BUKU TEKS MATHEMATICS FORM 3 (DLP)

Chapter 7 Plans and Elevations

MIND TEST 7.2c

1. The diagram below shows a right prism with square ABCD on a horizontal plane. ABNKJGF
is a uniform cross section of the prism.

(a) Draw to full scale,

(i) the plan of the prism. EH M

(ii) the elevation on a vertical plane parallel to F I N 6 cm
AB as viewed from X. 1 cm G L
C
(iii) the elevation on a vertical plane parallel to 5 cm B 5 cm
BC as viewed from Y. D
Y
(b) This right prism was originally a cuboid with a 1 cm J 2 cm K Î
dimension of 5 cm u 5 cm u 6 cm. A right prism A 5 cm
GJKNMLIH has been removed from the cuboid.
Calculate Î

(i) the volume of the removed prism X

(ii) the ratio of the volume of the right prism
GJKNMLIH to the volume of the remaining
right prism

2. The diagram below shows a combination of a right prism and a triangular pyramid on a
horizontal plane. AF and BG are vertical.

(a) Draw to full scale,

(i) the plan of the combined prism

(ii) the elevation on a vertical plane parallel to F
AC as viewed from X
G
(iii) the elevation on a vertical plane parallel to 4 cm E
BD as viewed from Y
3 cm

(b) Measure the lengths of CD, CG and DG on the A D CHAPTER 7
plan, elevation as viewed from X and elevation
as viewed from Y. Î 6 cm B 2 cm ÎY
X
(c) Use another way to calculate the lengths CD, CG
and DG of the original object. Is your answer the C
same as the answer in question (b)? Explain.

(d) Which orthogonal projections show the actual

values of ‘AEF, ‘AFE, ‘BCG, ‘BGC, ‘BCD
and ‘BDC?

191

Dynamic Challenge 5 cm
5 cm
Test Yourself 4 cm

1. The diagram on the right shows a combination of a cylinder
and a cone placed on a horizontal table. State whether the
following statements are true or false regarding the orthogonal
projections of the combined object.
(a) The plan is a circle of diameter 4 cm with a dot in the
centre of the circle.
(b) The elevations of the orthogonal projections from all
directions are congruent.
(c) The length of the hypotenuse of the cone on the side
elevation is less than 5 cm.
(d) There is no curved surface on the front elevation.

2. The diagram below shows the plan and the elevation of a combined object. Describe the original

design of the combined object. 1 cm

4 cm

1 cm 0.5 cm
2 cm Elevation
3 cm

Plan

CHAPTER 7 Skills Enhancement

1. The diagram on the right shows a right prism with D ÎF 3 cm E
square ABHG placed on a horizontal plane. ABCD is
a uniform cross section of the prism. G C 4 cm
(a) Draw to full scale, A
(i) the plan of the prism H
Î 5 cm Y
(ii) the elevation as viewed from X X B

(iii) the elevation as viewed from Y

(b) Measure the length of AD and size of ‘ A DC on
the projection plane which is the uniform cross
section of the prism.

192

Chapter 7 Plans and Elevations

2. The diagram on the right shows a combination of a H
right prism and a cuboid placed on a horizontal plane. G
AD, FG, BC and KJ are vertical.
(a) Draw to full scale, 3 cm J
(i) the plan of the object 2 cm
F E
(ii) the elevation as viewed from X K
4 cm I
(iii) the elevation as viewed from Y 6 cm Y
DC Î
(b) Calculate the volume, in cm3, of the combined
solid. A 4 cm B
X
Î

3. The diagram on the right shows the plan of a 9 cm 7 cm
combination of a cuboid and a right cylinder.
If the heights of the cuboid and the cylinder are
5 cm, calculate the volume of the combined
solid in cm3.

6 cm

4. The diagram on the right shows the plan of a
combination of a cube and a semi-cylinder. Given that
the circumference of the plan of the semi-cylinder is
11 cm and the height of the semi-cylinder is equal
to the length of the side of the cube, calculate the
volume of the combined solid, in cm3.

5. The diagram below shows the plan and the front elevation of a prism. Calculate the volume of CHAPTER 7
the prism in cm3.

3 cm 8 cm 4 cm
3 cm Plan 6 cm
Front elevation

193

Self Mastery

1. The diagram on the right shows a right prism E 6 cm H
L 1 cm
with rectangle ABCD placed on a horizontal 5 cm I
G
plane. AF, BK and JG are vertical. J
C
(a) Draw to full scale, F
Y
(i) the plan of the prism 4 cm K
D

(ii) the elevation on a vertical plane parallel Î

to AB as viewed from X AÎ 3 cm B
X
(iii) the elevation on a vertical plane parallel
to BC as viewed from Y

(b) The object should be reinforced so that the base of the object is equal to the shape of the
plan. What is the volume of the new object to be added?

(c) If the cost of 1 cm3 of the new object is RM2.20, calculate the total cost to build the entire
combined object.

2. The diagram below shows the plan, front elevation and side elevation of a hollow cuboid. The
hollowed section is a right cylinder.

(a) Sketch the three-dimensional shape of the object.

(b) Calculate the volume of the object.

Side elevation Front elevation G/F
G/H F/E H/E

5 cm 5 cm

CHAPTER 7 B/A 2 cm C/D A/D 1 cm 2 cm 1 cm B/C
0.5 cm 0.5 cm E/D F/C

45°

3 cm

H/A 4 cm G/B
Plan

194

Chapter 7 Plans and Elevations CHAPTER 7

S T EM

Aim: To build a study hut.
Instructions:
1. Carry out this activity in groups.
2. Propose the construction of a covered study hut in

an area of 5 m × 5 m.
3. Your proposal should consider the following criteria:

(a) Maximum use of sunlight during the day.
(b) Good air circulation.
(c) Eco-friendly and conducive.
(d) Minimum construction cost.
4. Prepare a report using multimedia applications.

P ROJ EC T

My dream house
1. Draw your dream house with an appropriate scale using scale drawing.
2. Draw the plan, front elevation and side elevation of the house.
3. Build a model of your dream house based on the scale drawing, the plan and the elevations

drawn.
4. Decide the building materials required from various

sources based on the size of the house that you want
to build.
5. Calculate the estimated cost to build your dream
house.
6. Show your model house and present your project.

195

CONCEPT MAP

Plans and elevations

Orthogonal projections Plans and elevations

Object

Object

CHAPTER 7 XÎ Y

ÎHorizontalElevation from YElevation from X
plane

45°

Plan

SELF-REFLECT

At the end of this chapter, I can:

1. Draw orthogonal projections.

2. Compare and contrast between objects and the corresponding orthogonal
projections.

3. Draw the plan and elevations of an object to scale.

4. Synthesise the plan and elevations of an object and sketch the object.

5. Solve problems involving plans and elevations.

196

Chapter 7 Plans and Elevations

EXPLORING MATHEMATICS

1. Prepare 15 pieces of cubes with the side
of 5 cm as shown in the diagram. You can
also use the Rubik’s cube.

2. Use the cubes to form a combined object 5 cm
according to your creativity.

3. Draw the plan and elevations of the
combined object that you have created.

4. The group that builds the most creative combined object is the winner.

CHAPTER 7

197

CHAPTER Loci in Two

8 Dimensions

What will you learn? < <<

8.1 Loci

8.2 Loci in Two Dimensions

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‡ .QRZOHGJH DERXW ORFL DOORZV RQH WR HVWLPDWH RU

predict the path of the moving points based on
certain conditions.
‡ 7KH FRQFHSW RI ORFL LV XVHG LQ FRQVWUXFWLRQ
engineering drawings, aviation, satellite
movements and so on.

National badminton champion Datuk Lee
Chong Wei currently holds the record for
the fastest smash since September 2015 when he
did the shot with a speed of 408 kilometres per
hour (km/h). He won the 2015 Hong Kong Open
Badminton Championships which was held at the
Hung Hom Coliseum. According to Badminton
World Federation (BWF), the speed of the shot
made by Chong Wei was recorded and measured
using Hawk Eye technology that has been adopted
in several major tournaments since September 2015.
Do you know that the movement of a shuttlecock
follows certain conditions?

198

EExxpplloorriinngg EErraa

Apollonius (260 – 190 BC) was an ancient Greek
mathematician who was very interested in studying
problems of loci. He had conducted research on
various forms of loci such as the straight lines and
certain curves. However, the most outstanding
Greek mathematician in studying loci was Pappus
(290 AD – 350 AD). Pappus’s loci materials are
still being researched by mathematicians today.

http://bukutekskssm.my/Mathematics/F3/
ExploringEraChapter8.pdf

< WORD B A N K

‡ equidistant ‡ EHUMDUDN VDPD
‡ circle ‡ EXODWDQ
‡ arc ‡ OHQJNRN
‡ curve ‡ OHQJNXQJ

‡ locus ‡ ORNXV

‡ loci ‡ ORNXV ORNXV

‡ perpendicular bisector ‡ SHPEDKDJL GXD

VDPD VHUHQMDQJ

‡ angle bisector ‡ SHPEDKDJL GXD
VXGXW

199

8.1 Locus

What do you understand about loci? LEARNING
STANDARD
In the picture on the right, a piece
of coloured sticker is pasted on the Recognise loci in real
tyre of a bicycle. life situations and hence
explain the meaning of
What is the shape generated locus.
by the sticker when the bicycle is
pedalled?

The shape formed by the sticker is a circle as shown in the diagram on the
right. Does this shape comply with certain conditions?

The picture below shows a ball being kicked by a football player. The BULLETIN
movement of a point on the ball yields a curve.
Malaysia’s football fans
will always remember
‘Super Mokh’, the late
Datuk Mokhtar Dahari,
ZKR ¿UHG D VROLG
40-metre shot against
England Squad 3 in
1978.

CHAPTER 88 The picture on the right shows a rocket being launched.
The movement of a point on the rocket will produce a
straight line.

A locus is a trace or trajectory formed by a set of points in a plane or three-dimensional space
WKDW VDWLV¿HV FHUWDLQ FRQGLWLRQV
200

Chapter 8 Loci in Two Dimensions

Brainstorming 1 In groups

Aim: To identify two-dimensional loci in daily life situations.

Materials: Situation cards.

Steps:

1. Each group is given several situation cards that show activities involving movements in
daily activities as shown below.

Situation A Situation B

Throwing a ball into the net. A durian falling from
Situation C a tree.

Situation D

An airplane landing. The moving tip of the wiper on
the windshield.

2. Discuss in the group and sketch the locus of a point on the object involved in the given CHAPTER 8
situations.

3. Present the loci sketch and compare your answers with other groups.

Discussion:
'LVFXVV ¿YH RWKHU PRYHPHQWV LQ GDLO\ DFWLYLWLHV WKDW FDQ EH FDWHJRULVHG DV ORFL

From Brainstorming 1, it is found that:
The shapes of two-dimensional loci can be seen in the form of straight lines, arcs and curves.
201

Example 1 Ɣ
Point C is drawn on a blade of a revolving fan as shown in the C
diagram. Elaborate and sketch the locus of point C.

Solution:
This locus is a circle.

Ɣ C

MIND TEST 8.1a
1. Explain and sketch the locus of point C on each object in the following diagrams.

(a) A ball centred at C rolling along an (b) Point C on a swinging pendulum.
inclined plane.

Ɣ
C

C Ɣ

(c) Point C on a spinning yo-yo. (d) Point C on the shoe of a child who is
playing on a slide at the playground.

CHAPTER 88 C C

2. State and sketch the locus of a point on
(a) a coconut falling from a tree
(b) a moving car on a straight road
(c) a leaping frog

202

Chapter 8 Loci in Two Dimensions

Example 2 N The side PQ is
RQ being rotated
The diagram on the right shows a pole MN. A rectangular Board 360° around the
board PQRS is attached to the pole where PQRS is pole MN.
movable. If the side PQ is rotated 360° around MN, what Pole
is the three-dimensional shape formed?
SP
Solution: M

N FLASHBACK

Q Cylinders, spheres,
cones, prisms and
The shape formed when the side PQ is pyramids are examples
rotated 360° around pole MN is a right of three-dimensional
cylinder. shapes.

P

M

Example 3 N
The diagram on the right shows a pole MN. A semicircular R
board PQR is attached to the pole where PQR is movable.
If PQR is rotated 360° around MN, what is the Board Q The side PQR
three-dimensional shape formed? is being rotated
P 360° around the
Solution: M pole MN.

N BULLETIN

The shape formed when the semicircular This three-
board is rotated 360° around pole MN is dimensional
a sphere. shape is
known as a
M frustum.

MIND TEST 8.1b CHAPTER 8

1. Sketch the three-dimensional loci when the two-dimensional shaded shapes are rotated 360°
around pole ST.

(a) T (b) T (c) T (d) T

S SS S

203

8.2 Loci in Two Dimensions

:KDW LV WKH ORFXV RI SRLQWV WKDW DUH RI FRQVWDQW GLVWDQFH IURP D ¿[HG SRLQW"

Brainstorming 2 In pairs LEARNING
STANDARD
Aim: To determine the locus of points that are of constant distance
IURP D ¿[HG SRLQW Describe the locus of
points that are of constant
Materials: Blank paper, a pencil and a ruler. GLVWDQFH IURP D ¿[HG
point.

Steps:

1. 0DUN D ¿[HG SRLQW O on a sheet of paper (Diagram 1).

2. Measure 5 cm from the point O and mark =.

3. Repeat step 2 as many times as possible (Diagram 2).

= = = = ==
== = =
== = =
= =
5 cm = ==
O == O =
= =
== =
=
= =
=
=
=

Diagram 1 Diagram 2

4. Note the location of the points marked with = (Diagram 2).

5. 5HSHDW VWHSV WR ZLWK GLIIHUHQW GLVWDQFHV IURP WKH ¿[HG SRLQW O.

6. Are the resulting geometric shapes the same as the shape obtained in step 4? Explain.

Discussion:
What is the geometric shape generated by the location of the dots =? Explain.

CHAPTER 88 From Brainstorming 2, it is found that:

3RLQWV PDUNHG DW WKH VDPH GLVWDQFH IURP D ¿[HG SRLQW O forms a circle.

In general, The locus of a point that is equidistant from a ¿[HG SRLQW is a circle centred
204 DW WKDW ¿[HG SRLQW.

Chapter 8 Loci in Two Dimensions

Brainstorming 3 In pairs

Aim: 7R FRQVWUXFW ORFXV RI SRLQWV WKDW DUH RI FRQVWDQW GLVWDQFH IURP D ¿[HG SRLQW

Materials: Dynamic software

Steps:

1. Start with 1HZ 6NHWFK.

2. Select &RPSDVV 7RRO and draw a circle.

3. Select 3RLQW 7RRO and mark.

4. Open 'LVSOD\ menu and select 7UDFH 3RLQW
followed by $QLPDWH 3RLQW.

5. Observe the animation of the movement
generated.

Discussion:
What is the geometric shape generated from the movement of the marked point?

From Brainstorming 3, it is found that:
$ SRLQW WKDW DOZD\V PRYHV DW WKH VDPH GLVWDQFH IURP D ¿[HG SRLQW IRUPV D FLUFOH

+RZ GR \RX FRQVWUXFW D ORFXV RI SRLQWV WKDW DUH RI FRQVWDQW GLVWDQFH IURP D ¿[HG SRLQW"

Example 4 CHAPTER 8
Construct a locus of point P ZKLFK LV DOZD\V FP IURP D ¿[HG SRLQW O.

Solution: Locus of 3
1. Mark point O.
O
2. Set the gap of the compasses at 3 cm. 3 cm
3. Construct a circle of radius 3 cm centred at the point O.

205

:KDW LV WKH ORFXV RI SRLQWV WKDW DUH HTXLGLVWDQW IURP WZR ¿[HG SRLQWV"

Brainstorming 4 In pairs LEARNING
STANDARD
Aim: To determine the locus of points that are equidistant from two
¿[HG SRLQWV Describe the locus
of points that are of
Materials: Plain paper, a compasses, a ruler and a pencil. HTXLGLVWDQW IURP WZR ¿[HG
points.

Steps:

1. 0DUN WZR ¿[HG SRLQWV P and Q which are 8 cm apart (Diagram 1).

2. Using the compasses, mark the intersection, 4.5 cm from point 3 and point Q (Diagram 2).

3. Repeat step 2 with distances more than 4.5 cm from point P and point Q (Diagram 3).

8 cm 4.5 cm QP Q
PQ from P
and Q.

P

Diagram 1 Diagram 2 Diagram 3

4. Note the location of the intersecting marks in Diagram 3. Scan the QR code or visit
http://bukutekskssm.my/
5. Repeat steps 1 to 3 with different distances between point P Mathematics/F3/Chapter8
and point Q. LocusfromTwoFixedPoint.
Are your answers the same as the answer in step 4? mp4 to watch a video that
describes the locus of the
Discussion: points that are equidistant
What is the shape formed by the location of the intersecting IURP WZR ¿[HG SRLQWV.
marks? Explain.

CHAPTER 88 From Brainstorming 4, it is found that:

7KH ORFDWLRQ RI WKH LQWHUVHFWLQJ PDUNV WKDW DUH HTXLGLVWDQW IURP ¿[HG SRLQWV P and Q form a
straight line through the midpoint of PQ.

In general, The locus of a point that is equidistant from WZR ¿[HG SRLQWV is the
perpendicular bisector RI WKH OLQH FRQQHFWLQJ WKH WZR ¿[HG SRLQWV

206

Chapter 8 Loci in Two Dimensions

Brainstorming 5 In pairs FLASHBACK

Aim: 7R FRQVWUXFW ORFXV RI SRLQWV WKDW DUH HTXLGLVWDQW IURP WZR ¿[HG The line AB is known as a
points. bisector.

Materials: Dynamic software B
Steps:
1. Start with 1HZ 6NHWFK. P Ɣ
ƔR

2. Select 6WUDLJKWHGJH 7RRO to draw a line segment. Select A
7H[W 7RRO to label point A and point B.

3. Select &RQVWUXFW menu to construct the midpoint of Locus which is
the line segment. equidistant from
point A and point B
4. Mark both lines and midpoint segments with 6HOHFWLRQ
$UURZ 7RRO.

5. Select &RQVWUXFW menu to construct a perpendicular A Ɣ Ɣ B
line (Diagram 1).

Discussion: Diagram 1
What is the geometric shape formed? Explain.

From Brainstorming 5, it is found that:

The locus that is equidistant IURP WZR ¿[HG SRLQWV A and B is a straight line perpendicular
to the straight line $% and it passes through the midpoint of AB.

+RZ GR \RX FRQVWUXFW WKH ORFXV RI SRLQWV WKDW DUH HTXLGLVWDQW IURP WZR ¿[HG SRLQWV"
Example 5
Construct the locus of point P WKDW LV HTXLGLVWDQW IURP WZR ¿[HG SRLQWV 0 and 1.

0 1

Solution: Locus
of P
1. Mark two small arcs using a pair of compasses with the gap set 0 CHAPTER 8
at more than half of the length of 01 from the point 0. 1

2. With the compasses set at the same gap, mark the intersecting Q
arcs of point 1.

3. Connect the two points of intersection with a straight line.

Example 6

The diagram on the right shows an equilateral triangle PQR. Determine P R
the locus of point X that is equidistant from point P and point R.

207

Solution: Q
Locus
Locus of point X that is equidistant from point P and point R is the of X
perpendicular bisector of the line connecting point P and point R.
PR

What is the locus of points that are of constant distance from a straight line?

Brainstorming 6 In pairs LEARNING
STANDARD
Aim: To determine the locus of points that are of constant distance
from a straight line. Describe the locus of
points that are of constant
Materials: Square grid paper, a ruler, a pencil. distance from a straight
line.
Steps:
1. Draw a straight line 01 (Diagram 1).

2. Mark a point =, which is 3 units from the line 01 (Diagram 2).

}= = = = = = = = =
3 units
}0 1 0 1 0 1
3 units
= = = = = = = = =

Diagram 1 Diagram 2 Diagram 3

3. Repeat step 2 with as many = points as possible (Diagram 3).

4. Note the location of the = points in Diagram 3. What do you think about the location of
the = points?

5. Repeat steps 1 through 4 with a different unit distance.

6. Repeat steps 1 through 4 with the straight line 01 drawn vertically.

Discussion:

What is your conclusion about the location of the points marked equidistantly from the straight
line?

CHAPTER 88 From Brainstorming 6, it is found that:

The locus of points that are equidistant from the line 01 is a pair of straight lines parallel
to 01.

In general, The locus of points that are of constant distance from a straight line are
208 straight lines parallel to that straight line.

Chapter 8 Loci in Two Dimensions

Example 7

The diagram on the right shows a line AB drawn on a square grid with

sides of 1 unit. On the diagram, draw the locus of the point X which always

moves at 3 units from the line $%. A B

Solution: Locus
of X
The locus of point X moving 3 units
from the line AB is a pair of lines parallel B
to AB and 3 units from AB.

A

Locus
of X

What is the locus of points that are equidistant from two parallel lines?

Brainstorming 7 In groups LEARNING
STANDARD
Aim: To determine the locus of points that are equidistant from two
parallel lines. Describe the locus of
points that are equidistant
Materials: Plain paper, compasses, a ruler and a pencil. from two parallel lines.

Steps:
1. Draw two straight lines PQ and 01 that are parallel (Diagram 1).
2. Using compasses, mark the point of intersection from point P and point 0.
3. Repeat steps 2 for point Q and point 1 (Diagram 2).
4. Connect all the intersection points marked by drawing a straight line (Diagram 3).

P QP QP Q

0 10 10 1

Diagram 1 Diagram 2 Diagram 3 CHAPTER 8

5. Describe the nature of the straight line that connects all the points of intersection
(Diagram 3).

Discussion:
1. Repeat steps 1 to 4 by drawing two vertical straight lines and two inclined straight lines.

Ensure that each pair of lines is parallel.

2. Do you get the same result as in step 4?

209

From Brainstorming 7, it is found that:

(a) The locus of points that are equidistant from two parallel lines PQ and 01 is a straight
line.

(b) The locus is parallel to the straight lines PQ and 01 and it passes through the midpoints
of the lines PQ and 01.

In general, The locus of points that are equidistant from two parallel lines is a straight
line parallel to and passes through the midpoints of the pair of parallel
lines.

Example 8 A B
D
The diagram on the right shows the rectangle, ABCD drawn on a C
square grid with sides of 1 unit. Describe and draw the locus of X B
which is equidistant from the lines AB and DC.
Locus
Solution: A of X

The locus of point X that is equidistant from the line AB
and DC is a line parallel to AB and DC and is 3 units from
the lines AB and DC.

DC

What is the locus of points that are equidistant from two intersecting lines?

Brainstorming 8 In groups LEARNING
STANDARD
Aim: To determine the locus of points that are equidistant from two
intersecting lines. Describe the locus
of points that are of
Materials: A square grid paper, a ruler, a pencil and a protractor. equidistant from two
intersecting lines.

CHAPTER 88 Steps:
1. Draw [-axis and \-axis on a Cartesian plane on the grid paper (Diagram 1).

2. Mark the coordinates of equal value pairs. For example, (0, 0), (–2, –2), (4, 4) and so on
(Diagram 2).

3. Connect all the points with a straight line. Measure D, E, F and G using a protractor
(Diagram 3).

210

Chapter 8 Loci in Two Dimensions

\ \ \

4 4= 4=
2 =
=
– 4 –2 O 2=
–2 = 2=
–4 E=
– 4 –2 =O= 2 4 D
[ = –2 [ F =O 2 [
24
= –4 –2 =G 4

= –4 = –2

=

= –4

Diagram 1 Diagram 2 Diagram 3

Discussion:

1. What is your conclusion about the values of D, E, F and G which are the angles
formed at the intersection of the [-axis and \-axis?

2. What is the relationship between the straight line that connects equal value pairs of
coordinates to the values of D, E, F and G?

From Brainstorming 8, it is found that:

(a) D = E = F = G = 45$.

(b) The straight line that connects equal value pairs of coordinates bisects the angle of
intersection between the [-axis and \-axis.

In general, The locus of points that are equidistant from two intersecting lines
is the angle bisector of the angles formed by the intersecting lines.

How do you construct a locus of points that are equidistant from two intersecting lines?

Example 9 Q

Construct the locus of point X that is equidistant from two straight P
lines PQ and 31 intersecting at P.

Solution: 1 CHAPTER 8

1. By using a pair of compasses, draw an arc from the point P P Q Locus
which cuts through the straight lines PQ and 31. A1 of ;
A2 1
2. Mark the points of intersection between the arc and the straight
lines PQ and 31 as A1 and A2 respectively.

3. Construct intersecting mark from A1 and A2.

4. Draw a straight line joining the intersecting mark constructed
in step 3 and the point P.

211

Brainstorming 9 In pairs

Aim: To construct locus of a point that is equidistant from two intersecting straight lines.

Materials: Dynamic software

Steps:
1. Start with 1HZ 6NHWFK.
2. Select 6WUDLJKWHGJH 7RRO to draw lines AB and BC intersecting at point B.
3. Use 7H[W 7RRO to label point A, followed by point B and then point C (point of intersection

must be marked on the second turn).

4. Mark all three points A, B and C with 6HOHFWLRQ $UURZ 7RRO. (Diagram 1)
5. Select the &RQVWUXFW menu to construct the bisector of the angle ($QJOH ELVHFWRU)

between the two intersecting lines. (Diagram 2)

ƔA ƔA

BƔ BƔ

ƔC ƔC

Diagram 1 Diagram 2

Discussion:

What is your conclusion about the locus of points that are equidistant from two intersecting
lines?

From Brainstorming 9, it is found that:

The locus of a point that is equidistant from the two straight lines AB and BC intersecting
at the point B is a straight line that bisects ABC.

Example 10 A B
C
CHAPTER 88 The diagram on the right shows a square ABCD. Describe and draw the locus
of a point which moves at the same distance from the straight lines AB and AD.

Solution: A D
B
The locus of a point which moves at the same distance
from the line AB and AD is a straight line which bisects
the angle BAD.

DC

212

Chapter 8 Loci in Two Dimensions

MIND TEST 8.2a

1. The diagram shows a straight line PQ of 5 cm. 3 Q
5 cm

(a) X is a point that is always 3 cm from point P. Describe the locus of point X completely.
(b) Y is a point that is always 4 cm from point Q. Describe the locus of point < completely.

2. The diagram on the right shows a square ABCD drawn on a square A E B

grid with sides of 1 unit. P, Q, R, S and T DUH ¿YH SRLQWV WKDW PRYH LQ

the square ABCD. Using the letters in the diagram, state the locus of

point HF
(a) P moving equidistantly from points A and D

(b) Q moving equidistantly from points B and D

(c) R moving such that it is always 4 units from the line BC DC
(d) S moving equidistantly from the straight lines AB and BC G

(e) T moving such that it is always 4 units from the line EG

3. The diagram on the right shows the straight line, CD & 6 cm D
which is 6 cm long. 7 is a point that is always 1.5 cm from
the straight line, CD.

(a) Draw the locus of point T.

(b) Describe completely, the locus of point T.

4. Construct the locus of point Y for a given situation.

(a) It is equidistant from the (b) <& = <' (c) 34< = RQY
straight lines PQ and PR.
P
& R
Q

3

R D Q

5. The picture on the right shows a running track. An athlete CHAPTER 8
always practises by running two lanes away from lane 4
of the track. Draw the locus of the athlete’s run.

213

+RZ GR \RX GHWHUPLQH WKH ORFXV WKDW VDWLV¿HV WZR RU PRUH FRQGLWLRQV"

The intersection of two or more loci can be determined by constructing LEARNING
HDFK VSHFL¿HG ORFXV LQ WKH VDPH GLDJUDP STANDARD

Example 11 A Determine the locus that
VDWLV¿HV WZR RU PRUH
The grid on the right shows a square ABCD drawn on a square conditions.
grid with sides of 1 unit. Points X and Y are two points that move
inside the square ABCD. On the grid, B

(a) draw the locus of a moving point X which is constantly 7
units from A

(b) draw the locus of a moving point Y which is equidistant
from the lines AB and CD

(c) mark all points of intersection of locus of X and locus of Y
with the symbol €

DC

Solution:

AB
Locus of X
Locus of Y

Intersection of the locus of
X and locus of Y.

DC

CHAPTER 88 Example 12 P Q R
W 0 S
The diagram on the right shows four combined squares V T
with sides of 2 cm. X and Y are two moving points inside the U
square PRTV. On the diagram,

(a) draw the locus of a moving point X which is always 2 cm
from point 0

(b) draw the locus of a moving point Y which is equidistant from
line PR and line 39

(c) mark all points of intersection of locus of X and locus of Y
with the symbol €

214

Chapter 8 Loci in Two Dimensions

Solution:

P QR
Locus of X

W 0S
Locus of Y

VU T

MIND TEST 8.2b AB

1. In the grid on the right, the rectangle ABCD DC
represents a part of a lake. ABCD is drawn on a
square grid with sides of 1 unit. Points V and W \ ƔF
represent the trips of boat V and boat W. On the
grid, 6 CHAPTER 8

(a) draw the locus of boat V which always ƔE
moves 5 units from point D
4
(b) draw the locus of boat W which is 3 units
from line %& ƔG

(c) mark the intersection of the paths of boat V 2
and boat W with the symbol 8

2. The diagram on the right shows the Cartesian
plane marked with four points E, F, G and H.
Faruk is at the same distance from [-axis and
\-axis. Faruk’s location is also less than 5 units
from the centre of O. Which of the points E, F,
G and H is Faruk’s location?

Ɣ[

O 2 4H

215

3. The diagram on the right shows the Cartesian plane. y
Point F always moves 3 units from the x-axis while 4
point G always moves 4 units from the origin. Mark 3
all the points of intersection between the locus of F 2
and the locus of G with the symbol €. 1

O 1 2 34 x

How do you solve problems involving loci? LEARNING
STANDARD
Example 13
A clinic will be built in a village. The clinic should be equidistant Solve problems involving
from house P and house Q, as well as 600 metres away from the loci.
highway AB. Determine the possible location of the clinic.
(scale 1 cm = 600 metres)

Ɣ Q
Ɣ P

AƔ ƔB

Solution: Highway

CHAPTER 8 Understanding the problem

The clinic is equidistant from P and Q. Therefore the locus is the bisector of the straight line
connecting points P and Q. The clinic is 600 metres from the highway AB. There are two lines
parallel to the highway AB.

Planning a strategy
To draw using a pair of compasses and a ruler.

216

Chapter 8 Loci in Two Dimensions

Implementing the strategy Making a conclusion

Ɣ House Q Two locations that are marked with the
symbol 8 satisfy the requirements of
the location to build the clinic.

House P Ɣ Location of clinic

600 m 8
AƔ
600 m
600 m
Highway ƔB
600 m

8

Location of clinic

MIND TEST 8.2c

1. The diagram on the right shows square PQRS with P Q
sides of 6 cm. Two semicircles with centres M and M ON
N are drawn inside square PQRS. M and N are the
midpoints of PS and QR. On the diagram, shade the
UHJLRQ WKDW VDWLV¿HV WKH IROORZLQJ FRQGLWLRQV.

(a) The locus of point X which moves such that
XM р 3 cm and more than 3 cm from line SR.

(b) The locus of Y which moves such that
YM у 3 cm and YN у 3cm.

(c) Describe the intersection between locus of X and S 6 cm R

locus of Y.

2. The diagram on the right shows a rectangular P Q
IHQFHG XS JUDVV ¿HOG PQRS measuring 8m
6 m = 8 m. A goat is tied at point Q with a CHAPTER 8
7-metre long rope.

Shade the region that is reachable by the goat.

S R
6m
217

3. Khalid draws the plan for a treasure hunt Jalan Bahagia 1 cm
on a square grid with a scale of 1 cm to 1 1 cm
metre.
On the diagram, draw P

(a) the location of the treasure if it is 3 m
DZD\ IURP WKH ÀDJSROH P

(b) the location of the treasure if it is 5 m
from Jalan Bahagia

Then, mark the possible locations of the
treasure with the symbol €.

Dynamic Challenge

Test Yourself

1. The diagram below shows an equilateral triangle ABC. S is a point on line AB. X and Y are two
moving points in the diagram. On the diagram,
(a) draw the locus of point X such that AX = AS
(b) draw the locus of point Y such that Y is equidistant from AC and BC
(c) mark all the intersection points for locus of X and locus of Y with the symbol 8

A ƔS B

CHAPTER 8 C

2. The diagram below shows a regular pentagon MNPQR. X and Y are two moving points inside
the pentagon. On the diagram,
(a) draw the locus of point X such that RX = XN
(b) draw the locus of point Y such that RY = RQ
(c) mark all the intersection points for locus of X and locus of Y with the symbol 8

M

RN

QP

218

Chapter 8 Loci in Two Dimensions

3. The picture below shows the triangular forest area PQR. Point X and point Y are two loci that
describe the location of a helicopter that crashed. On the diagram,
(a) draw the locus of point X such that it is equidistant from lines QR and QP
(b) draw the locus of point Y such that YP = PR
(c) mark the possible location for the helicopter with the symbol 8

R

QP

Skills Enhancement

1. The diagram on the right is drawn on a square grid with sides of 1 unit. Point X, point Y and

point Z are three points which move in the square. Q
(a) X is a point which moves such that it is equidistant P R

from points Q and C. Using the letters in the diagram,

state the locus of point X.

(b) On the diagram,

(i) draw the locus of point Y which moves such that A BS
it is equidistant from the straight lines PD and
DT

(ii) draw the locus of point Z which moves such that
it is always 5 units from point S

(c) Mark the location of all the intersection points for D CT
locus of Y and locus of Z with the symbol 8.

2. The diagram below shows a rhombus MNOP. Point X and point Y are two points that move CHAPTER 8
within the rhombus. On the diagram,
(a) draw the locus of point X which moves equidistantly from the straight lines PM and PO
(b) draw the locus of point Y which moves such that YP = PO
(c) mark the location of all the intersection points for locus of X and locus of Y with the
symbol 8

MN

PO

219

Self Mastery

1. The diagram below shows two semicircles, PKLT and QNMS centred at R, with diameters of
8 cm and 4 cm respectively. KNR and RML are arcs of circles centred at P and T respectively.

KL

VI

I N Ɣ III Ɣ M V

II IV

P QR S T

Based on the diagram above, state

(a) the point which is 2 cm from R and 4 cm from P

(b) the point which is more than 2 cm from R and 4 cm from T

(c) the location of a moving point X in the diagram such that it is less than 4 cm from P and
more than 2 cm from R

(d) the location of a moving point Y in the diagram such that YR Ͻ 2 cm and YP Ͻ 4 cm.
(e) the location of a moving point Z in the diagram such that ZT Ͼ 4 cm, ZP Ͼ 4 cm and

ZR Ͼ 2 cm

2. In the diagram below, SLMQ, PKLR, QNKS and RMNP are arcs of circles with radii of 4 cm
and centred at P, Q, R and S respectively.

S R

I Ɣ II
L

ƔK III M Ɣ

IV N V
Ɣ

CHAPTER 8 PQ

Based on the diagram above, state

(a) the location of a moving point X in the diagram such that XS Ͻ 4 cm, XP Ͻ 4 cm and
XQ Ͼ 4 cm

(b) the location of a moving point Y in the diagram such that YR Ͼ YP

(c) the location of a moving point Z in the diagram such that ZP Ͻ 4 cm, ZQ Ͻ 4 cm, ZR Ͻ 4 cm
and ZS Ͻ 4 cm

220

Chapter 8 Loci in Two Dimensions CHAPTER 8

3. The diagram below shows a square PQRS with sides of 4 cm and a circle centred at O with
radius of 1 cm. Point X and point Y are two points that always move inside the square PQRS.

PQ
AB
O
DC

SR
4 cm

Describe the possible movement of the loci of point X and point Y for the following points of
intersection:
(a) B and D
(b) A and C

P ROJ EC T
When we look at a clock to tell the time, we can see that the tip of the hour hand always
moves with the same pattern, that is always equidistant from the centre of the clock.

Why is the shape of a circle selected to represent the movement of time on a clock? Gather
information about the relationship between hours, minutes and seconds and the shape of a
circle. Create a report with illustrations using multimedia applications.

221

CONCEPT MAP

Loci in Two Dimensions
A set of points whose locations satisfy certain conditions.

The locus of The locus of The locus of The locus of The locus of
points that are points that are points that are points that are points that are
equidistant from equidistant from
of constant equidistant of constant two parallel two intersecting
distance from a IURP WZR ¿[HG distance from a
lines. lines.
¿[HG SRLQW points. straight line.

Circle Perpendicular A pair of A straight Angle
bisector parallel lines line bisector
Locus (a) Locus
Locus Locus Locus
(b)

Locus Locus

CHAPTER 8 SELF-REFLECT

At the end of this chapter, I can:
1. Recognise loci in real life situations and hence explain the meaning of locus.
2. 'HVFULEH WKH ORFXV RI SRLQWV WKDW DUH RI FRQVWDQW GLVWDQFH IURP D ¿[HG SRLQW
3. 'HVFULEH WKH ORFXV RI SRLQWV WKDW DUH HTXLGLVWDQW IURP WZR ¿[HG SRLQWV
4. Describe the locus of points that are of constant distance from a straight line.
5. Describe the locus of points that are equidistant from two parallel lines.
6. Describe the locus of points that are equidistant from two intersecting lines.
7. 'HWHUPLQH WKH ORFXV WKDW VDWLV¿HV WZR RU PRUH FRQGLWLRQV
8. Solve problems involving loci.

222

Chapter 8 Loci in Two Dimensions

EXPLORING MATHEMATICS

We can sketch an ellipse using the following steps:

1. Tie two nails with a string (one nail on each end of the string).

2. 3ODFH D VKHHW RI SDSHU RQ D ÀDW SLHFH RI ERDUG

3. Fix the two nails onto the piece of board but do not pull the string too tightly. The two
nails are called the foci.

4. We can begin to sketch an elliptical shape by using the tip of a pencil to pull the string
WLJKWO\ DQG GUDZ RQ WKH SDSHU ¿UVW IURP RQH QDLO WR WKH VHFRQG QDLO DQG WKHQ IURP WKH
VHFRQG QDLO EDFN WR WKH ¿UVW QDLO 7KH FXUYH GUDZQ E\ WKH SHQFLO LV DQ HOOLSVH

5. Observe the elliptical shape formed.

1 Nail Nail 3

String 4

2

SMART MIND CHAPTER 8

Why is an ellipse also
known as a locus?

223

CHAPTER Straight Lines

9

What will you learn?
9.1 Straight Lines

WWhhyy ddoo yyoouu lleeaarrnn tthhiiss cchhaapptteerr??
‡ 7KH FRQFHSW RI VWUDLJKW OLQHV LV ZLGHO\ XVHG LQ

WKH FRQVWUXFWLRQ RI YDULRXV JHRPHWULF VKDSHV
such as squares, triangles and kites.
‡ 7KH FRQFHSW RI VWUDLJKW OLQHV LV XVHG LQ
HQJLQHHULQJ DUFKLWHFWXUH FRQVWUXFWLRQ PDSSLQJ
VFLHQFHV VSRUWV DQG VR RQ

Normally, every building is built vertically.
Some buildings such as the Leaning Tower
of Teluk Intan which was built in 1885, became
inclined due to the soil structure. Although inclined
and over 100 years old, the Leaning Tower of Teluk
Intan is still standing strong and is a landmark of
Teluk Intan. The leaning tower was declared a
national heritage in 2015.

224

EExxpplloorriinngg EErraa

Euclid was a Greek mathematician. He had
conducted a lot of research about straight lines and
geometry such that he was known as the founder of
geometry.
7KH ¿HOG RI JHRPHWU\ ZDV QDPHG (XFOLGHDQ
Geometry to commemorate Euclid’s contributions
WR WKH IXQGDPHQWDO SULQFLSOHV RI JHRPHWU\.

KWWS EXNXWHNVNVVP P\ 0DWKHPDWLFV )
([SORULQJ(UD&KDSWHU SGI

WORD B A N K

‡ straight line ‡ JDULV OXUXV

‡ SDUDOOHO OLQH ‡ JDULV VHODUL

‡ YHUWLFDO GLVWDQFH ‡ MDUDN PHQFDQFDQJ

‡ KRUL]RQWDO GLVWDQFH ‡ MDUDN PHQJXIXN

‡ JUDGLHQW ‡ NHFHUXQDQ

‡ D[LV ‡ SDNVL

‡ LQWHUFHSW ‡ SLQWDVDQ

‡ VLPXOWDQHRXV ‡ SHUVDPDDQ VHUHQWDN
equation

‡ LQWHUVHFWLRQ SRLQW ‡ WLWLN SHUVLODQJDQ

225

9.1 Straight Lines

What is the equation of a straight line? LEARNING
STANDARD
,Q )RUP \RX KDYH OHDUQW KRZ WR GUDZ WKH JUDSK RI OLQHDU IXQFWLRQ
and non-linear functions by constructing a table of values of related Make connection between
functions. the equation, y = mx + c,
and the gradient and
yyyy y-intercept, and hence
make generalisation about
OxOxOxOx the equation of a straight
line.

'LDJUDP 'LDJUDP 'LDJUDP Diagram 4

(DFK RI WKH DERYH JUDSK LV GUDZQ EDVHG RQ D VSHFL¿F IXQFWLRQ 7KH IXQFWLRQ LV DOVR DQ HTXDWLRQ
IRU WKH UHODWHG JUDSK

&DQ \RX GL൵HUHQWLDWH D JUDSK RI OLQHDU IXQFWLRQ DQG D JUDSK RI QRQ OLQHDU IXQFWLRQ" 'LVFXVV

Brainstorming 1 In groups FLASHBACK

Aim: 7R GHWHUPLQH WKH UHODWLRQVKLS EHWZHHQ HTXDWLRQ \ P[ F with The gradient, m, of a
gradient and y LQWHUFHSW straight line that connects
two points (x1 , y1) and
Materials: *UDSK SDSHU OLQHDU IXQFWLRQ FDUGV (x2 , y2)
P = –y–2––––y–1
Steps:
1. *HW LQWR IRXU JURXSV x2 – x1
2. (DFK JURXS LV JLYHQ D FDUG ZULWWHQ ZLWK WZR OLQHDU IXQFWLRQV or
P = – –y–-i–n–te–r–c–e–p–t

x-intercept

*URXS *URXS *URXS *URXS


y x + 6 y = 2x + 6 y = 5x – 10 y = 4x – 8
y = –2x – 4
y = – 4x + 8 y ± x y = –2x + 2

CHAPTER 9 3. &RPSOHWH WKH WDEOH RI YDOXHV EHORZ IRU HDFK JLYHQ IXQFWLRQ

x ± ± ±
y

4. %DVHG RQ WKH WDEOH RI YDOXHV GUDZ WKH JUDSKV RI WKH IXQFWLRQV.

226

Chapter 9 Straight Lines

5. )URP WKH JUDSK RI WKH IXQFWLRQ FDOFXODWH WKH JUDGLHQW DQG VWDWH WKH y LQWHUFHSW.
6. &RPSDUH WKH YDOXHV RI JUDGLHQW DQG \ LQWHUFHSW IURP WKH JUDSK ZLWK WKH YDOXHV LQ WKH

function card.
Discussion:
1. &RPSDUH \RXU ¿QGLQJV LQ VWHS ZLWK OLQHDU IXQFWLRQ \ P[ F. What is your conclusion"
2. 3UHVHQW \RXU ¿QGLQJV $UH \RXU ¿QGLQJV WKH VDPH DV WKH RWKHU JURXSV¶ ¿QGLQJV"

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW

(a) )RU D OLQHDU IXQFWLRQ, \ P[ F P is the gradient and F is the y LQWHUFHSW RI WKH VWUDLJKW
line.

E 7KH JUDSK RI OLQHDU IXQFWLRQ, \ P[ F is a straight line.

In general, Equation of straight line
y = mx + c

Gradient y LQWHUFHSW

Brainstorming 2 In pairs

Aim: 7R SURGXFH D JUDSK RI OLQHDU IXQFWLRQ

Materials: Dynamic software

Steps:
1. Start with 1HZ VNHWFK
2. Select JUDSK icon
3. Select SORW QHZ IXQFWLRQ and enter the required equation of straight line (Diagram 1).

7KH ¿UVW JUDSK RI VWUDLJKW OLQH \ = 2[ ± Diagram 1 CHAPTER 9

227

4. &OLFN VWUDLJKWHGJH WRRO DQG PDUN WZR SRLQWV RQ WKH FRQVWUXFWHG JUDSK RI VWUDLJKW OLQH

5. &OLFN PHDVXUH and then click VORSH (Diagram 2).
7KH JUDGLHQW YDOXH ZLOO EH GLVSOD\HG 'LDJUDP

Diagram 2 'LDJUDP

6. 5HSHDW VWHSV WR WR GUDZ DQG GHWHUPLQH WKH JUDGLHQW RI WKH JUDSK RI VWUDLJKW OLQH IRU
function \ –2[ + 8.
(Diagram 4)

K(x) = –2[ + 8
J(x) = 2[ ±

7KH VHFRQG JUDSK RI VWUDLJKW Diagram 4 TIPS
OLQH y = –2x + 8
Relationship between the
value m and the form of a
straight line graph.
If m Ͼ 0

y

7. Straight lines that are parallel to the x-axis and y-axis.

$ GLVSOD\HG H[DPSOH RI VWUDLJKW OLQHV VXFK DV

(a) y = 4 (b) x = 6

x

y y= 4 = 4 If m Ͻ 0
y

CHAPTER 9 x = 6x = 6 x

228

Chapter 9 Straight Lines

Discussion:

1. &RPSDUH WKH IRUPV RI JUDSK UHVXOWLQJ IURP G\QDPLF VRIWZDUH ZLWK WKH IRUPV RI JUDSK IURP

Brainstorming 1.

2. 0DNH D FRQFOXVLRQ IRU WKH YDOXHV RI P and F of the equation of straight line in the form

\ P[ F 'LVFXVV WKH VKDSH RI WKH JUDSK ZKHQ

(a) P LV SRVLWLYH (b) P is negative

(c) SDUDOOHO WR x D[LV (d) SDUDOOHO WR y D[LV

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW

D 7KH JUDSK RI OLQHDU IXQFWLRQ \ P[ F is a straight line.
E 7KH JUDSK RI IXQFWLRQ \ K LV D VWUDLJKW OLQH SDUDOOHO WR x D[LV
(c) 7KH JUDSK RI IXQFWLRQ [ K LV D VWUDLJKW OLQH SDUDOOHO WR y D[LV

Example 1

Determine the gradient and y LQWHUFHSW RI WKH VWUDLJKW OLQH SMART MIND

(a) y = 2x E y = –2x + 12 What is the y-intercept of
a straight line that passes
Solution: through the origin?

D &RPSDUH y = 2x ZLWK y = P[ F; QU I Z
P = 2 and F
Thus, gradient = 2 and y LQWHUFHSW What is the value of the
gradient for the straight
E *LYHQ y = –2x + 12 line
(a) y = x
· ² \ ± 2² [ 1² 2 'LYLGH E\ VR WKDW WKH (b) y = –x

FRH൶FLHQW RI y is +1. BULLETIN

In the equation
&RPS\D =UH– y—= 2 –[ —+2 4x + 4 with \ P[ F; y = mx + c, WKH FRHI¿FLHQW
of y is +1.
P = – —2 and F = 4


Thus, gradient = – —2 and y LQWHUFHSW


Example 2

State the value of K IRU WKH VWUDLJKW OLQH JUDSK EHORZ 6WDWH UHDVRQV IRU \RXU DQVZHU

(a) y (b) y

[ = K

(–2, 6) 6 (0, 6) P \ = K 8 P(4, 8)
Q

4 (4, 4) CHAPTER 9

–2 O x Q(4,0)
O 4x

229

Solution: (b) K = 4 because the straight line x = 4 is
always 4 units from the y D[LV
(a) K = 6 because the straight line y = 6
is always 6 units from the x D[LV

MIND TEST 9.1a

1. Determine the gradient and y LQWHUFHSW RI WKH IROORZLQJ VWUDLJKW OLQHV

(a) y x + 5 (b) y = 2x – 7 (c) y = –x + 4
(f) – 4y = –2x + 5
(d) 2y = 8x H y = –x + 18

2. State the value of N and of K IRU HDFK VWUDLJKW OLQH JUDSK JLYHQ

(a) y (b) [ = N y [ = K

4 \ = N

± O 4x

Ox
–2 \ = K

What is the relationship between the equations of straight LEARNING
STANDARD
lines in the form ax + by = c, —x + —y = 1 and y = mx + c?
a b Investigate and interpret
the equations of straight
Brainstorming 3 In groups lines in other forms such
as ax + by = c and
Aim: 7R GHWHUPLQH WKH UHODWLRQVKLS EHWZHHQ WKH HTXDWLRQV RI VWUDLJKW —ax + —by = 1, and change
F —D[ —\
lines in the form of D[ E\ = + E = 1 and \ P[ F to the form of y = mx + c,
and vice versa.

Materials: *UDSK SDSHU VWUDLJKW OLQH HTXDWLRQ FDUGV

Steps:
1. *HW LQWR IRXU JURXSV
2. (DFK JURXS LV JLYHQ D FDUG ZLWK WKUHH HTXDWLRQV RI D VWUDLJKW OLQH ZULWWHQ RQ LW

*URXS *URXS *URXS *URXS

CHAPTER 9
4x – 2y = –8 ± x + 4y = –12
2x y = 6 —(–x2—) + —4y = 1 —4x + — ±y— = 1 –x – 4y = 4
— x + —2y = 1 —(–x4—) + —(–y—1) = 1
y = – — 2 x + 2 y = 2x + 4 y = — 4 x ±
y = – —14 [ – 1

230

Chapter 9 Straight Lines

3. 'HWHUPLQH WKH FRUUHVSRQGLQJ YDOXH RI y when x = 0 and the TIPS
FRUUHVSRQGLQJ YDOXH RI x when y = 0 for each equation.
A straight line graph can
([DPSOH x y = 6 When y be drawn by plotting at
When x 2[ = 6 least two points.
x0 y = 6
y 2x = 6 TIPS
2 0 y = 6 x
y=2 Gradient = – —y-i—nt—erc—ep—t
x-intercept
4. 'UDZ D VWUDLJKW OLQH JUDSK IRU HDFK HTXDWLRQ.
5. )URP WKH JUDSK VWDWH WKH x LQWHUFHSW DQG y LQWHUFHSW DQG

GHWHUPLQH WKH JUDGLHQW RI WKH JUDSK.

Discussion:
1. :KDW LV \RXU FRQFOXVLRQ DERXW WKH UHODWLRQVKLS EHWZHHQ WKH x LQWHUFHSW ZLWK WKH y LQWHUFHSW

DQG WKH JUDGLHQWV RI WKH WKUHH VWUDLJKW OLQH JUDSKV"
2. :KDW LV \RXU FRQFOXVLRQ DERXW WKH UHODWLRQVKLS EHWZHHQ WKH HTXDWLRQV RI VWUDLJKW OLQH LQ

GL൵HUHQW IRUPV"

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW

(a) The values of x LQWHUFHSW DQG y LQWHUFHSW DQ G WKH JUDGLHQW IRU WKHVH WKUHH VWUDLJKW OLQHV DUH WKH
same.
(b) Equations of straight line in the forms of D[ + E\ = F, —D[ + —E\ = 1 and y = P[ + F SURGXFH WKH
VDPH VWUDLJKW OLQH JUDSK LI WKH YDOXHV RI x LQWHUFHSW DQG y LQWHUFHSW DUH WKH VDPH

In general,

Straight line equation can also be written in the form of ax + by = c and —xa + —y = 1; D 
and E  b

How do you change the equation of straight line in TIPS
—ax —by
the form of ax + by = c, + = 1 to the form of For the straight line
y = mx + c and vice versa? —[ + —y = 1,
D E

Example 3 D = x-intercept
a&nKdDQ\J H PWK[H HT FX.DWLRQ RI VWUDLJKW OLQH EHORZ WR WKH IRUP RI ²Dx ²Ey E = \ -intercept

y

(a) 2x y E x – 5y = 15 E y
Solution: E
x + =1
D

(a) 2x y = 12 O Dx

(i) 2x y = 12 Divide by (ii) 2x y = 12
812 —122x– + —1 2y– = —1122 12 to get
a value y = –2x + 12 'LYLGH E\ VR CHAPTER 9
—6[ + —4y = 1 of 1. 8 ² y± ²–2±x 1²2 WKDW WKH FRH൶FLHQW
of y is +1.

y = – —2 x + 4


231

E x – 5y = 15 DISCUSSION CORNER

L x – 5y = 15 LL x – 5y = 15 Among the three forms
–5y ± x + 15 of equation of straight
8 15 —1 5x– – —155y– = —1155 line that you have
8 (–5) —(––55–y)– = —(±– —5x) + —(–155—) learned, which is the
—5[ – —\ = 1 easiest form to know
y = — 5 x ± the gradient value, the
y-intercept value and
Example 4 the x-intercept value of a
straight line? Discuss.

&KDQJH WKH HTXDWLRQ RI VWUDLJKW OLQH EHORZ WR WKH IRUP RI D[ E\ F and \ P[ F.

(a) —[ + —\ = 1 (b) – —[ + —\ = 1


Solution:

(a) —[ + —\ = 1 (b) – —[ + —\ = 1 TIPS

( i) — [ + — \ = 1 – 4x + 2y = 8
(i) – —2[ + —4\ = 1 can also be written as
— x –+ — 6 –y = 1 Equate the —–4–2x—(+4)–2—y = 1 4x – 2y = – 8
denominator. –4x + 2y = 1(8)
–4x + 2y = 8 QU I Z
x + 6y = 1(18) –2x + y = 4
x + 6y = 18 What is the value of the
x + 2y = 6 gradient for the straight

(ii) —x + —\ = 1 (ii) – —2x + —4\ = 1 line – —2x – —2\ = 1?


—y = – —x + 1 —4y = —2x + 1 0XOWLSO\ E\
4 so that the
= ² y ± ²–x² ± = 4 —44y– = —x(24—) + 1(4)
y = – —21 x y = 2x + 4 FRH൶FLHQW y
is +1.

Example 5

&KDQJH WKH HTXDWLRQ RI WKH IROORZLQJ VWUDLJKW OLQHV WR WKH IRUP RI D[ E\ F and ²D[ ²E\ 1.
(a) y = –2x + 8 (b) y x + 6

Solution: (b) y x + 6

(a) y = –2x + 8 (i) y x + 6
(i) y = –2x + 8 ± x + y = 6
2x + y = 8

CHAPTER 9 (ii) y = –2x + 8 (ii) y x + 6
2x + y = 8 ± x + y = 6
8 6 —±6 —[ + —6y = —66
8 8 —28[– + —8y = —88
– —2[ + —6\ = 1
—4[ + —8\ = 1

232

Chapter 9 Straight Lines

MIND TEST 9.1b

1. Write the equation of the following straight lines in the form of —Dx + —\ = 1 and \ P[ F.
E

D [ – 4y = 24 (b) 7[ + 2y = 28 (c) 5[ ± y = 15 (d) –2[ y

2. Write the equation of the following straight lines in the form of D[ E\ = F and \ P[ F.

(a) —x + ²\ = 1 ( b) – —[ + — \ = 1 (c) — [– + —\ = 1 (d) —2[– – —\ = 1

—D[ —\
3. Write the equation of the following straight lines in the form of D[ E\ = F and + E = 1.

(a) \ = 2x + 6 (b) \ x – 12 (c) \ = –x + 5 (d) \ = –2x – 4

What is the relationship between the points on a straight LEARNING
line and the equation of the line? STANDARD

Diagram 1 and Diagram 2 show two straight lines drawn on a Investigate and make
&DUWHVLDQ SODQH EDVHG RQ WKH HTXDWLRQ RI VWUDLJKW OLQHV x + 2y = 4 and inference about the
x – y ± relationship between the
points on a straight
line and the equation of
the line.

y y [ ± \ ±
R
2 2
1 R(1, 2) Q(–1, 2) 1
P(–2, 1)
± O Q(2, 1)
–1 ± ± ± O x
P(4, 0) –1
x


[ 2\ 4

Diagram 1 Diagram 2

([DPLQH WKH SRVLWLRQ RI SRLQWV P, Q and R in Diagram 1 and Diagram 2. What can you say about
WKH SRLQWV P, Q and R and the straight line drawn"

Left Right

x + 2y = 4
(a) Diagram 1 }

(i) 6XEVWLWXWH SRLQW P(4, 0) }

CHAPTER 9
(ii) 6XEVWLWXWH SRLQW Q(2, 1) (iii) 6XEVWLWXWH SRLQW R(1, 2)

/HIW 5LJKW /HIW 5LJKW /HIW 5LJKW

x + 2y =4 x + 2y =4 x + 2y =4
= 4 + 2(0) = 2 + 2(1) = 1 + 2(2)
=4 Equal =4 Equal =5 Not Equal

233

(b) Diagram 2 Left Right

x – y ±
CHAPTER 9
(i) 6XEVWLWXWH SRLQW P(–2, 1) (ii) 6XEVWLWXWH SRLQW Q(–1, 2) (iii) 6XEVWLWXWH SRLQW R
}
/HIW 5LJKW /HIW 5LJKW /HIW 5LJKW
}x – y ± x – y ± x – y ±

± ± ± ± ±

± Equal ± Equal ± Not Equal

)URP WKH DERYH DFWLYLW\ LW LV IRXQG WKDW

(a) 3RLQWV RQ D VWUDLJKW OLQH RU SRLQWV WKDW WKH VWUDLJKW OLQH SDVVHV WKURXJK ZLOO VDWLVI\ WKH HTXDWLRQ
of a straight line.

(b) Points that do not lie on a straight line will not satisfy the equation.

Example 6

1. 'HWHUPLQH ZKHWKHU SRLQW P lies on the given straight line.

(a) \ [ + 2 3 E [ ± 2\ = 12, P(– 4, 2)
(c) — [ + — \ 1, P(6, –2) (d) 2\ = –5[ ± 7, P

Solution: E [ ± 2\ = 12, P(– 4, 2)
(a) \ [ + 2 3

} } }}
Left Right }}Left Right

/HIW 5LJKW /HIW 5LJKW

[ + [ ± 2\ = 12

± 4) – 2(2)

=8 = –16

Equal Not Equal

Thus, P(2, 8) lies on the Thus, P(– 4, 2) does not lie on the
straight line \ [ VWUDLJKW OLQH [ ± 2y = 12

(c) —[ + —\ = 1 , P(6, –2) (d) 2\ = –5[ – 7, P
Right
Left Right
Left
}

}

/HIW 5LJKW /HIW 5LJKW

— [ + — \ =1 2\ –5[ –7

± ±
= (—6 ) + (—– 2–) =6 = –27

Not Equal

= 1 Equal Thus, P GRHV QRW OLH RQ WKH
straight line 2\ ±5x – 7.
Thus, P(6, –2) lies on the

straight line — [ + — y = 1.


234

Chapter 9 Straight Lines

Example 7

7KH GLDJUDP VKRZV D VWUDLJKW OLQH x + 5y = 15. Given that O is the y
origin, determine the value of
N P(2, q)
(a) K (b) N (c) T (d) gradient of the straight line
S olu tion: x + 5y = 15 x
OK
(a) K is the x LQWHUFHSW (b) N is the y LQWHUFHSW
Thus, \ = 0 Thus, [ = 0 TIPS
x + 5y = 15 x + 5y = 15
N) = 15 Ƈ For points that lie
K) + 5(0) = 15 on x-axis its value of
K = 15 5N = 15 y-coordinate is zero.
K = 1— 5 N = 1—55
Ƈ For points that lie on
K = 5 N y-axis its value of
x-coordinate is zero.
(c) 3 (2, q) LV D SRLQW RQ WKH (d) Gradient of the straight
straight line x + 5y = 15. line x + 5y = 15 FLASHBACK
Thus, P = – —yx —LLQQWW—HHUUFF—HHSS—WW
Gradient, P
x + 5y = 15 Gradient = – — 5 P = – —y-—int—er—ce—pt–
q) = 15
x-intercept
6 + 5T = 15
5T = 15 – 6 SMART MIND
5q
Do the coordinates
T = — 5 (–3, –3) lie on the
straight line \ = [?

MIND TEST 9.1c

1. 'HWHUPLQH ZKHWKHU WKH IROORZLQJ SRLQWV OLH RQ WKH VWUDLJKW OLQH y = 2x + 16.

(a) M(– (b) N(1, 18) (c) P(–8, 0) (d) Q(–5, 8)

2. 'HWHUPLQH ZKHWKHU WKH IROORZLQJ SRLQWV OLH RQ WKH VWUDLJKW OLQH x y = 12.

(a) M(0, 4) (b) N ± F P(15, – 6) (d) Q(– 4, 8)

3. 'HWHUPLQH ZKHWKHU WKH IROORZLQJ SRLQWV OLH RQ WKH VWUDLJKW OLQH ²2[ ² y

(a) M(2, 0) (b) N(–2, 12) (c) P ± G Q(0, 6)

4. The diagram shows two straight lines, \ = [ + 2 and y
2[ \ 6
2x y = 6. Given that O is the origin, determine the \ [ + 2

value of

(a) K (b) N (c) Q (0, K) CHAPTER 9

NO x
Q

235

What do you understand about the gradients of parallel LEARNING
lines? STANDARD
You have learnt that the gradient of a straight line is the ratio of vertical
Investigate and make
GLVWDQFH WR KRUL]RQWDO GLVWDQFH DQG WKH FRUUHVSRQGLQJ DQJOHV RI WKH inference about the
SDUDOOHO OLQHV DUH HTXDO. gradients of parallel lines.

Brainstorming 4 In pairs

Aim: 7R GHWHUPLQH WKH UHODWLRQVKLS EHWZHHQ JUDGLHQWV RI VWUDLJKW OLQHV ZLWK SDUDOOHO OLQHV.
Steps:

1. ([DPLQH WKH VWUDLJKW OLQHV EHORZ WKDW ZHUH GUDZQ EDVHG RQ WKH HTXDWLRQ RI D VWUDLJKW OLQH
with the same gradient of P = 2.

y y y
\ 2[ 2 \ 2[ 4 \ 2[

4 4%

2

θ x θ x θ x
–1 O –2 O y O2

Diagram 1 Diagram 2 'LDJUDP
y

\ 2[ ± 2 \ 2[ ± 4

O1 x O2 x
–2 θ

Diagram 4 –4 θ
Diagram 5

2. Based on Diagram 1 to Diagram 5, calculate the value θ.


4 4 4 FLASHBACK
eθ
2 θe 2 2 θe y
eθ 2 'LDJUDP eθ 2
1 1 θ
Diagram 1 Diagram 2 Diagram 4 Diagram 5 x

CHAPTER 9 tan θ = —12 tan θ = —yx
θ ƒ

3. Are the values of θ IRU WKH ¿YH GLDJUDPV WKH VDPH"

236

Chapter 9 Straight Lines

4. 7KH JUDSKV LQ 'LDJUDP WR 'LDJUDP DUH FRPELQHG DV EHORZ TIPS

y y = 2x + 4 PQ and RS are parallel
y = 2x + 2 lines because they have
corresponding angles.
4 y = 2x
y = 2[ – 2 QS

2 y = 2[ – 4 ĮĮ

x

–2 –1 12 PR
–2

–4

Discussion:
1. What is the connection between the values of θ ZLWK WKH ¿YH VWUDLJKW OLQHV DERYH"
2. Are the straight lines y = 2x + 4, y = 2[ + 2, y = 2[ \ = 2x – 2 and y = 2x – 4 SDUDOOHO" :K\"
3. :KDW DUH WKH FRQQHFWLRQV EHWZHHQ WKH JUDGLHQWV DQG WKH SDUDOOHO OLQHV"
4. $UH \RXU ¿QGLQJV WKH VDPH DV WKRVH RI WKH RWKHU JURXSV"

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW

The straight lines y = 2x + 4, y = 2x + 2, y = 2x, y = 2x – 2 and y = 2x – 4 DUH SDUDOOHO EHFDXVH
they have the same gradient, that is P DQG WKH VDPH FRUUHVSRQGLQJ DQJOH WKDW LV ƒ

In general, Straight lines WKDW KDYH WKH VDPH JUDGLHQWV DUH SDUDOOHO

Example 8

Determine whether the straight line y x LV SDUDOOHO WR WKH VWUDLJKW TIPS
line 6x – 2y

Solution: To determine the gradient

y x + 5 value of a straight line,
&RPSDUH ZLWK y = P[ F
*UDGLHQW 6x – 2y change the equation of
–2y = –6x the given straight line to
the form \ = P[ + F.

—––22—y = —(––6—2[) + —(– 2—)

y x – —2 CHAPTER 9

*UDGLHQW

Equal

The gradients of both straight lines are equal, thus \ x LV SDUDOOHO WR x – 2y

237

Example 9

Determine whether the straight line y x LV SDUDOOHO WR WKH VWUDLJKW OLQH y x ±

Solution: 6y x ± DISCUSSION CORNER
\ = — 6x– – —6
y x + 8 Will two parallel lines
&RPSDUH ZLWK y = P[ + F \ = — 2 x – — 2 intersect? Discuss.
*UDGLHQW Gradient = —21

Not equal

The gradients of both straight lines are not equal. So, \ x + 8 LV QRW SDUDOOHO WR 6y x ±

Example 10

Given that the straight line 4x y LV SDUDOOHO WR WKH VWUDLJKW OLQH x + K\ = 2, calculate the
value of K.

Solution:

,I ERWK VWUDLJKW OLQHV DUH SDUDOOHO WKHQ WKH JUDGLHQWV DUH HTXDO

)RU x y = 18 )RU x + K\ = 20 )RU
y = – 4x + 18 – —4 = – —2
K\ = –2x + 20
y = – — 4 x + 6 K
Gradient = – —4 y = – —K2 x + —20 K = 2 × — 4
K K = — 2
Gradient = – —K2

MIND TEST 9.1d

1. 'HWHUPLQH ZKHWKHU WKH IROORZLQJ SDLUV RI VWUDLJKW OLQHV DUH SDUDOOHO

D y = –6x DQG y + 2x = 14 (b) 2x y DQG x + 6y = 12

(c) y = 2x + 1 and 8x – 4y = 5 (d) y ± x DQG x + 2y = 12

2. Determine the value of N IRU HDFK RI WKH IROORZLQJ SDLUV RI SDUDOOHO OLQHV

(a) y ± x + 4 and y + N[ = 14 (b) N[ + 2y = 7 and 6x + 2y = 15

(c) 8y = 5x + 1 and N[ ± y = 8 G x + N\ = 4 and 2x + y

3. 7KH GLDJUDP RQ WKH ULJKW VKRZV D SDUDOOHORJUDP S y \ = 1–2[ + 6 R
PQRS. Given that the straight line PQ LV SDUDOOHO K[ + y = 10
to SR and the straight line PS LV SDUDOOHO WR 45 2\ = 5x – 4
and O is the origin, calculate the values of K and N
CHAPTER 9



P 6 \ = N[ + 12 Q x
O

238

Chapter 9 Straight Lines

How do you determine the equation of a straight line? LEARNING
STANDARD
The equation of a straight line \ P[ F can be determined by the
IROORZLQJ VWHSV Determine the equation
1 Determine the value of gradient, P. of a straight line.

2 'HWHUPLQH D SRLQW ZKLFK WKH VWUDLJKW OLQH SDVVHV WKURXJK RU D SRLQW ZKLFK OLHV RQ WKH VWUDLJKW
line.

3 Substitute the gradient, P, the x-coordinate and y FRRUGLQDWH IURP WKH SRLQW LQWR WKH HTXDWLRQ
\ P[ F to determine the value of F, that is, the y LQWHUFHSW

x-coordinate value

y-coordinate value \ P[ F

Gradient value

4 Substitute the gradient value and y LQWHUFHSW YDOXH VSHFL¿HG LQ WKH HTXDWLRQ RI WKH VWUDLJKW OLQH
\ P[ F.

Determine the equation of a straight line when the gradient and a point on the straight
line are given

Example 11
'HWHUPLQH WKH HTXDWLRQ RI D VWUDLJKW OLQH ZLWK D JUDGLHQW RI ²12 DQG SDVVHV WKURXJK SRLQW P(6, 8).
Solution:

P = —2൫ Given P(6, 8), thus x = 6, y = 8

Substitute the values of P [ and y into y = P[ F to determine the value of F.

8 = —12 (6) + F QU I Z

F Determine the equation
of a straight line with a
F ± gradient of 0 and passes
F=5

Therefore, the equation of the straight line is y = —12 x + 5. through point P(1, 5).

MIND TEST 9.1e

1. 'HWHUPLQH WKH HTXDWLRQ RI D VWUDLJKW OLQH ZLWK WKH JLYHQ JUDGLHQW DQG SDVVHV WKURXJK SRLQW P
given.

(a) Gradient = 2, P E Gradient ± P(– 6, 4) CHAPTER 9
(c) Gradient = — 2 , P(12, 5) (d) Gradient = – — 1 , P(4, – 6)

Determine the equation of a straight line that passes through two points

:KHQ WZR SRLQWV RQ D VWUDLJKW OLQH DUH JLYHQ WKH JUDGLHQW RI WKH VWUDLJKW OLQH FDQ EH FDOFXODWHG
Hence the equation of the straight line can be determined.

239

Example 12 TIPS

'HWHUPLQH WKH HTXDWLRQ RI D VWUDLJKW OLQH WKDW SDVVHV WKURXJK SRLQW You can also substitute
P ± DQG SRLQW Q(2, –7). the value of point Q,
where x = 2 and y = –7
Solution: and P = – 4
into y = P[ + F to calculate
P = — – ±—7 –—± 5– = — – — 12 – = —–1 —2 = – 4 the value of c and thus
)RU SRLQW P(–1, 5), x = –1, y = 5. determine the equation of
the straight line.
Substitute the value of P [ and y into \ P[ F to determine the
value of F. QU I Z

5 = (– 4)(–1) + F Determine the equation of
5=4+F a straight line that passes
F = 5 – 4 through the points P(– 4, 4)
F = 1 and Q(5, –5).
Therefore, the equation of the straight line is y = – 4x + 1.

Example 13 y S
M(0, 4) A(2, 4)
The diagram on the right shows straight lines PQ P
and RS. Given that straight line PQ LV SDUDOOHO WR Q
the x D[LV DQG VWUDLJKW OLQH RS LV SDUDOOHO WR WKH
y D[LV GHWHUPLQH O N(2, 0) x
(a) the equation of the straight line PQ R

(b) the equation of the straight line RS

Solution:

(a) Gradient of straight line PQ with (b) Gradient of straight line 56 with A(2, 4)

A(2, 4) and M(0, 4) and N(2, 0).

P = —24 —–– 04– = —20 = 0 P ²24 ––²20± ²04 8QGH¿QHG
y LQWHUFHSW = 4
Thus, the equation of the straight line PQ is The gradient of the straight line RS LV XQGH¿QHG
y = 0(x) + 4 and is always 2 units from the y D[LV
y=4 Hence, the equation of the straight line RS is

x = 2.

MIND TEST 9.1e

CHAPTER 9 1. 'HWHUPLQH WKH HTXDWLRQ RI WKH VWUDLJKW OLQH WKDW SDVVHV WKURXJK WKH JLYHQ SDLU RI SRLQWV.

(a) K(0, 2), L(6, 0) (b) R(–2, 0), S(0, 8) (c) T ± U(5, 7)
(d) G(– 4, –2), H(8, 6) (e) M ± N(1, 5) (f) P ± Q(4, – 6)

240