a parallelogram by finding slopes of opposite sides.

4 Triangle ABC has vertices A(3, 6), B(¡1, ¡2) and C(7, 4).

a Use slopes to prove that angle BAC is a right angle.

b Find the midpoint of BC.

c Find the equation of the tangent to the circle through A, B and C at point A.

5 Use midpoints to find the fourth vertex of the given parallelograms:

a b c

EXERCISE 2E.1

A( 1, 4)�

B(1, 1)

C

D

E

P(3, 2)�Q(11, 6)�

A B C

A(3, 0) B(2, 1)�

D C(8, 2)�

P( 1, 4)�

S(4, 0)

Q( 2, 5)�

R

W( 1, 5)� X

Y(3, 2)�Z(0, 4)

Use midpoints to find the fourth

vertex of the given parallelogram:

Since ABCD is a parallelogram, the diagonals bisect each other.

) the midpoint of DB is the same as the midpoint of AC, and if D is (a, b),

a+ 2

2=

¡1 + ¡2

2and

b+ 4

2=

3 + ¡1

2

) a+ 2 = ¡3 and b+ 4 = 2

) a = ¡5 and b = ¡2

) D is (¡5, ¡2)

A( 1, 3)� B(2, 4)

D C( 2, 1)� �

Example 16


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6 OABC is a parallelogram. You may assume

that the opposite sides of the parallelogram

are equal in length.

a Find the coordinates of B.

b Find the midpoints of AC and OB.

c

The perpendicular bisector of two points A and B divides the plane into two regions. On one

side of the line points are closer to B than to A, and vice versa on the other side.

We observe that the midpoint of line seg-

ment AB must lie on the perpendicular bi-

sector of AB.

Points on the perpendicular bisector of AB are equidistant to A and B.

Note: We could have let P on the perpendicular bisector be (x, y) and as AP = PB thenp(x¡ ¡1)2 + (y ¡ 2)2 =

p(x¡ 3)2 + (y ¡ 4)2

) (x+ 1)2 + (y ¡ 2)2 = (x¡ 3)2 + (y ¡ 4)2

) x2 + 2x+ 1 + y2 ¡ 4y + 4 = x2 ¡ 6x+ 9 + y2 ¡ 8y + 16

) 4y = ¡8x+ 20or y = ¡2x+ 5

x

y C( , )b c B

A( , 0)a

PERPENDICULAR BISECTORS

A

B

M

the perpendicular

bisector of AB

Find the equation of the perpendicular bisector of AB for A(¡1, 2) and B(3, 4).

M is

µ¡1 + 3

2,

2 + 4

2

¶i.e., M(1, 3)

slope of AB is4 ¡ 2

3 ¡ ¡1=

2

4= 1

2

) slope of perpendicular is ¡21

) equation of perpendicular bisector isy ¡ 3

x¡ 1= ¡2 f gusing M(1, 3)

) y ¡ 3 = ¡2(x¡ 1)

) y ¡ 3 = ¡2x+ 2

) y = ¡2x+ 5

A(-1'\\2)

B(3'\\4)

Mperpendicularbisector AB

Example 17

What has this confirmed about paral-

lelograms from your results in ?b


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1 Find the equation of the perpendicular bisector of AB for:

a A(

A(

3

4

,

,

¡¡

3

2

)

)

and

and

B(1, ¡1) b A(1, 3) and B(¡3, 5)

c dA( B(3 4, ,1 4) )and B(¡3, 6)

2 Two Post Offices are located at P(3, 8) and Q(7, 2)

on a Council map. What is the equation of the line

which should form the boundary between the two

regions being serviced by the Post Offices?

3 The diagram alongside shows the location of three

Post Offices and the Voronoi diagram of regions of

closest proximity. The Voronoi edges are the per-

pendicular bisectors of AB, BC and CA respectively.

Find:

a the equations of the Voronoi edges

b the coordinates of the Voronoi vertex.

4 The perpendicular bisector of a chord of a circle

passes through its centre.

Find the centre of a circle passing through points

P(5, 7), Q(7, 1) and R(¡1, 5) by finding the per-

pendicular bisectors of PQ and QR and solving them

simultaneously.

5 Revisit the Opening Problem on page to find where the oval should be located.68

6 Triangle ABC has vertices as shown.

Find:

a the coordinates of P, Q and R, the midpoints of

AB, BC and AC respectively

b the equation of the perpendicular bisector of

i AB ii BC iii AC

c the coordinates of X, the point of intersection

of the perpendicular bisector of AB and the per-

pendicular bisector of BC.

d Does the point X lie on the perpendicular bisec-

tor of AC?

e

f What is special about the point X in relation to the vertices of the triangle ABC?

P(3, 8)

Q(7, 2)

EXERCISE 2E.2

A(��B(5��

C(1��

What does your result from suggest aboutthe perpendicular bisectors of the sides of atriangle?

d

A(3, 5)

C(1, )��

B(7,1)

XR

P

Q


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RATIO OF DIVISIONF


7 By considering the figure alongside:

a Find the equations of the perpendicular

bisectors of OA and AB.

b Find, using a the x-coordinate of S.

c Show that RS is perpendicular to OB.

d Copy and complete:

The perpendicular bisectors of the sides

of a triangle .......

A and B where the coordinates of A and B are

known.

To avoid confusion we agree to adopt the following convention:

So, in our original figure, P divides AB in the ratio 3 : 2 whereas

P divides BA in the ratio 2 : 3.

How do we find the coordinates of P?

Consider:

If AP : PB = r : s then

AM : MN = r : s also fsimilar ¢’sg) x

P¡ x

1: x

2¡ x

P= r : s

)xP

¡ x1

x2

¡ xP

=r

s

) sxP

¡ sx1

= rx2

¡ rxP

) rxP

+ sxP

= rx2

+ sx1

) (r + s)xP

= rx2

+ sx1

) xP

=rx

2+ sx

1

r + s

A FORMULA FOR ‘RATIO OF DIVISION’

P divides AB converts to the ratio AP : PB.

So, P divides AB in the ratio r : s means that AP : PB = r : s.

1

2

A

P

B

M N

x1 xP x2x

y

A(2 , 2 )a c

R( , 0)b B(2 , 0)bx

y

P Q

S

A

P

B

We P anotice that divides AB in certain ratio invol-ving parts and parts.3 2

Suppose is located on line passing through pointsP a


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Likewise, if we project onto the y-axis yP

=ry

2+ sy

1

r + s:

Thus,

1 For A(¡1, 2), B(3, 0), C(4, ¡3) and D(¡7, 2), find the coordinates of:

a P which divides AB in the ratio 1 : 2

b Q which divides BD in the ratio 2 : 3

c R which divides AD in the ratio 3 : 5

d S which divides DA in the ratio 4 : 1

e T which divides CA in the ratio 3 : 7.

2 The line through T and P is perpendicular

to MN and T divides MN in the ratio 2 : 3.

Find the equation of TP.

3 For the given figure find:

a the coordinates of the midpoints P, Q and R

b X on AP which divides AP in the ratio 2 : 1Y on BQ which divides BQ in the ratio 2 : 1Z on CR which divides CR in the ratio 2 : 1:

c What conclusion can you draw from b?

(AP, BQ and CR are called medians of ¢ABC.)

d Find the average of the x-coordinates of A, B and

C. Likewise average the y-coordinates. Conclusion?

if A(x1, y

1) and B(x

2, y

2) are two points and P divides AB in theµ

rx2+ sx

1

r+ s,ry

2+ sy

1

r+ s

¶:ratio then the coordinates of arer s: P

EXERCISE 2F

For A(¡3, 2) and B(5, ¡1) find P on AB such that P divides AB in

the ratio 2 : 5.

P divides AB in 2 : 5

) P is

µ2( )5 + 5(¡3)

2 + 5,

2(¡1) + 5( )2

2 + 5

¶i.e., P is (¡5

7 , 87 )

Example 18

B's coordinatesA's coordinates

A( 2, 7)� Q C(10, 2)

R P

B(3, 1)�

T

P

N(7, 13)

M(3, 5)


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THE DISTANCE

FROM A POINT TO A LINE

G


Suppose that N is the foot of the perpendicular from P to

the line l.

If M is any point on the line (other than at N), then ¢MNP

is right angled with hypotenuse MP and so MP > NP.

Consequently, NP is the shortest distance from P to line l.

So,

To find the shortest distance from a point P to a line l we follow these steps:

Step 1: Find the slope of the line l and then the slope of PN.

Step 2: Find the equation of line segment PN.

Step 3: Solve simultaneously the equations of line l and line PN.

Step 4: Write down the coordinates of N.

Step 5: Find distance PN using the distance formula.

Find the distance from P(7, ¡4) to the line with equation 2x+ y = 5.

The slope of 2x+ y = 5 is ¡21

) the slope of NP is 12

) the equation of NP isy ¡ ¡4

x¡ 7= 1

2

) 2(y + 4) = 1(x¡ 7)

) 2y + 8 = x¡ 7

or x¡ 2y = 15

So, we now solve simultaneously

½2x+ y = 5

x¡ 2y = 152x+ y = 5 £2

x¡ 2y = 15 ......) 4x+ 2y = 10

) x¡ 2y = 15

5x = 25

) x = 5

FINDING THE DISTANCE

Example 19

When we talk about the distance from point to line we actually mean thefrom the point to the line.

a a shortest dis-

tance

l

N

M

P

the distance from a point to a line is the distance

from the point to the foot of the perpendicular from

P to the line. l

N

P

|{z

}

distance

required

N

P(7, 4)�

2 + = 5x y


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and when x = 5, 2(5) + y = 5

) 10 + y = 5

) y = ¡5

) N is (5, ¡5) So PN =p

(7 ¡ 5)2 + (¡4 ¡ ¡5)2

=p

22 + 12

=p

5 units

1 Find the distance from:

a (7, ¡4) to y = 3x¡ 5 b (¡6, 0) to y = 3 ¡ 2x

c (8, ¡5) to y = ¡2x¡ 4 d (¡10, 9) to y = ¡4x+ 3

e (¡2, 8) to 3x¡ y = 6 f (1, 7) to 4x¡ 3y = 8

2

a y = 3x+ 2 and y = 3x¡ 8 b 3x+ 4y = 4 and 3x+ 4y = ¡16

3 A straight water pipeline passes through two

points with map references (3, 2) and (7, ¡1)

respectively. The shortest spur pipe from the

pipeline to farm P is PN. Find:

a the coordinates of N

b the length of the pipeline spur from N to

P given that the grid reference scale is 1unit ´ 0:5 km.

Consider Example 19 again and suppose Q is a moving point on the line and has variable

x-coordinate t.

On the line 2x+ y = 5 with x = t,y = 5 ¡ 2t

so Q is (t, 5 ¡ 2t).

d =p

(t¡ 7)2 + (5 ¡ 2t¡ ¡4)2

=p

(t¡ 7)2 + (9 ¡ 2t)2

=pt2 ¡ 14t+ 49 + 81 ¡ 36t+ 4t2

=p

5t2 ¡ 50t+ 130

EXERCISE 2G.1

USING QUADRATIC THEORY

SHORTEST DISTANCES BY OTHER METHODS (EXTENSION)

A(3, 2) P(9, 7)

B(7, 1)�

N

P(7, 4)�d

2 + = 5x y

Q(t, .....)

Using the distance formula,

Find the distance between the parallel lines: ( : find a point on one of the lines andfind the distance from this point to the other line.)

Hint


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Now d2 is minimised when t =¡b2a

fquadratic theoryg

=¡(¡50)

10= 5

) dmin =p

125 ¡ 250 + 130

=p

5 units

1 Use the quadratic theory method to find the distance from:

a (7, ¡4) to y = 3x¡ 5 b (¡6, 0) to y = 3 ¡ 2x

c (¡2, 8) to 3x¡ y = 6 d (9, 0) to 3x+ 2y = 10

The distance from a point to a line can also be found using the point to line distance

formula. This is:

the distance from (h, k) to the line Ax+By + C = 0

is .d =jAh +Bk +C jp

A2 +B2

² the equation of the line is in the general form with zero (0) on the RHS

²

This formula provides us with the ‘quickest’ way of calculating these distances.

Use the distance from a point to a line formula to check the distance from

P(7, ¡4) to the line with equation 2x+ y = 5 obtained in Example 19.

First we write 2x+ y = 5 as 2x+ y ¡ 5 = 0

) d =j2( )7 + (¡4) ¡ 5jp

22 + 12fA = 2, B = 1, C = ¡5g

=j14 ¡ 4 ¡ 5jp

5

=j5jp

5

=5p5

=p

5 units

Example 20

EXERCISE 2G.2

POINT TO LINE DISTANCE FORMULA (EXTENSION)

Notice that:

As is positive

is minimised whenis minimum.

d d

d

\

�2

the vertical lines enclosing the numerator read as “the of ” or

“the of ”.

This ensures that the calculated distance is .

modulus

absolute value

positive

��


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a (3, 2) to 2x+ 5y + 6 = 0 b (¡1, 4) to 4x¡ 3y = 4

c (¡1, 0) to 5x¡ 12y + 1 = 0 d (¡1, ¡2) to 2x¡ y = 7

e (3, ¡3) to 3x¡ y = 6 f (2, ¡3) to 4x¡ 3y = 17


a (¡1, 5) to y = 4x¡ 3 b (2, ¡7) to y = 3 ¡ 2x

c (3, ¡1) to y = ¡2x¡ 4 d (¡1, ¡5) to y = ¡1 ¡ 4x

4 Find the distance between the parallel lines:

a 3x+2y = 5 and 3x+2y+1 = 0 b 2x¡ y + 6 = 0 and 2x¡ y = 11

c 5x+ 12y + 3 = 0 and 5x+ 12y = 7 d 3x+ 4y = 6 and 3x+ 4y = ¡4

Find the equation of the two lines which are parallel to 2x+ y = 11 and are

2p

5 units from it.

Parallel lines have the same slope

) they are of the form 2x+ y + k = 0

Nowj2( )0 + ( )11 + kjp

4 + 1= 2

p5

)jk + 11jp

5= 2

p5

) jk + 11j = 10

) k + 11 = §10

) k = ¡1 or ¡21

) the lines have equations

½2x+ y ¡ 1 = 02x+ y ¡ 21 = 0

Find the distance from (3, ¡4) to the line with equation y = 3x¡ 2:

We put y = 3x¡ 2 in the form 3x¡ y ¡ 2 = 0

) d =j3( )3 ¡ (¡4) ¡ 2jp

9 + 1fas A = 3, B = ¡1, C = ¡2g

=j9 + 4 ¡ 2jp

10

= 11p10

units

Example 21

Example 22

2p

5 units

2p

5 units

2 + = 11x y (0, 11)

is our choice

of any point

on the given line


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5 Find the equations of the two lines which are parallel to the line with equation:

a 3x+ 4y = 6 and 2 units from it b x¡ 4y = 10 andp

17 units from it

c x+ y = 5 and 4p

2 units from it

6 Find k given that:

a the distance from (k, ¡3) to the line with equation 3x¡ 2y+ 6 = 0 isp

13 units

b the distance from (5, k) to the line with equation x¡ y = 4 isp

8 units

c 3x+ 4y = 10 and 5x+ 12y + k = 0 are equidistant from O(0, 0)

d x+ y = k and x¡ y + 7 = 0 are equidistant from A(1, ¡2).

7 To prove ‘the distance from a point to line’ formula you could follow these steps:

Step 1:

First show that the distance from O(0, 0)

to Ax+By +C = 0 isjCjpA2 +B2

by finding the area of the shaded triangle

using two different altitudes and then

equating the areas.

Step 2:

Now consider the given figure:

Translate (h, k) and the line

Ax+By + C = 0 by the same amount

so that (h, k) moves to O(0, 0).

What is the equation of line 2?

Now use step 1 to find d.

Example 23

Find k given that the distance from (2, k) to the line with equation

3x+ y = 7 is 2p

10 units.

The distance from (2, k) to 3x+ y ¡ 7 = 0 is 2p

10 units

)j3( )2 + (k) ¡ 7jp

9 + 1= 2

p10

)jk ¡ 1jp

10= 2

p10

2p

10) jk ¡ 1j = 20

) k ¡ 1 = §20

) k = 1 § 20

) k = 21 or ¡19(2, )k

3 + 7 = 0x y �

x

y

Ax By C� � � 0

d

d

( , )h kline 2

line 1

x

y

Ax By C� � � 0

d ( )0,AC

�

( ),0BC

�


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CIRCLES IN THE CARTESIAN PLANEH


A circle is a set of all points which are equidistant from a fixed point (called its centre).

CP is constant for all positions of P

and CP is a radius of the circle.

If C is (2, 5) and P is (x, y) on a circle of radius 3 units then

CP = 3

)p

(x¡ 2)2 + (y ¡ 5)2 = 3 fusing the distance formulagand so (x¡ 2)2 + (y ¡ 5)2 = 9:

This equation connects the x and y coordinates for all points on the circle and so must be the

equation of the circle.

If we expand the brackets we get:

x2 ¡ 4x+ 4 + y2 ¡ 10y + 25 = 9 which simplifies to

x2 + y2 ¡ 4x¡ 10y + 20 = 0

So the equation of the circle could be written as:

² (x¡ 2)2 + (y ¡ 5)2 = 9 in centre-radius form or

² x2 + y2 ¡ 4x¡ 10y + 20 = 0 in expanded or general form.

In general,

Proof:

If a circle has centre (a, b) and radius r units and

P(x, y) is any point on it, then

CP = r

)p

(x¡ a)2 + (y ¡ b)2 = r fdistance formulag) (x¡ a)2 + (y ¡ b)2 = r2

CENTRE-RADIUS FORM

(x¡a)2 +(y¡b)2 = r2 is the equation of a circle with centre (a, b) and radius r

and is in form.centre-radius

x

y

r

P( , )x y

C( , )a b�

C

P

centre


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What do the following equations represent?

a (x¡ 5)2 + (y ¡ 2)2 = 16 b (x+ 2)2 + (y ¡ 1)2 = 7

c x2 + y2 = 5

a (x¡ 5)2 + (y ¡ 2)2 = 42 is a circle centre (5, 2), radius 4:

b (x+ 2)2 + (y ¡ 1)2 = 7 can be written as (x¡ ¡2)2 + (y ¡ 1)2 = (p

7)2

which is a circle centre (¡2, 1), radiusp

7:

c x2 + y2 = 5 can be written as (x¡ 0)2 + (y ¡ 0)2 = (p

5)2

which is a circle centre (0, 0), radiusp

5:

1 State the coordinates of the centre and find the radius of the circle with equation:

a (x¡ 2)2 + (y ¡ 5)2 = 16 b (x¡ 3)2 + (y + 1)2 = 7

c (x+ 3)2 + (y + 4)2 = 11 d (x+ 5)2 + y2 = 1

e (x¡ 4)2 + y2 = 8 f x2 + y2 = 3

Write down the equation of the circle with:

a centre (¡3, 1) and radiusp

5 units b centre (0, ¡2) and radius 3 units

a equation is (x¡ [¡3])2 + (y ¡ 1)2 = (p

5)2

i.e., (x+ 3)2 + (y ¡ 1)2 = 5

b equation is (x¡ 0)2 + (y ¡ [¡2])2 = 32

i.e., x2 + (y + 2)2 = 9

2 Write down the equation of the circle with:

a centre (2, 4) and radius 5 units

b centre (¡1, ¡5) and radiusp

7 units

c centre (5, ¡1) and radiusp

13 units

d centre (4, ¡2) and radius 6 units

e centre (0, 4) and radiusp

17 units

f centre (¡3, 0) and radiusp

3 units

g centre (0, 0) and radiusp

6 units

EXERCISE 2H.1

Example 24

Example 25


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Find the equation of the circle:

a with centre (1, ¡3) and touching the x-axis

b with diameter ends A(3, ¡1) and B(¡1, ¡5).

a From the diagram r = 3 units

) equation is

(x¡ 1)2 + (y + 3)2 = 9

bThe centre is

µ¡1 + 3

2,

¡5 + ¡1

2

¶i.e., C is (1, ¡3)

) r =p

(3 ¡ 1)2 + ((¡1) ¡ (¡3))2

=p

22 + 22

=p

8 units

) equation is (x¡ 1)2 + (y + 3)2 = 8

3 Find the equations of the following circles, giving your answer in the form

(x¡ a)2 + (y ¡ b)2 = r2:

a centre (3, ¡2) and touching the x-axis

b centre (¡4, 3) and touching the y-axis

c centre (¡2, ¡3) and touching the y-axis

d centre (5, ¡2) and touching the x-axis

e centre (5, 3) and passing through (4, ¡1)

f centre (¡2, 4) and passing through (3, 1)

g ends of a diameter (¡2, 3) and (6, 1)

h ends of a diameter (0, ¡4) and (¡4, ¡6)

i radius 3 and concentric with (x¡ 3)2 + (y + 4)2 = 1

j radiusp

7 and concentric with (x+ 3)2 + (y ¡ 2)2 = 5.

4 What do the following equations represent in 2-D coordinate geometry?

a (x+ 2)2 + (y ¡ 7)2 = 5 b (x+ 2)2 + (y ¡ 7)2 = 0

c (x+ 2)2 + (y ¡ 7)2 = ¡5

In order to find theequation of a circlewe need to know itscentre and radius.

Example 26

3

(1, 3)�

y

x

A(3, 1)�

B(��

C r


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5 Consider the shaded region inside the circle, centre (a, b), radius r units.

Let P(x, y) be any point inside the circle.

a Show that (x¡ a)2 + (y ¡ b)2 < r2.

b What region is defined by the inequality

(x¡ a)2 + (y ¡ b)2 > r2?

c Without sketching the circle with equation (x+ 2)2 + (y ¡ 3)2 = 25, determine

whether the following points lie on the circle, inside the circle or outside the circle:

i A(2, 0) ii B(1, 1) iii C(3, 0) iv D(4, 1)

6 T is a mobile telephone tower in a remote country town and its region of reception is

defined by the inequality (x¡ 3)2 + (y¡ 2)2 6 40. David’s car has broken down on

the highway at grid reference D(9, 5).

a Write down the grid reference of the mobile telephone tower.

b What is the radius of the region of reception for this tower if each grid unit is equal

to 5 km?

c If there are no other mobile telephone towers in the near vicinity, will David be

able to use his mobile telephone to ring for roadside assistance?

Find k if (k, 2) lies on the circle with equation (x¡ 2)2 + (y ¡ 5)2 = 25.

Since (k, 2) lies on the circle, x = k and y = 2 satisfy the equation

) (k ¡ 2)2 + (2 ¡ 5)2 = 25

) (k ¡ 2)2 + 9 = 25

) (k ¡ 2)2 = 16

) k ¡ 2 = §4 and ) k = 6 or ¡2:


a (3, k) lies on the circle with equation (x+ 1)2 + (y ¡ 2)2 = 25

b (k, ¡2) lies on the circle with equation (x+ 2)2 + (y ¡ 3)2 = 36

c (3, ¡1) lies on the circle with equation (x+ 4)2 + (y + k)2 = 53.

(x¡ 2)2 + (y + 3)2 = 7,

x2 ¡ 4x+ 4 + y2 + 6y + 9 = 7

) x2 + y2 ¡ 4x+ 6y + 6 = 0

which is of the form x2 + y2 +Ax+By +C = 0 with A = ¡4, B = 6, C = 6:

THE GENERAL FORM

Notice that for

Example 27

y

x

( , )a b�

P( , )x y�

x2 + y2 +Ax + By + C = 0


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In fact all circle equations can be put into this form, called the general form,

i.e., the general form of the equation of a circle is x2 + y2 +Ax + By + C = 0:

Often we are given an equation in general form and need to find the centre and radius of the

circle.

We can do this by either

² ‘completing the square’ on the x and y terms, or

² using a formula.

Before attempting to find the centre and radius of a circle given in general form it is essential

to make the coefficients of x2 and y2 be 1 (if they are not already 1).

For example, if 2x2 + 2y2 + 4x+ 8y ¡ 3 = 0 then x2 + y2 + 2x+ 4y ¡ 32 = 0.

Find the centre and radius of the circle with equation

x2 + y2 + 6x¡ 2y ¡ 6 = 0 by ‘completing the square’.

x2 + y2 + 6x¡ 2y ¡ 6 = 0

) x2 + 6x + y2 ¡ 2y = 6

) x2 + 6x+ 32 + y2 ¡ 2y + 12 = 6 + 32 + 12 fcompleting the squaresg) (x+ 3)2 + (y ¡ 1)2 = 16

) the circle has centre (¡3, 1) and radius 4 units

1 Use ‘completing the square’ to find the centre and radius of:

a x2 + y2 + 6x¡ 2y ¡ 3 = 0 b x2 + y2 ¡ 6x¡ 2 = 0

c x2 + y2 + 4y ¡ 1 = 0 d x2 + y2 + 4x¡ 8y + 3 = 0

e x2 + y2 ¡ 4x¡ 6y ¡ 3 = 0 f x2 + y2 ¡ 8x = 0

g 2x2 + 2y2 + 8x+ 4y ¡ 3 = 0 h 3x2 + 3y2 ¡ 12y ¡ 8 = 0

2 a Use ‘completing the square’ on the circle with equation x2+y2+Ax+By+C = 0

to obtain

µx+

A

2

¶2+

µy +

B

2

¶2=A2

4+B2

4¡ C

b From a write down the centre and radius of the circle given in general form.

COMPLETING THE SQUARE

EXERCISE 2H.2

Example 28


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In question 2 you should have discovered that:

If a circle has equation x2 + y2 +Ax+By +C = 0 then

² its centre is

µ¡A2

,¡B2

¶and

² its radius isp

(x-coord. of centre)2 + (y-coord. of centre)2 ¡C

The above make it easy to find the centre and radius of a circle given in general form.

For example, for the circle given in Example 28,

i.e., x2 + y2 + 6x¡ 2y ¡ 6 = 0, its centre is ¡3, 1)

and r =p

(¡3)2 + 12 ¡ (¡6)

=p

9 + 1 + 6

=p

16

= 4 units

3 Use the ‘formulae’ to find the centre and radius of each circle given in question 1.

4 Under what conditions is x2 + y2 +Ax+By +C = 0 not the equation of a circle?

Explain your answer in each case.

Find k given that x2 + y2 ¡ 4x+ 6y + k = 0 represents a circle with radius

5 units.

Using ‘completing the square’ on

x2 + y2 ¡ 4x+ 6y + k = 0 we have

x2 ¡ 4x+ 22 + y2 + 6y + 32 = ¡k + 22 + 32

i.e., (x¡ 2)2 + (y + 3)2 = 13 ¡ kSo, 13 ¡ k = 52

) k = ¡12

Using ‘the formula’

x2 + y2 ¡ 4x+ 6y + k = 0 has centre (2, ¡3) and

r =p

(2)2 + (¡3)2 ¡ k =p

13 ¡ k units

Thusp

13 ¡ k = 5

) 13 ¡ k = 25

) k = ¡12

Example 29

(¡62 , ¡¡22 ) i.e., (


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a x2 + y2 ¡ 12x+ 8y + k = 0 is a circle with radius 4 units

b x2 + y2 + 6x¡ 4y = k is a circle with radiusp

11 units

c x2 + y2 + 4x¡ 2y + k = 0 represents a circle

d x2 + y2 ¡ 6x+ k = 0 represents a point.

Find the equation of the tangent to the circle with equation x2 + y2 ¡ 8x¡ 4 = 0at the point P(8, ¡2).

x2 + y2 ¡ 8x¡ 4 = 0 has centre (4, 0).

Now slope of CP is0 ¡ ¡2

4 ¡ 8=

2

¡4= ¡1

2

) slope of tangent is 21

) equation of tangent isy ¡ ¡2

x¡ 8= 2

y + 2 = 2(x¡ 8)

y + 2 = 2x¡ 16

i.e., 2x¡ y = 18

6 Find the equation of the tangent to the circle with equation:

a x2 + y2 + 6x¡ 10y + 17 = 0 at the point (¡2, 1)

b x2 + y2 + 6y = 16 at the point (0, 2)

c x2 + y2 ¡ 4x¡ 14 = 0 at the point (¡1, ¡3).

7 A circular lake in a park has its boundary defined

by the equation x2+y2¡24x¡16y+111 = 0.

A straight path meets the edge of the lake at grid

reference A(3, 4). Find the:

a diameter of the circular lake

b equation of the straight path.

8 One end of a diameter of a circle with equation x2 + y2 ¡ 4x+ 4y = 2 is (3, 1).

Find the coordinates of the other end of the diameter.

9 Find the coordinates of the centre and radius of each of the circles x2 + y2 = 4 and

(x+4)2+(y¡3)2 = 9. Find the distance between the centres of the circles and hence

deduce that the circles touch each other externally.

10 A circle is concentric with the circle with equation (x+2)2 +(y¡ 7)2 = 6 and passes

through the point (¡1, 2). Find the equation of the circle.

A

C(4, 0)

P(8, 2)�

Example 30


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11 Find the equation of the circle which is tangential to the x-axis at (2, 0) and touches

the line y = 6.

12 Find the equation of the chord of the circle x2+y2 ¡ 10x+9 = 0 which has midpoint

(3, 1).

13 Find k given that the circles x2 + y2 + 4x+ 6y + 4 = 0 and x2 + y2 ¡ 4x+ k = 0touch each other externally.

14 Show that the circles with equations x2+y2 = 5 and x2+y2¡6x+12y = 35 touch

each other internally and find the coordinates of their point of contact.

Some of the technology currently available does not allow us to graph the circle x2+y2 = 4directly from this equation.

The two functions y =p

4 ¡ x2 and y = ¡p4 ¡ x2 must be entered and graphed

separately.

Note: x2 + y2 = 4

) y2 = 4 ¡ x2) y = §p

4 ¡ x2y =

p4 ¡ x2 is the top half of the circle

and y = ¡p4 ¡ x2 is the bottom half.

Rearrange (x+ 2)2 + (y ¡ 4)2 = 10 to make y the subject of

the formula.

(x+ 2)2 + (y ¡ 4)2 = 10

) (y ¡ 4)2 = 10 ¡ (x+ 2)2

) y ¡ 4 = §p10 ¡ (x+ 2)2

) y = 4 §p10 ¡ (x+ 2)2

Example 31

USING TECHNOLOGY TO GRAPH CIRCLES

point of contact

external touching internal touching

x

y

= ¡p

4 ¡ x 2y

=p

4 ¡ x 2y2

2


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1 Rearrange each of the following to make y the subject of the formula:

a x2 + y2 = 16 b x2 + y2 = 7

c (x¡ 1)2 + y2 = 12 d (x+ 2)2 + y2 = 5

e x2 + (y ¡ 3)2 = 9 f x2 + y2 ¡ 4x+ 2y ¡ 4 = 0

g x2 + y2 + 8x¡ 4y + 4 = 0 h x2 + y2 + 6x+ 4y = 28

To graph (x+ 2)2 + (y ¡ 4)2 = 10 we rearrange to get y = 4 §p10 ¡ (x+ 2)2:

Enter the formulae Y1 = 4 +p

10 ¡ (X + 2)2 and Y2 = 4 ¡p10 ¡ (X + 2)2

Now graph these two functions.

Enter (x+ 2)2 + (y ¡ 4)2 = 10:

2 a Use your graphics calculator or the graphing package to graph the circles given in

question 1.

b Find where the circles cut the coordinate axes.

How do we determine whether a line is a tangent to a circle, cuts it twice or misses it

altogether?

Find the point of intersection of the line y = 2 ¡ x and the circle

(x¡ 2)2 + (y ¡ 3)2 = 9:

These meet where (x¡ 2)2 + ([2 ¡ x] ¡ 3)2 = 9 f greplace byy x[2 ]¡i.e., (x¡ 2)2 + (¡x¡ 1)2 = 9

) x2 ¡ 4x+ 4 + x2 + 2x+ 1 = 9

EXERCISE 2H.3

USING A GRAPHICS CALCULATOR

USING A GRAPHING PACKAGE

ALGEBRAIC DETERMINATION

Example 32

GRAPHING

PACKAGE TI

C

HP

S

THE INTERSECTION

OF LINES AND CIRCLES

I

tangent

chord

no intersection


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) 2x2 ¡ 2x+ 5 = 9

) 2x2 ¡ 2x¡ 4 = 0

) x2 ¡ x¡ 2 = 0 ......(¤)

) (x+ 1)(x¡ 2) = 0

) x = ¡1 or 2

When x = ¡1, y = 2 ¡ (¡1) = 3 and when x = 2, y = 2 ¡ 2 = 0

so, they meet at (¡1, 3) and at (2, 0).

To check this we graph the line and circle:

Note:

Using the algebraic technique we will always

obtain a quadratic equation (see Example 32 ¤)

which could have 2, 1 or 0 solutions.

If ax2 + bx+ c = 0 and ¢ = b2 ¡ 4ac then:

² if ¢ > 0, we have two real solutions (indicating 2 points of intersection)

² if ¢ = 0, we have a repeated solution (indicating a tangential case)

² if ¢ < 0, we have no solutions (indicating that the line misses the circle)

1 Find where the following lines meet the given circles:

a y = ¡x+ 6 and (x¡ 4)2 + (y ¡ 2)2 = 8

b y = ¡2x+ 6 and (x¡ 2)2 + (y + 3)2 = 10

c y = ¡2x¡ 4 and (x+ 3)2 + (y ¡ 2)2 = 5

d y = 2x¡ 9 and (x¡ 5)2 + (y + 4)2 = 10

e y = 3x+ 1 and (x¡ 5)2 + (y ¡ 6)2 = 10

f y = 3x+ 2 and (x¡ 4)2 + (y ¡ 3)2 = 5

2 Use your graphing package or graphics calculator to find where the line y = 4 ¡ xmeets the circle (x¡ 3)2 + (y ¡ 2)2 = 16:

[If using a graphics calculator you will need to write the circle as y = 2§p16 ¡ (x¡ 3)2

and enter Y1 = 4 ¡ X

Y2 = 2 +p

16 ¡ (X ¡ 3)2

Y3 = 2 ¡p16 ¡ (X ¡ 3)2 and plot these functions.]

3 Use technology to check your answers to question .1

EXERCISE 2I

x

y

2

2

(2, 3)

( 1, 3)�

GRAPHING

PACKAGE

TI

C

HP

S


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WHERE TWO CIRCLES MEETJ


4

a the equation of the straight section of highway between A and B

b the equation of the boundary of the region of reception of the mobile telephone

tower T

c the grid reference of the points where Peter and Sue enter and exit the mobile

telephone reception region

d the length along the highway in which their phone will operate

e the time they have to use their telephone if they are travelling at 100 kmph.

We can find the point(s) of intersection of two circles by algebraic means.

Find where the circles (x¡ 1)2 + (y ¡ 4)2 = 36 and (x¡ 6)2 + (y + 1)2 = 26meet.

(x¡ 1)2 + (y ¡ 4)2 = 36

) x2 ¡ 2x+ 1 + y2 ¡ 8y + 16 = 36

) x2 + y2 ¡ 2x¡ 8y ¡ 19 = 0

) x2 + y2 = 2x+ 8y + 19

(x¡ 6)2 + (y + 1)2 = 26

) x2 ¡ 12x+ 36 + y2 + 2y + 1 = 26

) x2 + y2 ¡ 12x+ 2y + 11 = 0

) x2 + y2 = 12x¡ 2y ¡ 11

Now, the circles meet where 2x+ 8y + 19 = 12x¡ 2y ¡ 11 fequating x2 + y2gi.e., ¡10x+ 10y + 30 = 0

or ¡x+ y + 3 = 0 f gdividing by 10

or y = x¡ 3 (¤)

Substituting [ 3]x y¡ for into the first circle equation gives

(x¡ 1)2 + ([x¡ 3] ¡ 4)2 = 36

) (x¡ 1)2 + (x¡ 7)2 ¡ 36 = 0

) x2 ¡ 2x+ 1 + x2 ¡ 14x+ 49 ¡ 36 = 0

) 2x2 ¡ 16x+ 14 = 0

Example 33

T

A

B

T a ( , )

a

a

A ( , ) B ( ,

is mobile telephone tower with grid reference

in remote country town. The region of reception for the

tower has radius km. AB is straight section of highway

between two homesteads, at and at ). The

scale is grid unit represents km.

11

3240 56

: :2 4 7

0 01 1

Peter and Sue are driving from to along the highway

and wish to use their mobile telephone. Find:

B A


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) x2 ¡ 8x+ 7 = 0 ......(¤¤)

) (x¡ 1)(x¡ 7) = 0

i.e., x = 1 or 7

But when x = 1, y = ¡2 and when x = 7, y = 4

) the circles intersect at (1, ¡2) and (7, 4).

Note: ² The linear equation ¤ satisfies ‘where the circles meet’.

So, this must be the equation of the common chord in the case of intersection

or the common tangent in the case of touching circles.

² The quadratic at ¤¤ could have:

I 2 real solutions indicating intersecting circles (¢ > 0)

I 1 repeated solution indicating touching circles (¢ = 0)

I 0 real solutions indicating non-intersecting circles (¢ < 0).

1

a x2 + y2 ¡ 2x¡ 6y = 0 and x2 + y2 ¡ 4x¡ 2y = 0

b x2 + y2 + 2x¡ 6y + 5 = 0 and x2 + y2 ¡ 4x+ 6y = 7

c (x+ 3)2 + y2 = 25 and (x¡ 5)2 + (y + 4)2 = 25

d (x+ 1)2 + (y ¡ 3)2 = 20 and (x¡ 2)2 + (y + 6)2 = 50

e (x+ 3)2 + y2 = 18 and (x¡ 2)2 + (y + 5)2 = 8

f x2 + (y ¡ 3)2 = 9 and x2 + y2 ¡ 8x+ 2y + 16 = 0

2 Use technology to check your answers to question .1

3 Paul’s Pizza Parlour is located at map

reference P(7, 6) and will deliver only

within a 5 km radius of the shop.

Pizza Heaven is located at Q(16, 9) and

will deliver in an 8 km radius. It is

known that a highway AB passes di-

rectly through the two points where the

delivery region boundaries meet. Find:

a the equations which define the two

delivery region boundaries

b

c the equation of the highway AB.

EXERCISE 2J

the map references of the points onthe highway where the two deliveryregion boundaries meet

Find where the following circle pairs intersect. If they meet, state either the equation ofthe common chord or the equation of the common tangent.

GRAPHING

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Q

P

A

B

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DISCUSSION

THE RECTANGULAR HYPERBOLAK


A contract painter estimates that it will take him 36 days to paint the walls of a hostel if he

works alone. How long would it take to paint the walls if two people are available and work

at the same rate? What about the time for three people, four people, etc.?

This situation was described in a previous course as inverse variation. We can tabulate and

graph the time (T ) against the number of people painting (n).

n 1 2 3 4 6 9 12 18 36

T 36 18 12 9 6 4 3 2 1

Note:

Now consider the situation of manufacturingenvelopes with area 36 cm2:

Firstly, consider xy = 36 where x can take any value; positive, zero or negative.

What would the graph look like?

Of course, xy = 36 can be written as y =36

xand graphs of these functions are called

rectangular hyperbolae.

Whether in the form xy = 36 or y =36

xwe can see that neither x

nor y can be zero.

RECTANGULAR HYPERBOLAE

��

��

��

�

��

��

T (days)

n (painters)

x cm

y cm

If each envelope is cm by cm, then .

A table of values identical to those for the painter problem

could be formed and the graph drawn.

However, on this occasion we have a continuous graph,

not discrete points, as any positive value of is possible.

x y xy

x

= 36

�

�

�

The time taken depends on the number of

painters . So, is the independent variable.

It is not sensible to join these points with a

curve. Why?

Did you notice that for all tabled

values?

Tn n

nT� �=36

��

��

��

�

��

��

y

x

xy��


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INVESTIGATION 2 THE FAMILY OF CURVES


1 On the same set of axes draw the graphs y =1

x, y =

2

x, y =

4

x:

2 The functions in 1 are members of the family y =A

x

y=Ax

where A is 1, 4 or 8.

Describe the effect changes in the A values have on the graphs for A > 0.

3 Repeat 1 for y =¡1

x, y =

¡2

x, y =

¡4

x:

4 Comment on shape changes in 3.

5 Explain why no graph exists when x = 0.

6 Consider the function y =10

x.

a Find y values for x = 1000, 100, 10, 5, 2, 1, 0:5, 0:1, 0:01

b Find y values for x = ¡1000, ¡100, ¡10, ¡5, ¡2, ¡1, ¡0:5, ¡0:1, ¡0:01

c Draw a sketch graph of y =10

x.

d Without calculating new values, sketch the graph of y =¡10

x.

From the above investigation you should have discovered that:

The use of oris recommended.

a graphing package

graphics calculator

What to do:

² All graphs of relations of the form y =A

x

I have lines of symmetry y = x and y = ¡xI are point symmetric about the origin O. (This means that under

a 180o rotation about O, they map onto themselves.)

² All graphs of relations of the form y =A

xare asymptotic to the x

and y-axes. This means that the further we move away from the origin,

the closer the graph approaches these lines without ever reaching them.

² y

x

y AAx� �, 0

y

x

y AAx� �, 0

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1 On the same set of axes draw graphs of y =8

xand y = ¡ 8

x:

2 Draw the graph of y2 =16

x2:

3 Determine the equations of the following rectangular hyperbolae:

a b c

d e f

4 Find the coordinates of xy = 4 which are closest to the origin (0, 0).

5 A cyclist travelling at 24 km/h travels from home to a friend’s house in 4 hours.

a How is this information related to the equation st = 96?

b How long would it take to travel the same distance at 32 km/h?

6 Four scouts on an expedition calculate that their remaining food supplies will last them

for 9 more days. How long would this same amount of food last if there were 6 scouts

on the expedition?

7 Sally receives $8 pocket money per week and is saving it all to buy a new pair of shoes.

She calculates that it will take her 15 weeks to save enough money. If her father had

increased her pocket money to $10 each week, and she continued to save all of her

pocket money, how long would it have taken her to save up for the shoes?

EXERCISE 2K.1

y

x

(4, 2)

y

x

( , )��

y

x

( 3, 1)� �

y

x

( ,5)��

y

x

(2, 6)�

y

x

(4, )��

² y

x

xy 5�

xy 1�

xy 1�

xy 3�

xy 3�

xy 5�

y

x

xy 5��

xy 5��

xy 1��

xy 1��

xy 3��

xy 3��

xy A�Families of curves of the form


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8 Peter noticed that his 15 Zebra finches took 6 days to empty their seed container. He

added 3 more Zebra finches to the cage. How long would it take the Zebra finches to

consume the same quantity of seed as they did before the 3 were added?

9 “Squeaky Clean Window Cleaners” can clean all windows of a building in 5 days when

they use a team of 8 people. This year the client wants the job completed in 4 days.

How many cleaners are required?

10 A platoon of 30 soldiers are provisioned for 16 days. After 3 days it was necessary

to evacuate 3 sick soldiers. How much longer will the provisions now last (after the 3days)?

11 Find where the line:

a y = x+ 4 meeets y = ¡ 3

xb y = x¡ 5 meets y =

6

x

c y = 2x¡ 1 meets y =10

xd y = ¡2x+ 2 meets y =

4

x

e y = 2x+ 8 meets y = ¡ ¡8 4

x xf y = 2x+ 1 meets y =

12 Use your graphics calculator or computer graphing package to find where the line:

a y = 2x¡ 3 meets y =4

xb y = 4x+ 3 meets y = ¡10

x

c y = 3x¡ 2 meets y =6

xd y = 3x+ 5 meets y =

12

x

e 3x¡ 5y = 10 meets y =5

xf 3x+ 2y = 6 meets y =

10

x

13 a

i y =10

xii y = ¡2x¡ 9

Find where the line y = x¡ 3 meets y =4

x.

y = x¡ 3 meets y =4

xwhen x¡ 3 =

4

x

) x2 ¡ 3x = 4

) x2 ¡ 3x¡ 4 = 0

) (x¡ 4)(x+ 1) = 0

) x = 4 or x = ¡1

Thus y = x¡ 3 meets y =4

xat (4, 1) and (¡1, ¡4).

Example 34

GRAPHING

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graphics calculator computer graphing package

,


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INVESTIGATION 3 FUNCTIONS OF THE FORM y=A

x¡h+k


b Is the line a tangent to the curve? At first glance it certainly appears to be.

c Confirm your answer by using:

i your technology ii algebraic methods as in Example 34.

1 On the same set of axes draw the graphs of the functions

y =4

x, y =

4

x¡ 2and y =

4

x+ 3.

2 The functions in 1 are members of the family y =4

x¡ h where h is 0, 2 or ¡3.

What effect has changing values of h on the members of this family?


y =4

x, y =

4

x+ 3 and y =

4

x¡ 2:


x+ k where k is 0, 3 or ¡2.

What effect has changing values of k on the members of this family?


y =4

x, y =

4

x¡ 2+ 3 and y =

4

x+ 1¡ 5.


x¡ h + k where h = 0, k = 0

or h = 2, k = 3 or h = ¡1, k = ¡5:

What transformation moves y =4

xonto y =

4

x¡ h + k?

7 What does y =7

xbecome under a translation of [2, ¡3]?

From the investigation you should have discovered that:

² y =A

xmoves to y =

A

x¡ h under a horizontal translation of [h, 0]

² y =A

xmoves to y =

A

x+ k under a vertical translation of [0, k]

² y =A

xmoves to y =

A

x¡ h + k under a translation of [h, k].

What to do:

y= Ax¡h

+ kFUNCTIONS OF THE FORM

The use of oris recommended.

a graphing package

graphics calculator

TI

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So, to graph y =A

x¡ h + k we could

imagine x = h and y = k to be the

axes on which we would graph y =A

x:

1 Using a translation technique, sketch the graph of:

a y =3

x¡ 1+ 2 b y =

4

x¡ 3+ 1 c y =

1

x+ 2+ 3

d y =8

x+ 2+ 1 e y =

6

x+ 2¡ 4 f y =

2

x+ 4¡ 1

g y =3

x¡ 3¡ 2 h iy y= =

5 5

x x¡ 2¡ ¡2 2


a y =¡4

x¡ 2+ 1 b y =

¡3

x+ 2¡ 2 c y =

¡6

x+ 3+ 2

d y =¡2

x¡ 3¡ 2 e y =

10

2x+ 1+ 3 f y =

¡8

2x¡ 1¡ 2

EXERCISE 2K.2

Draw the graph of y =6

x¡ 2+ 3 using a translation technique.

y =6

x¡ 2+ 3 is the translation of y =

6

xby [2, 3].

So, to graph y =6

x¡ 2+ 3 we sketch y =

6

xon new x-axis

x = 2 and new y-axis y = 3.

Check:

Cuts the y-axis when x = 0

) y = 6¡2 + 3

= ¡3 + 3= 0 X

Example 35

x

y

ky �

hx �

khx

Ay �

��

x

Ay �

y

x

y��

x��


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Real world models of rectangular hyperbolic functions exist. In the following exercise we

will examine models and then use technology to assist us.

An experimental breeding colony of wallabies is set up and the size of the colony

at time t years (0 6 t 6 8) is given by N = 20 ¡ 400t¡ 10

:

a What was the original size of the colony?

b What is the size of the colony after i 3 years ii 8 years?

c How long would it take for the colony to reach a size of 100?

d Draw the graph of N against t, 0 6 t 6 8, using only a, b and c.

N = 20 ¡ 400

t¡ 10

a When t = 0, N = 20 ¡ 400¡10 = 20 + 40 = 60 wallabies.

b i When t = 3, N = 20 ¡ 400¡7 + 77 wallabies.

ii When t = 8, N = 20 ¡ 400¡2 = 20 + 200 = 220 wallabies.

c When N = 100, d

100 = 20 ¡ 400

t¡ 10

) 80 =¡400

t¡ 10

) t¡ 10 = ¡5

) t = 5

i.e., it would take 5 years.

1 In order to remove noxious weeds from his property a farmer sprays a weedicide over

it. The chemical is slow acting and the number of weeds per hectare remaining after t

days is given by N = 20 +100

t+ 2weeds per hectare.

a How many weeds per hectare were alive before spraying the weedicide?

b How many weeds were alive after 8 days?

c How long would it take for the number of weeds still alive to reach 40 per hectare?

d Sketch the graph of N against t, using only a, b and c.

e Is the chemical spraying program going to eradicate all weeds?

APPLICATIONS OF RECTANGULAR HYPERBOLAE

EXERCISE 2K.3

Example 36

8642

200

150

100

50

N (wallabies)

t (years)


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2 The current remaining in an electrical circuit t seconds after it is switched off, is given

by I =1000

t+ 4amps.

a What current flowed in the circuit at the instant before it was switched off?

b Find the current after i 6 seconds ii 20 seconds.

c How long would it take for the current to reach 10 amps?

d Graph I against t, using only a, b and c.

e Use a graphics calculator or graphing package to graph

Y1 = 1000=(X+4): If using a graphics calculator , graph the function and use trace

to check your answers to a b c, and .

3 A helicopter on its helipad climbs vertically to a height h metres above the pad so that

at time t minutes, h(t) = 2500

µ1 ¡ 1

t+ 1

¶.

a Check that h(0) = 0:

b Find the height reached after i 9 minutes ii 20 minutes.

c Use technology to help you sketch the graph of h against t, (t > 0) and check

your answers to a and b.

d Is there an altitude beyond which the helicopter cannot go? Explain.

4 On a wet day a cyclist was travelling at a constant speed along a bitumen road.

He braked suddenly and his speed afterwards was given by S = 20t+ 1

¡ 4 ms¡1,

for t > 0.

a How fast was he travelling at the instant when the brakes were applied?

b How long did it take for him to reach a speed of 10 ms¡1?

c How long did it take for him to become stationary?

d Graph S against t for the braking interval.

e The inequality t > 0 should be replaced with a more appropriate one. What should

it be?

A simple rational function is of the form y =ax+ b

cx+ dwhere a, b, c and d are constants, and c 6= 0.

y = 3 +4

x+ 2

= 3

µx+ 2

x+ 2

¶+

4

x+ 2

=3x+ 6 + 4

x+ 2

=3x+ 10

x+ 2

SIMPLE RATIONAL FUNCTIONS

Notice that the rectangular hyperbola

GRAPHING

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insert the denominator here

to balance


which from the definition above is a simple rational function.

Consequently,

all simple rational functions have rectangular hyperbolic graphs.

1

a b y = ¡4

xunder translation [2, ¡5]

c y =6

xunder translation [4, 0] d y = ¡3

xunder translation [0, ¡3]

The question now arises:

“If we are given a simple rational function y =ax+ b

cx+ d, how can we draw a sketch

graph and give details of its special features, such as asymptotes?”

If we can write y =ax+ b

cx+ din the form y = k +

A

x¡ h we can easily sketch the

graph of the original relation.

We can do this algebraically.

if y =2x+ 3

x+ 4, then y =

2(x+ 4) ¡ 5

x+ 4

Under translation [2, 5], y =4

xbecomes y =

4

x¡ 2+ 5

) y =4

x¡ 2+ 5

µx¡ 2

x¡ 2

¶

) y =4 + 5(x¡ 2)

x¡ 2

i.e., y =5x¡ 6

x¡ 2

Example 37

EXERCISE 2K.4

For example,

Find the equation, in the form y =ax+ b

cx+ d, of the image of

y =4

xunder translation [2, 5]

Find the equation, in the form y =ax+ b

cx+ d, of the image of:

y =5

xunder translation [1, 4]


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) y =2(x+ 4)

x+ 4¡ 5

x+ 4

) y = 2 ¡ 5

x+ 4

i.e., y =¡5

x+ 4+ 2

From which, the asymptotes are: x = ¡4 and

y = 2:

Knowing one point on the graph enables us to

draw a quick sketch of it.

The y-intercept could be this point.

When x = 0, y = 34 .

Note: The x-intercept could also be found as a check.

Consider y =4x¡ 3

x¡ 2.

a Write the function in the form y =A

x¡ h + k:

b Write down the equations of its asymptotes.

c Find its y-intercept.

d Sketch its graph.

a y =4x¡ 3

x¡ 2

) y =4(x¡ 2) + 5

x¡ 2

) y = 4 +5

x¡ 2

d

b So, the vertical asymptote

is x = 2 and the horizontal

asymptote is y = 4:

c When x = 0, y = 32 = 112

) the y-intercept is 112 :

Example 38

y = 2

x = ��

x

y

y = 4

x = 2

x

y

1 Qw_

y = 2

x = ��

x

y

Er_


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AREAS OF POLYGONSL


2 For each of the following simple rational functions:

i write in the form y =A

x¡ h + k

ii write down the equations of its asymptotes

iii find the y-intercept

iv sketch the graph.

v Use technology to sketch the graph and check your answers to iii and iv.

a y =2x+ 3

x+ 1b y =

x+ 3

x¡ 2c y =

3

x¡ 2

d y =4x¡ 3

x¡ 1e y =

5x¡ 1

2x+ 1f y =

¡4

x+ 2

3 To convert y =2x¡ 1

3 ¡ x into the required form we first multiply the numerator and

denominator by ¡1, so it becomes y =¡2x+ 1

x¡ 3and then proceed as before.

a Convert y =2x¡ 1

3 ¡ x into the form y =A

x¡ h + k:


c Find the y-intercept.

d Find the x-intercept.

e Sketch the graph.

f Use technology to sketch the graph and check your answers to c and d.

The simplest polygon is a triangle. Consider

the triangle with vertices O(0, 0), P(2, 5) and

Q(6, 1). How can we find the area of this

triangle?

Our method is to project the points P and Q

onto the x-axis.

The area of ¢OPQ is then calculated using:

area ¢OPP0 + area trapezium PQQ0P0 ¡ area ¢OQQ0

= 12 (2 £ 5) +

¡5+12

¢£ 4 ¡ 12 (6 £ 1)

= 5 + 12 ¡ 3

= 14 units2

The above method, although easy to follow, is tedious and especially so if we have many

triangles which require area finding.

x

y P(2, 5)

Q(6, 1)

P0 Q02 46

5

1

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INVESTIGATION 4 TO FIND THE AREA OF A TRIANGLE


From the investigation, you should have discovered that:

For triangle OAB where A(a, b) and B(c, d) are two of its vertices,

the area is 12(ad¡ bc) or 1

2 (bc¡ ad), whichever one is positive,

or using modulus notation area = 12 jad¡ bcj :

Checking our original example area = 12 j2 £ 1 ¡ 5 £ 6j

= 12 j¡28j

= 12 £ 28

= 14 units2

Note: This formula is true no matter which quadrants A

and B are in.

1 Find the area of triangle AOB given that:

a A(1, 4) and B(7, 3) b A(¡2, 5) and B(5, 3)

c A(¡3, ¡4) and B(4, 5) d A(7, ¡2) and B(¡1, 4)

Notice the use ofthe modulus here.This ensures thatthe answer for thearea is positive.

What to do:

1

a

b

c

2 1

3

1 2

4

5

Using the diagram alongside, find

the area of: triangle AOP

triangle BOQ

trapezium ABQP.

Use the results from to find the area of trian-

gle OAB.

In the diagram alongside the same triangle has

been drawn except the points A and B have

been interchanged. Repeat steps and to

find the area of triangle OAB.

Compare your two results for the area of trian-

gle AOB.

Write down a general formula for the area of

any triangle AOB where A( , ) and B( , ).a b c d

x

x

Q

P

P

Q

A( , )a b

B( , )c d

B( , )c d

A( , )a b

EXERCISE 2L.1


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Note: A diagram showing actual points is essential as sometimes the areas of the

triangles are not all added.

For example: Area of ¢PQR

= area ¢OPQ + area ¢OQR¡ area ¢OPR

2 Find the area of triangle PQR with vertices:

a P(1, 7), Q(3, 2) R(6, 4) b P(0, 5) Q(4, 1) R(¡2, ¡1)

c P(4, 0) Q(¡1, ¡3) R(¡2, 5) d P(0, ¡5) Q(¡4, ¡3) R(¡2, 7)

3 By creating triangles, use the above method to find the area of the following polygons:

a A(¡2, 4), B(3, 1), C(1, ¡2) and D(¡4, ¡1)

b A(2, 5), B(6, 7), C(8, ¡1) and D(5, ¡3)

c P(6, ¡1), Q(2, ¡5), R(¡4, ¡4), S(¡3, 2) and T(2, 7)

Consider the triangle with vertices A(1, 2),

B(3, 6) and C(6, 4).

We will now find the area of this triangle pro-

jecting the points onto the x-axis to create

three trapezia.

Find the area of triangle PQR for P(¡3, 2) Q(4, 1) and R(1, ¡3).

Area ¢PQR

= area ¢OPQ + area ¢OQR + area ¢ORP

= 12 j(¡3 £ 1) ¡ (2 £ 4)j + 1

2 j(1 £ 1) ¡ (4 £ ¡3)j+12 j(¡3 £ ¡3) ¡ (2 £ 1)j

= 12 j¡11j + 1

2 j13j + 12 j7j

= 112 + 13

2 + 72

= 312

= 1512 units2

Example 39

AREA FINDING USING TRAPEZIA

PQ

R x

y

y

x

5

5

A

B

C

P��

Q� ��

R��

x

y


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The area of ¢ABC is calculated using

area ABB0A0 + area BCC0B0 ¡ area ACC0A0

=

µ2 + 6

2

¶£ 2 +

µ6 + 4

2

¶£ 3 +

µ2 + 4

2

¶£ 5

= 8 + 15 ¡ 15

= 8 units2

1 Use the method of creating trapezia described above to find the area of ¢ABC for:

a A(2, 4), B(5, 6) and C(6, 1)

b A(3, ¡1), B(7, ¡2) and C(4, ¡6)

2 Using the method of creating trapezia, show

that the area of ¢PQR is

[bc+ de+ fa] ¡ [ad+ cf + be]

2units2:

3 The area of quadrilateral ABCD could be

found using area ABB0 0 0A + area BCC0B¡ area ADD0A0 ¡ area DCC0D0

Find the area.

4

Using the method of question 3, show that

the area of quadrilateral PQRS is given by

[bc+ de+ fg + ah] ¡ [ad+ cf + eh+ bg]

2

5

That is in question :2 and in question :4

y

x

A' B' C'

B

A

C

2

2

64

35

EXERCISE 2L.2

y

x

P( , )a b

Q( , )c d

R( , )e f

P' Q' R'

A(1, 4)

B(5, 7)

C(6, 3)

D(3, 1)

A' D' B' C'

y

x

y

x

P( , )a b

Q( , )c d

R( , )e fS( , )g h

a

c

e

a

b

d

f

b

adbc

cfde

beaf

a

c

e

g

a

b

d

f

h

b

adbc

cfde

eh

bg

fg

ah

Milo showed that the numerators of the answers to questions and could be found by

writing down the coordinates of the points, listed in cyclic order around the figure, with

the first point mentioned being repeated at the end.

2 4

j j

jj


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Explain how to use these diagrams to obtain area formulae for finding areas of triangles

and quadrilaterals.

In the previous Exercise we have discovered a method for finding the area of a polygonal

region.

Milo’s method is:

Step 1: Plot the vertices of the polygonal region.

Step 2:

Step 3:

Step 4: Subtract the smaller of these two numbers from the larger and halve the result.

This is the required area.

So, area =46 ¡ 30

2= 8 units2:

Milo’s method gives us a quick and easy calculation of the area of any polygonal region.

��

�

��

�

��

��

��

��

�

MILO’S METHOD FOR AREAS OF POLYGONS

Find the area of pentagon ABCDE.

By Milo’s method:

area =78 ¡ (¡23)

2= 50:5 units2

��

��

�

��

��

�

� �

��

��

��

��

��

��

��

y

x

A��B��

C��

D��

E��

Example 40

Find the diagonal products of the numbers, both onthe left and on the right, and add them. For example,for the triangle ABC with vertices A( ), B( ),C( that we started with:

1 2 3 66 4

, ,, )

List the coordinates of the points in two columnsand in cyclic order (clockwise or anticlockwise), butinclude the first point second time at the end.a


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1 Use Milo’s method to find the area of:

a b c

d e f

2

Consider the line x+ y = 3.

All points on this line satisfy the rule that the

sum of the coordinates is 3, i.e., x+ y = 3.

��

��

��

��

��

��

��

��

��

� ��

��

��

��

��

��

� ��

��

��

��

��

��

��

��

��

A��

C��

E��D��

X��

P

farm X

farm Y

Y

farm

Z

A farmer subdivides his farm into three areas, each to be farmed by one of his three children.

The separating boundaries are the straight road XY and fence BD.

Each grid unit is m.100

a

b

c

d

e

Find the equation of boundary fence

AE.

Find the equation of the roadway

connecting town X to town Y.

Find the coordinates of corner post P.

Find the area of each farmlet in

hectares.

If council rates for the whole

property are $ per annum, what

portion of the rates should be met by

each child?

1375

LINEAR INEQUALITIESM

REGIONS OF THE PLANE y

x

3

3

x y� ��

EXERCISE 2L.3


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However, this line divides the plane into two regions or ‘half planes’:

² one representing all points where the sum of the coordinates is less than 3,

i.e., x+ y < 3

² the other representing all points where the sum of the coordinates is greater than 3,

i.e., x+ y > 3.

We can check this by substituting additional points on either side of the line.

Click on the icon to see the region being ‘built up’.

This program randomly selects points in the 2-dimensional plane, and tests

them to see if they satisfy the inequality.

If they do then they are retained, but if they do not they disappear.

To graph a region given by a linear inequality (inequation) we follow these steps:

Step 1: Replace the inequality sign (>, 6, > or <) by an equals sign.

This line forms the boundary of the region and we plot it on a set of axes.

Step 2: If the original inequality is > or 6 we use a full boundary line indicating

that the points on it are included in our region. If it contains > or < we

use a dashed line which indicates that points on it are not included.

Step 3: Choose a point in the plane not on the boundary line (often (0, 0) is a good

choice) and use it to test which side of the boundary line is included in the

region described by the inequality.

PLOTTING REGIONS

Sketch the region defined by 3x+ 4y > 12.

3x+ 4y > 12 has boundary line 3x+ 4y = 12.

3x+ 4y = 12 cuts the x-axis when y = 0) 3x = 12) x = 4 i.e., x-intercept is 4.

3x+ 4y = 12 cuts the y-axis when x = 0) 4y = 12) y = 3 i.e., y-intercept is 3.

y

x4

3

3 + 4 =12x y

Example 41

REGION

PLOTTER

x y >+ 3

x y <+ 3

x y+ = 3

All points in thisregion have andcoordinates whosesum is greater than 3.

e.g. (4, 2) as4 + 2 > 3

x yAll points in thisregion have andcoordinates whosesum is less than 3.

e.g. (0, 0) as0 + 0 < 3

x y3

3

x

y


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Note: ² The region defined by 3x+ 4y 6 12 is:

² Likewise 3x+ 4y > 12 is: and 3x+ 4y < 12 is:

1 Sketch the region of the plane defined by:

a x > 0 b x < 0 c y > 0

d y 6 0 e x > 2 f 1 6 x 6 3

g 0 6 y 6 4 h x > 0, y > 0 i x 6 0, y > 0


a x+ 3y 6 6 b x+ y > 5 c 2x+ 3y > 15

d x¡ y 6 6 e 2x¡ y > 4 f 5x+ 4y 6 20

g 7x+ 2y > 14 h 3x¡ 2y 6 12 i 4x¡ 3y > 15

Check your answers using the region plotter or graphics calculator.

3 P(x, y) is closer to A(3, 0) than it is to B(¡1, 4), that is AP < BP.

a Use the distance formula to show that x¡ y > ¡1:

b On a set of axes show A, B and shade the possible positions of point P.

c What is the relationship between A, B and the boundary line x¡ y = ¡1?

To find the required region we substitute a point which is clearly on one side of the

boundary line,

for example, if x = 0 and y = 0,3 £ 0 + 4 £ 0 > 12 is false,

) (0, 0) does not lie in the region.

Hence the region defined by 3x+ 4y > 12 is

shaded in the graph alongside.

EXERCISE 2M.1

REGION

PLOTTER

y

x4

3 3!\+\4@\}\12

3!\+\4@\=\12

y

x4

3

3!\+\4@\=\12

y

x4

3

3!\+\4@\=\12

y

x4

3

3!\+\4@\=\12


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4 Find the inequality which defines the following regions:

a b c

d e f

A region may be defined in terms of several inequalities involving the variables x and y.

Each inequality may be called a constraint.

We therefore need to be able to graph the region which satisfies all of the constraints.

This provides us with a feasible region of possible solutions called a simplex.

Each corner point of the simplex is called a vertex.

For example, the simplex where

x > 2

y > 3 and

x+ y 6 8

is shaded alongside.

y

x

5

2

y

x2

��

y

x

2

�3

y

x

(2, 2)

y

x

��

��

y

x

2

2

CONSTRAINTS AND FEASIBLE REGIONS

Graph the feasible region (simplex) defined by the following inequalities and mark

the coordinates of each vertex:

x > 0, y > 0, x+ y 6 8 and x+ 2y 6 12.

Example 42

LINEAR

PROGRAMMING

PACKAGE

A, B and C arethe vertices ofthe simplex.

y

x

x y� � 8

x � 2

y � 3

feasible regionor simplex

A

B

C4

4

8

8


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The constraints are: x > 0y > 0

x+ y 6 8x+ 2y 6 12

Boundary lines x-intercept y-intercept

x = 0 y axis

y = 0 x axis

x+ y = 8 8 8x+ 2y = 12 12 6

) the simplex is:

Note: The vertex (4, 4) can be checked using substitution:

x+ y = 4 + 4 = 8 X and x+ 2y = 4 + 8 = 12 X

If a vertex is not a simple lattice point, finding its exact

value by solving simultaneously may be necessary.

When graphing feasible regions, it is a good idea to shade in pencil and reduce the area

shaded as you account for more constraints. For example, the phases in drawing the region

above would be:

In the above example, the feasible region is a bounded region. The feasible region may also

be unbounded as in the next example.

x

y

x y

x y

>

>

6

6

0

0

+ 8

+ 2 12

and

and

and

x

y

x y

>

>

6

0

0

+ 8

and

and

x

y

>

>

0

0

andx > 0

x

y

x

y

x

y

8

8x

y

8

6

812

y

x

10

(0, 6)

(4, 4)

3

3(0, 0) (8, 0)

x y� � 8

x y� �2 12


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Graph the simplex defined for the following inequalities and mark the coordinates of

each vertex: x > 0, y > 0, x+ 4y > 12, 5x+ y > 10, x+ y > 6:

The constraints are x > 0, y > 0, x+ 4y > 12, 5x+ y > 10, x+ y > 6

Boundary lines x-intercept y-intercept

x = 0 y-axis

y = 0 x-axis

x+ 4y = 12 12 35x+ y = 10 2 10x+ y = 6 6 6

) the simplex is

1 Graph the feasible region (simplex) defined by the following inequalities and mark the

coordinates of each vertex (corner point):

a x+ 4y 6 12, x+ y 6 6, x > 0, y > 0

b x+ 2y 6 12, 3x+ 2y 6 24, x > 0, y > 0

c x+ 2y 6 8, 2x+ y 6 13, x > 0, y > 0

d x+ y 6 10, 2x+ y 6 10, x+ y 6 6, x > 0, y > 0

e x+ 2y 6 14, x+ y 6 8, 2x+ y 6 13, x > 0, y > 0.

2 Graph the simplex defined by the following inequalities and mark the coordinates of

each vertex:

a x+ y > 8, x+ 2y > 12, x > 0, y > 0

b 2x+ y > 8, 4x+ y > 12, 3x+ 10y > 30, x > 0, y > 0

c 4x+ y > 8, 2x+ y > 6, 5x+ 18y > 46, x > 0, y > 0

d 3x+ y > 12, 3x+ 2y > 18, x+ 4y > 16, x > 0, y > 0

e 4x+ 3y > 48, 2x+ 3y > 30, x+ 2y > 18, x > 0, y > 0.

EXERCISE 2M.2

Example 43

Simultaneous solution of5 + = 10 and + = 6x y x y

Simultaneous solution of+ 4 = 12 and + = 6x y x y

y

x

(0, 10)

(4, 2)

(12, 0)

(1, 5)

2 6

3

6

5 10x y� �

x y� � 6

x y� �4 12


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3

P

Challenge:

, Q and R are Post Offices within the region.

It is decided that each Post Office will serve

a part of the council region using the Voronoi

solution.

a Find the equations of the boundary limits

of the council region.

b Find the equations of the Voronoi edges.

c Find the Voronoi vertex.

d Find the constraints for the region to be serviced by the Post Office P.

In the previous section feasible regions (simplexes) were graphed, these being defined by a

number of linear inequalities. This process can now be extended to solve problems where

there are restrictions placed on resources or costs.

For example:

for the region illustrated we may be seeking all

ordered pair solutions such that x > 2, y 6 4,

x+ 3y > 6 and 2x+ y 6 12:

There are 14 integer solutions to this problem.

(2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3),

(3, 4), (4, 1), (4, 2), (4, 3), (4, 4), (5, 1),

(5, 2) and (6, 0).

A��

P��

R��

Q��

B�� C��

D��

PROBLEM SOLVING WITH REGIONS

Example 44

4

5

1 4

litre cans of base white paint are converted into lime green or pine green by

adding yellow tint and blue tint in different proportions.

For lime green we add units of yellow tint to unit of blue tint.

For pine green we add unit of yellow tint to units of blue tint.

If units of yellow tint and units of blue tint are available, state inequalities

connecting the number of cans of lime green paint that can be made, and the

number of cans of pine green paint that can be made.

What are the different possible combinations of cans of lime green and pine

green paint that can be made up given the restrictions?

If each can of lime green or pine green paint is to be sold for a profit of

$ , how many of each should be made in order to maximise the profits?

1

15 12

20

x y, ,

a

b

The illustrated region shows the boundary

limits of local council.a

They are:

5431

3

1

6622

44

22

y

x

2!\+\@ = 12

!\+\3@ = 6

! = 2

@ = 4


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Now x > 0, y > 0 fsince there will not be a negative number of cansgUnits of yellow:

x cans of lime green and y cans of pine require 5x+ y units,

but

The boundary line has -intercept and -intercept .5 + = 15 3 15x y x ythere are only 15 units available ) 5x+ y 6 15.

Units of blue:

x cans of lime green and y cans of pine require x+ 4y units,

but

The boundary line has -intercept and -intercept .x y x y+ 4 = 12 12 3there are only 12 units available ) x+ 4y 6 12:

ordered pair lime green pine green

(0, 0) 0 0

(0, 1) 0 1

(0, 2) 0 2

(0, 3) 0 3

(1, 0) 1 0

(1, 1) 1 1

(1, 2) 1 2

(2, 0) 2 0

(2, 1) 2 1

(2, 2) 2 2

(3, 0) 3 0

1

Food Carbohydrate Protein Vitamins

Fight-n-fit 30 units 30 units 100 units

Supalite 10 units 30 units 200 units

EXERCISE 2M.3

15

x

y

3

3

12

Type Number of tins Units of yellow Units of blue

lime green x 5 1pine green y 1 4

15 12

Since we are only interested in integer solutions (we are not part fillingcans), the possible combinations of lime green and pine green paint are:

a

The best possible combination is cans of lime green and cans of pinegreen as cans are made in total and this gives a profit of $ $ .

2 24 4 20 = 80£

b

Each week an athlete must consume at least 170 units of carbohydrate and 1400 units

of vitamins, but no more than 330 units of protein.

Two varieties of special food used by super athletes areavailable. The breakdown of each tin of food is shown.


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If x tins of Fight-n-fit and y tins of Supalite are purchased each week:

a list the inequalities connecting x and y

b graph the feasible region (simplex) defined by these inequalities

c

d

list the possible combinations of cans of Fight-n-fit and Superlite which may be

purchased each week.

2

Vitavim capsules contain 4 units of vitamin A, 1 unit of vitamin B and 1 unit of

vitamin C.

Vegefit capsules contain 2 units of vitamin A, 4 units of vitamin B and 1 unit of

vitamin C.

If John is to consume x capsules of Vitavim and y capsules of Vegefit:

a list the constraints involving x and y

b graph the simplex defined by these constraints

c

d

list the possible combinations of Vitavim and Vegefit capsules that John may take.

3 A manufacturer makes two drawer filing cabinets and bedside units with a single drawer.

The filing cabinet uses two handles and one lock while the bedside unit requires one

handle and two locks. The manufacturer has 18 handles and 22 locks.



c

d

list the possible combinations of furniture items which could be manufactured.

4



c

d

list the possible combinations of round and square piers that may be used in this

foundation.

If Fight-n-fit costs $ per can and Superlite costs $ per can, how many of each

should be purchased each week to minimise costs?

5 3

If Vitavim capsules cost cents each and Vegefit capsules cost cents each,

what combination should John take of each to minimise costs?

10 50

If filing cabinets return a profit of $ each and bedside units return a profit of

$ each, what combination should be manufactured to maximise profits?

10080

If round piers cost $ to use and square piers cost $ to use, what combi-

nation should be used to minimise costs?

30000 20000� �

A a a

a

civil engineer is designing foundation for bridge with round and square concrete

piers. Each round pier gives three units of rigidity and units of load strength. Each

square pier gives six units of rigidity and four units of load strength. The final structure

needs units of rigidity and units of load strength.

If there is room for maximum of piers and represents the number of round piers

and the number of square piers:

12

72 88

16 xy

John’ diet requires him to consume at least units of vitamin and units of vitamineach day However he must not on any day consume more than units of vitamin C.

s A B. ,

28 2112

The manufacturer has contract to supply minimum of items of furniture.If represents the number of filing cabinets and the number of bedside units:

a a 12x y


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1 Find the distance between P(¡4, 7) and Q(¡1, 3).

2 Determine the midpoint of the line segment joining K(3, 5) to L(7, ¡2).

3 Find the equation of the line in slope-intercept form, through:

a (2, ¡1) with slope ¡3 b (3, ¡2) and (¡1, 4)

4 Find the equation of the line in general form, through:

a (1, ¡5) with slope 23 b (2, ¡3) and (¡4, ¡5)

5 Find where the following lines cut the axes:

a y = ¡32x+ 7 b 5x¡ 3y = 12

6 Does (2, ¡5) lie on the line with equation 3x+ 4y = ¡14?

7 Find the equivalent polar coordinates for a point with Cartesian coordinates:

a (5, 2) b (¡1, 3)

8 Find the equivalent Cartesian coordinates for a point with polar coordinates (3, 50o).

9 Find the distance between the points with polar coordinates (4, 30o) and (6, 70o).


a through P(1, ¡5) which is parallel to the line with equation 4x¡ 3y = 6

b through Q(¡2, 1) which is perpendicular to the line with equation y = ¡4x+ 7:

11 State the equation of the line:

a parallel to 3x+ 5y = 7 through (¡1, 3)

b perpendicular to 2x¡ 7y = 5 through (4, 2).

12

a Find the equation of:i East Avenue

ii North Street

iii Diagonal Road.

b Where does Diagonal Road intersecti East Avenue

ii North Street?

13 If 3x+ ky = 7 and y = 3 ¡ 4x are the equations of two lines, find k if:

a the lines are parallel b the lines are perpendicular.

REVIEWNEXERCISE 2N

The Circular Gardens are bounded by East Ave-

nue and Diagonal Road. Diagonal Road intersects

North Street at C and East Avenue at D. Diagonal

Rd is tangential to the Circular Gardens at B.

B��

A��E��

X��

C

D

Diagonal Rd

East Ave

Nort

hS

tr


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14 If 5x¡ 7y = 8 and 3x+ ky = ¡11 are the equations of two lines, find the value

of k for which:

a the lines are parallel b the lines are perpendicular.

15 Find the coordinates of the point where the line through A(¡3, 2) and B(1, 7) meets

the line with equation 3x+ 2y = 6:

16 Determine the nature of the triangle KLM for K(¡5, ¡2), L(0, 1) and M(3, ¡4).

17 A(3, 2) and B(5, 7) are beach shacks and

y = 5 is a power line. Where should the

one power outlet be placed so that it is an

equal distance to both shacks and they pay

equal service costs?

18 Find k given that (¡3, k) is 7 units away from (2, 4).

19 A point T on the y-axis, is 3 units from the point A(¡1, 2). Find:

a the coordinates of T (there are two points T1, T2 say)

b the equation of the line AT1, given that T1 is above T2.

20 If P(x, y) is equidistant from A(¡1, 4) and B(3, ¡2):

a draw a sketch of the possible positions of P

b find the equation connecting x and y.

21 Use midpoints to find the fourth

vertex of the given parallelogram:

22 Use midpoints to find the fourth vertex, K, of parallelogram HIJK for H(3, 4),

I(¡3, ¡1), and J(4, 10).

23 Find the equation of the perpendicular bisector of the line segment joining P(7, ¡1) to

Q(¡3, 5).

24 Two primary schools are located at P(5, 12) and Q(9, 4) on a council map. If the Local

Education Authority wishes to zone the region so that children must attend that school

which is closer to their place of residence, what is the equation of the line which should

form this boundary?

25 The diagram alongside shows the location of three

junior football clubs and the Voronoi diagram of

regions of closest proximity. If children must play

for that club nearest to their place of residence,

find:

a the equations of the lines representing the

boundary divisions

b the coordinates of X.

A(3, 2)

B(5, 7)

y � 5

A( 5, 4)�

D( , 1)� �� C

B(2, 5)

B(6, 7)

C(2, 3)�

A( 2, 5)�X


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26 Given A(2, ¡3) and B(¡1, 5) find X if X divides AB:

a in the ratio 2 : 1 b in the ratio 2 : 5:

27 Given that P, Q and R are collinear, draw an accurate sketch of points P, Q and R such

that:

a P divides QR in the ratio 4 : 1 b R divides PQ in the ratio 4 : 7:

28 A is (¡1, 4) and B is (5, 1) and X lies on AB such that AX : XB = 2 : 1.

Find the equation of the perpendicular to AB through X.


a (¡2, 1) to y = x¡ 5 b (4, 3) to x+ 5y = ¡7

30 Find the distance between the parallel lines:

a y = 3x¡ 2 and y = 3x¡ 7 b 2x¡ y = 4 and 2x¡ y = ¡2

31 A straight rail line passes through two stations with

map references X(2, 5) and Y(8, ¡3) respectively.

A spur line is to be built from a smelting works

P(9, 4) to the line at T. To minimise costs the spur

line PT is to be as short as possible. Find:

a the coordinates of T

b the length of the spur line from P to T given

that the grid reference scale is 1 unit ´ 10 km.

32 Use the quadratic theory method to find the distance from:

a (0, 7) to y = 2x¡ 3 b (4, ¡2) to 2x+ 3y = 2

33 State the coordinates of the centre and find the radius of the circle with equation:

a (x¡ 4)2 + (y + 3)2 = 25 b (x¡ 5)2 + y2 = 7

34 Find the equations of the following circles:

a centre (¡2, 3), radius 4 b centre (4, 0), radiusp

7

35 Find the equation of the circle with ends of diameter A(¡1, 3) and B(1, ¡1).

36 Find the equation of the circle with centre (¡2, 5) and touching the x-axis.

37 Find the equation of a circle with centre (2, ¡1) and passing through the point (5, 0).

38 (4, 1) is one end of the diameter of a circle and the tangent through the point at the

other end of the diameter has equation 3x¡ y = 1. Determine the coordinates of the

centre of the circle.

39 A circle has centre A(4, ¡2) and a tangent at B(¡1, 2). Find:

a the radius b the slope of AB c the equation of the tangent at B

d the equation of the other tangent parallel to the tangent at B.

40 A ship’s radar system has a range of 100 km. If it is located at grid reference (4, 3), will

its radar system detect another ship located at grid reference (7, 6) given that the scale

is 1 unit ´ 20 km?

X(2, 5) P(9, 4)

Y(8, 3)�

T


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41 Find the coordinates of the centre and radius of the following circles:

a x2 + y2 ¡ 4x+ 8y ¡ 3 = 0 b x2 + y2 + 5x¡ 6y = 4

42 Find k given that the circle x2 + y2 ¡ 6x+ 4y + k = 0 has radius 5 units.

43 Find the equation of the chord of the circle x2 + y2 ¡ 10x+ 4y ¡ 11 = 0 which has

midpoint (3, ¡3).

44 Two circles have equations (x¡5)2+(y¡5)2 = 5 and x2+6x+y2¡2y¡35 = 0:

a Find the centre and radius of each circle.

b Show that the circles touch externally.

c Find the coordinates of the point of contact P of the touching circles.

45 Show that circles with equations x2 + y2 ¡ 2x¡ 10y + 6 = 0 and

(x¡ 7)2 + (y ¡ 2)2 = 5 touch each other and determine the coordinates of their point

of contact.

46 Two circles with equations (x¡2)2+(y¡3)2 = 40 and x2+y2¡6x+k = 0 touch

internally. Find k.

47 S is a ship at sea with grid reference (12:6, 2:1). Its

radar system has a range of 100 km. Another ship

X is sailing on a direct route from port P(0, 10) to

port Q(30, 0). Given that 1 grid unit represents 10km, find:

a the equation of the line representing X’s route

from P to Q

b the equation of the boundary of the region detected by the radar screen

(Note: 1 unit ´ 10 km)

c the grid references of the points where ship X enters and exits the radar screen.

48 Determine the equation of the following rectangular hyperbolae:

a b

49 If 3 men can paint a grain silo in 18 days, how long would it take 8 men, working at

the same rate, to paint 4 silos?

50 Find where the line:

a y = x+ 3 meets y =10

xb y = 2x+ 1 meets y =

¡2

x


a y =4

x¡ 2+ 1 b y =

¡3

x+ 2¡ 4

PX

S Q

y

x

��

y

x

��


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52 The number of birds in a breeding enclosure is given by N = 40 +240

12 ¡ t where t is

the number of weeks after the program was originally set up and 0 6 t 6 10.

a How many birds were used to start the breeding program?

b Determine the number of birds present after i 7 weeks ii 10 weeks.

c How long did it take for the number of birds to reach 100?

53

N = 10 +500

t+ 2, t > 0.

a How many koalas were there originally?

b How many koalas were there after i 8 days ii 20 days?

c How long did it take for the number of koalas to reach ?30

d Sketch the graph of N against t.

54 For the simple rational function y =3x+ 2

x¡ 4:

a Write the function in the form y =A

x¡ h + k:


c Find the y-intercept.

d Sketch the graph.

e Use technology to sketch the graph and check your answers to c and d.

55 Find the area of triangle ABC with vertices:

a A(2, 6) B(4, 1) C(5, 3) b A(0, ¡4) B(¡4, ¡7) C(¡2, 5)

56 Find the area of quadrilateral ABCD:


a x > 3 b y < ¡5 c 1 6 x < 4

58 Find the inequality which defines the following regions:

a b c

A��

B��C��

D��

x

y

y

x

3

5

y

x

��

�

y

x

��

��

A a .farmer was trying to eradicate koala colony from his property The number of koalasdays after he started the eradication program is given byt


0 5 25

75

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95

100

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a x+ 2y > 8 b 3x¡ y < 6 c 2x+ 3y 6 9

60 Graph the feasible region (simplex) defined by the following inequalities and mark the

coordinates of each vertex (corner point):

a x+ y > 6, 3x+ 4y 6 24, x > 0, y > 0

b x+ y > 4, x+ y 6 7, 2x+ 5y 6 20, x > 0, y > 0

61 An oil company operates refineries in Adelaide and Brisbane.

If the Adelaide refinery operates for x days and the Brisbane refinery operates for ydays:


b

c

graph the feasible region (simplex) defined by these inequalities.

62 A factory makes gas meters and water meters. Gas meters need 4 gears, 1 dial and 8minutes of assembly time for a profit of $20. Water meters need 12 gears, 1 dial and 4minutes of assembly time to yield a profit of $31. There are 60 gears, 9 dials and 64minutes of assembly time available for use in this production.

If the factory makes x gas metres and y water metres:


b graph the feasible region (simplex) defined by these inequalities

c

c

list the possible combinations of gas metres and water metres which could be pro-

duced.

If the oil company wishes to minimise running costs, for how many days should

each refinery operate?

What production combination gives the maximum profit?

The Adelaide refinery produces barrels of Lead Replacement barrels ofUnleaded and barrels of Diesel each day with running costs of

The Brisbane refinery produces barrels of Lead Replacement barrels ofUnleaded and barrels of Diesel each day with running costs of

The company has orders for barrels of Lead Replacement barrels of

Unleaded and barrels of Diesel.

3000 10001000 8000

1000 20003000 5000

18000 2600030000

,$ .

,$ .

,� ��


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0 5 25

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Date post:	05-Jan-2017
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SA Yr 11 geom & trig (gm)

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