2 The table below shows the values of the population of a type of bacteria recorded at 1 hour intervals. The
variables x and y are related by the equation y = pqx, where p and q are constants.
Jadual di bawah menunjukkan nilai-nilai bagi populasi sejenis bakteria yang direkod setiap 1 jam. Pemboleh ubah x dan y
dihubungkan oleh persamaan y = pqx, dengan keadaan p and q ialah pemalar.
Number of hours /Bilangan jam, x 1 23456
Population /Populasi, y 78 98 123 158 199 251
(a) Construct a table for the values of x and log10 y. Give your answer correct to two decimal places.
Bina satu jadual bagi nilai-nilai x and log10 y. Beri jawapan anda betul kepada dua tempat perpuluhan.
(b) Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 0.25 unit on the log10 y-axis, plot a graph of
log10 y against x. Hence, draw the line of best fit.
Menggunakan skala 2 cm kepada 1 unit pada paksi-x dan 2 cm kepada 0.25 unit pada paksi-log10 y, plotkan graf log10 y
melawan x. Seterusnya, lukis garis lurus penyuaian terbaik.
(c) Using the graph in (b), find the value of
Menggunakan graf di (b), cari nilai
(i) p,
(ii) q,
(iii) the expected number of population of the bacteria after 24 hours.
anggaran jumlah populasi bakteria selepas 24 jam.
(a) x 123456
log10 y 1.89 1.99 2.09 2.20 2.30 2.40
(b)
log10 y
2.50
2.25
2.00
1.75
1.50
1.25
1.00
0.75
0.50
0.25
x
0 123456
(c) y = pqx (ii) Gradient = log10 q
log10 y = log10 (pqx) 2.46––11.89 = log10 q
log10 y = log10 p + log10 qx
log10 y = log10 p + xlog10 q 0.102 = log10 q
log10 y = (log10 q)x + log10 p q = 100.102
(i) log10 p = 1.80
q = 1.265
p = 63.1
(iii) When x = 24
y = (63.1)(1.265)24
= 17 791
101
Review 6
Paper 1 Questions
1 The variables x and y are related by the equation xy
xy = 12x – 4x3. A straight line graph is obtained by
plotting y against x2 as shown in Diagram 1. (p, 17)
Pemboleh ubah x dan y dihubungkan oleh persamaan
xy = 12x – 4x3. Graf garis lurus diperoleh dengan memplot y 5
melawan x2 seperti yang ditunjukkan dalam Rajah 1.
x3
y O
G Diagram 3/ Rajah 3
Find the values of p and n.
Cari nilai p dan n. [3 marks/markah]
x2 4 The variables x and y are related by the equation
OH u ,
SPM y =x+ x2 where u is a constant. Diagram 4
Diagram 1/ Rajah 1 CLONE
(a) Convert the equation xy = 12x – 4x3 to the `17
linear form to obtain the straight line graph as P1Q19 shows a straight line graph obtained by plotting
shown in Diagram 1.
Tukarkan persamaan xy = 12x – 4x3 kepada bentuk (y – x) against 1 .
linear untuk memperoleh graf garis lurus seperti yang x2
ditunjukkan dalam Rajah 1.
Pemboleh ubah x dan y dihubungkan oleh persamaan
(b) State/Nyatakan
(i) the gradient of the straight line, y = x + u , dengan keadaan u ialah pemalar. Rajah 4
kecerunan garis lurus itu, x2
(ii) coordinates of G.
koordinat G. menunjukkan graf garis lurus yang diperoleh dengan
[3 marks/markah] memplot (y – x) melawan 1 .
2 Diagram 2 shows a straight line graph obtained x2
by plotting (y – x) against x2. y–x
Rajah 2 menunjukkan graf garis lurus yang diperoleh
dengan memplot (y – x) melawan x2. k , 6t
4
y–x
1
18 O x2
Diagram 4/ Rajah 4
Express k in terms of u and t.
Ungkapkan k dalam sebutan u dan t.
[3 marks/markah]
5 Diagram 5 shows a straight line graph obtained
x2 1.
SPM by plotting y against x
CLONE
x2
`18
P1Q13 Rajah 5 menunjukkan graf garis lurus yang diperoleh
–6 O
x2 1
Diagram 2/ Rajah 2 dengan memplot y melawan x .
Express y in terms of x. x2
y
Ungkapkan y dalam sebutan x.
[3 marks/markah] (6, 5)
3 The variables x and y are related by the equation
SPM y = 3x2 – n, where n is a constant. Diagram 3 1
CLONE x Ox
`16 shows a straight line graph obtained by plotting
P1Q16 –4
xy against x3.
Pemboleh ubah x dan y dihubungkan oleh persamaan Diagram 5/ Rajah 5
y = 3x2 – n , dengan keadaan n ialah pemalar. Rajah 3 Express y in terms of x. [3 marks/markah]
x Ungkapkan y dalam sebutan x.
menunjukkan graf garis lurus yang diperoleh dengan
memplot xy melawan x3.
102
Paper 2 Questions
1 Table 1 shows the values of two variables, x and y, obtained from an experiment. The variables x and y are
SPM related by the equation y = 2p , where p and t are constants.
CLONE tx
`14 Jadual 1 menunjukkan nilai-nilai bagi dua pemboleh ubah, x dan y, yang diperoleh daripada satu eksperimen. Pemboleh ubah x dan y
P2Q9
dihubungkan oleh persamaan y = 2p , dengan keadaan p dan t ialah pemalar.
tx
x 4 6 8 10 12 14
y 2.45 1.86 1.38 1.04 0.78 0.59
Table 1/ Jadual 1
(a) Based on Table 1, construct a table for the values of log10 y. Give your answer correct to two significant
figures.
Berdasarkan Jadual 1, bina satu jadual bagi nilai-nilai log10 y. Berikan jawapan anda betul kepada dua angka bererti.
[1 mark/markah]
(b) Plot log10 y against x, using a scale of 2 cm to 2 units on the x-axis and 2 cm to 0.1 unit on the log10 y-axis.
Hence, draw the line of best fit.
Plot log10 y melawan x, menggunakan skala 2 cm kepada 2 unit pada paksi-x dan 2 cm kepada 0.1 unit pada paksi-log10 y.
Seterusnya, lukis garis lurus penyuaian terbaik.
[3 marks/markah]
(c) Use the graph in (b) to find the value of
Gunakan graf di (b) untuk mencari nilai
(i) t, (ii) p.
[6 marks/markah]
2 Table 2 shows the values of two variables, x and y, obtained from an experiment. The variables x and y are
SPM related by the equation hx = ky + 2xy, where h and k are constants. daripada satu eksperimen. Pemboleh ubah x dan
CLONE Jadual 2 menunjukkan nilai-nilai bagi dua pemboleh ubah, x dan y, yang diperoleh
`15
P2Q11 y dihubungkan oleh persamaan hx = ky + 2xy, dengan keadaan h dan k ialah pemalar.
x 1.5 2.0 3.0 4.0 5.0 6.0
y 2.72 0.83 0.49 0.41 0.38 0.35
Table 2/ Jadual 2
(a) Based on Table 2, construct a table for the values of 1 and 1 .
x y
1 1
Berdasarkan Jadual 2, bina satu jadual bagi nilai-nilai x dan y .
[2 marks/markah]
(b) Plot 1 against 1 , using a scale of 2 cm to 0.1 unit on the 1 -axis and 2 cm to 0.5 unit on the 1 -axis.
y x x y
Hence, draw the line of best fit.
Plot 1 melawan 1 , dengan menggunakan skala 2 cm kepada 0.1 unit pada paksi- 1 dan 2 cm kepada 0.5 unit pada paksi- 1 .
y x x y
Seterusnya, lukis garis lurus penyuaian terbaik.
[3 marks/markah]
(c) Using the graph in (b), find the value of
Menggunakan graf di (b), cari nilai
(i) h, (ii) k.
[5 marks/markah]
3 Table 3 shows the values of two variables, x and y, obtained from an experiment. The variables x and y are
SPM related by the equation y = 3ax + b , where a and b are constants.
CLONE 6x
`16 Jadual 3 menunjukkan nilai-nilai bagi dua pemboleh ubah, x dan y, yang diperoleh daripada satu eksperimen. Pemboleh ubah x dan
P2Q10
y dihubungkan oleh persamaan y = 3ax + b , dengan keadaan a dan b ialah pemalar.
6x
x1 2 3 4 5 6
y 14.65 8.28 6.59 6.06 6.00 6.18
Table 3/ Jadual 3
103
(a) Based on Table 3, construct a table for the values of x2 and xy.
Berdasarkan Jadual 3, bina satu jadual bagi nilai-nilai x2 dan xy.
[2 marks/markah]
(b) Plot xy against x2, using a scale of 2 cm to 5 units on both axes. Hence, draw the line of best fit.
Plot xy melawan x2, dengan menggunakan skala 2 cm kepada 5 unit pada kedua-dua paksi. Seterusnya, lukis garis lurus
penyuaian terbaik.
[3 marks/markah]
(c) Using the graph in (b), find the value of
Menggunakan graf di (b), cari nilai
(i) a, (ii) b.
[5 marks/markah]
4 Table 4 shows the values of two variables, x and y, obtained from an experiment. A straight line graph is
SPM obtained by plotting y2 against 1 .
CLONE x x
`18
P2Q11 Jadual 4 menunjukkan nilai-nilai bagi dua pemboleh ubah, x dan y, yang diperoleh daripada satu eksperimen. Graf garis lurus
diperoleh dengan memplot y2 melawan 1 .
x x
x 1.25 1.43 2.00 2.5 4.00 5.00
y 4.29 4.26 4.18 4.12 3.94 3.84
Table 4/ Jadual 4
(a) Based on Table 5, construct a table for the values of 1 and y2 .
x x
1 y2
Berdasarkan Jadual 5, bina satu jadual bagi nilai-nilai x dan x .
[2 marks/markah]
(b) Plot y2 against 1 , using a scale of 2 cm to 0.1 unit on the 1 -axis and 2 cm to 2 units on the y2 -axis.
x x x x
Hence, draw the line of best fit.
Plot y2 melawan 1 , menggunakan skala 2 cm kepada 0.1 unit pada paksi- 1 dan 2 cm kepada 2 unit pada paksi- y2 .
x x x x
Seterusnya, lukis garis lurus penyuaian terbaik. (ii) express y in terms of x. [3 marks/markah]
ungkapkan y dalam sebutan x. [5 marks/markah]
(c) Using the graph in (b),
Menggunakan graf di (b),
(i) find the value of y when x = 3.6,
cari nilai y apabila x = 3.6,
H OTS Zone
1 Table 1 shows the values of two variables, x and y, obtained from an experiment. The variables x and y are
SPM related by the equation y – √p = pt , where p and t are constants.
CLONE x
`17
P2Q9 Jadual 1 menunjukkan nilai-nilai bagi dua pemboleh ubah, x dan y, yang diperoleh daripada satu eksperimen. Pemboleh ubah x dan
y dihubungkan oleh persamaan y –√p = pt , dengan keadaan p dan t ialah pemalar.
x
x 1.2 2.0 3.5 4.5 5.0 6.0
y 4.15 5.16 5.81 5.68 6.06 6.19
Table 1/ Jadual 1
(a) Plot xy against x, using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the xy-axis. Hence,
draw the line of best fit.
Plot xy melawan x, menggunakan skala 2 cm kepada 1 unit pada paksi-x dan 2 cm kepada 5 unit pada paksi-xy. Seterusnya,
lukis garis lurus penyuaian terbaik.
(b) Using the graph in (a), find
Menggunakan graf di (a), cari
(i) the values of p and t,
nilai p dan t,
(ii) the correct value of y if one of the values of y has been wrongly recorded during the experiment.
nilai betul bagi y jika satu daripada nilai y telah direkodkan salah semasa eksperimen. HO TS Applying
104
Chapter Learning Area: Geometry
7 Coordinate Geometry
Geometri Koordinat
7.1 Divisor of a Line Segment / Pembahagi Tembereng Garis
Smart Tip
Point that divides the line segment by the ratio m : n m n R Q = nx1 + mx2 , ny1 + my2
Titik yang membahagikan tembereng garis dengan nisbah m : n Q (x2, y2) m+n m+n
P
(x1, y1) m+n
Exercise 1 Solve each of the following.
Selesaikan setiap yang berikut.
TP 2 Mempamerkan kefahaman tentang pembahagi tembereng garis.
Example 1 1 Given P(–2, 5), Q and R(7, 2) 2 Given P(2, –1), Q and R(5, 8)
are collinear such that are collinear such that
Given P(–3, –9), Q and R(4, PQ : QR = 1 : 2. Find the PQ : QR = 2 : 1. Find the
5) are collinear such that coordinates of Q. coordinates of Q.
PQ : PR = 2 : 7. Find the Diberi P(–2, 5), Q dan R(7, 2) Diberi P(2, –1), Q dan R(5, 8)
coordinates of Q. adalah segaris dengan keadaan adalah segaris dengan keadaan
Diberi P(–3, –9), Q dan R(4, 5) PQ : QR = 1 : 2. PQ : QR = 2 : 1.
adalah segaris dengan keadaan Cari koordinat Q. Cari koordinat Q.
PQ : PR = 2 : 7. Cari koordinat Q.
m:n=1:2 m:n=2:1
Solution
2(–2) + 1(7) 2(5) + 1(2) 1(2) + 2(5) 1(–1) + 2(8)
Given/Diberi PQ : PR = 2 : 7,
Q = 1 + 2 , 1 + 2 Q = 2 + 1 , 2 + 1
then/maka
= 33 , 132 = 132 , 135
PQ : QR = 2 : 5
= (1, 4) = (4, 5)
m:n=2:5
nx1 + mx2 ny1 + my2
m+n m+n
Q = ,
5(–3) + 2(4) 5(–9) + 2(5)
= 2 + 5 , 2 + 5
= –77 , –735 = (–1, –5)
3 Given P(–5, 4), Q and R(5, –1) 4 Given the points P(–2, 2) and 5 Given the coordinates of P
are collinear such that Q(7, 8). Point R lies on the and Q are (2, –3) and (6, 5)
PQ : PR = 2 : 5. Find the straight line PQ such that respectively. If point R divides
coordinates of Q. 2PR = RQ. Find the PQ in the ratio of 1 : 3, find the
coordinates of point R. coordinates of point R.
Diberi P(–5, 4), Q dan R(5, –1)
adalah segaris dengan keadaan Diberi titik P(–2, 2) dan Q(7, 8). Diberi koordinat P dan Q masing-
Titik R terletak pada garis lurus PQ masing ialah (2, –3) dan (6, 5). Jika
PQ : PR = 2 : 5. dengan keadaan 2PR = RQ. Cari titik R membahagi PQ dalam nisbah
Cari koordinat Q. koordinat titik R. 1 : 3, cari koordinat bagi titik R.
Given PQ : PR = 2 : 5 Given 2PR = RQ PR : RQ = 1 : 3
Then PQ : QR = 2 : 3 PR 1 m:n=1:3
m:n=2:3 RQ = 2
m1 3(2) + 1(6) 3(–3) + 1(5)
3(–5) + 2(5) 3(4) + 2(–1) n = 2 or m : n = 1 : 2
R = 1 + 3 , 1 + 3
Q = 2 + 3 , 2 + 3 2(–2) + 1(7) 2(2) + 1(8)
= 142 , –4
= –55 , 150 R = 1 + 2 , 1 + 2 4
= (–1, 2) = 33 , 132 = (1, 4) = (3, –1)
105
Exercise 2 Solve the following problems.
Selesaikan masalah yang berikut.
TP 2 Mempamerkan kefahaman tentang pembahagi tembereng garis.
Example 2
It is given that F(p, 6), G(2, 3) and H(7, q) are collinear such that FG : FH = 1 : 3. Find the values of p and q.
Diberi bahawa F(p, 6), G(2, 3) dan H(7, q) adalah segaris dengan keadaan FG : FH = 1 : 3. Cari nilai p dan q.
Solution For x coordinate/Untuk koordinat x, For y coordinate/Untuk koordinat y,
Given/Diberi FG : FH = 1 : 3
then/maka FG : GH = 1 : 2 2p + 7 2(6) + 1(q)
m:n=1:2 3 = 2 1+2 =3
12 + q
nx1 + mx2 2p + 7 = 6
m+n 3 = 3
, ny1 + my2 = (x, y) 2p = –1 12 + q = 9
m+n
1 q = –3
2(p) + 1(7) 2(6) + 1(q) p = – 2
1 + 2 , 1 + 2 = (2, 3)
1 It is given that F(p, 4), G(1, 7) and H(6, q) are collinear such that FG : GH = 3 : 5. Find the values of p and q.
Diberi bahawa F(p, 4), G(1, 7) dan H(6, q) adalah segaris dengan keadaan FG : GH = 3 : 5. Cari nilai p dan q.
FG : GH = 3 : 5 For x coordinate, For y coordinate,
m:n=3:5
5(p) + 3(6) 5(4) + 3(q)
nx1+mx2, ny1 + my2 = (x, y) 3 + 5 = 1 3 + 5 = 7
m+ n m + n
5p + 18 20 + 3q
5(p) + 3(6) 5(4) + 3(q) 8 = 1 8 = 7
3 + 5 , 3 + 5 = (1, 7) 5p + 18 = 8 20 + 3q = 56
5p = –10 3q = 36
p = –2 q = 12
2 It is given that F(–4, q), G(2, –1) and H(p, 1) are collinear such that FG : GH = 2 : 1. Find the values of p and q.
Diberi bahawa F(–4, q), G(2, –1) dan H(p, 1) adalah segaris dengan keadaan FG : GH = 2 : 1. Cari nilai p dan q.
FG : GH = 2 : 1 For x coordinate, For y coordinate,
m:n=2:1
1(–4) + 2(p) 1(q) + 2(1)
nx1+mx2, ny1 + my2 = (x, y) 2 + 1 = 2 2 + 1 = –1
m+ n m + n q+2
–4 + 2p
1(–4) + 2(p) 1(q) + 2(1) 3 = 2 3 = –1
2 + 1 , 2 + 1 = (2, –1) –4 + 2p = 6 q + 2 = –3
2p = 10 q = –5
p = 5
3 A straight line passes through J(12, –13) and K(–9, 8).
Suatu garis lurus melalui J(12, –13) dan K(–9, 8).
(a) It is given that L(4, s) lies on the straight line JK, find the value of s.
Diberi bahawa L(4, s) terletak di atas garis lurus JK, cari nilai s.
(b) Point P divides the line segment JK in the ratio 2 : 4. Find the coordinates of P.
Titik P membahagi tembereng garis JK dalam nisbah 2 : 4. Cari koordinat P.
nx1 + mx2 nym1 ++ my2
m+n n
(a) Gradient of KL = Gradient of JK (b) P = ,
s − 89 = 8− (–13) 4(12) + 2(–9) 4(–13) + 2(8)
4 + –9 − 12 = 4 + 2 , 4 + 2
s −8 = 21 48 − 18 –52 + 16
13 –21
= 6 , 6
s−8
13 = –1 = 30 , – 366
6
s − 8 = –13
= (5, –6)
s = –5
106
7.2 Parallel Lines and Perpendicular Lines / Garis Lurus Selari dan Garis Lurus Serenjang
Smart Tip
(a) Parallel lines/Garis selari (b) Perpendicular lines/Garis serenjang
m1 = m2
m1 m2 = –1 m1
m1
m2 m2
Exercise 3 Determine whether each of the following pairs of lines are parallel or not.
Tentukan sama ada setiap pasangan garis lurus berikut adalah selari atau tidak.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 3 1 4x + 6y = 5, 2x = 6 − 3y
2y – x = 7, x – 2y = 5 4x + 6y = 5 1
Solution 2x = 6 − 3y 2
2y − x = 7 1
x − 2y = 5 2 From 1 , 6y = –4x + 5
From/Daripada 1 , From/Daripada 2 , –4x 5
y = 6 + 6
2y = x + 7 –2y = –x + 5 25
m1 = – 23 y = – 3 x + 6
x7 x5
y = 2 + 2 y = 2 − 2
11
m1 = 2 m2 = 2 From 2 , 3y = –2x + 6
Compare/Bandingkan –2x 6
y = mx + c y = 3 + 3
y = – 23x + 2
m1 = m2 m2 = – 23
m1 = m2
∴ The pair of straight lines are parallel.
Pasangan garis lurus adalah selari. ∴ The pair of straight lines are parallel.
2 3y – x = 6, 3x – 2y = 6 3 8y + 4x = 15, 2x + 4y = 17
3y − x = 6 1 8y + 4x = 15 1
3x − 2y = 6 2 2x + 4y = 17 2
From 1 , 3y = x + 6 From 1 , 8y = –4x + 15
1
y = – 84 x + 15
y = 3 x + 2 8
1 15
y = – 1 x + 8
m1 = 3 2
From 2 , 2y = 3x − 6
m1 = – 1
3 2
y = 2 x − 3
From 2 , 4y = –2x + 17
3
m2 = 2 y = – 24 x + 17
m1 ≠ m2 4
∴ The pair of straight lines are not parallel. y = – 12 x + 17
4
m = – 12
m1 = m2
∴ The pair of straight lines are parallel.
107
Exercise 4 It is given that each of the following pairs of straight lines are parallel. Find the value of p.
Diberi bahawa setiap pasangan garis lurus berikut adalah selari. Cari nilai p.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 4 1 px + 6y − 12 = 0, 2x − 3y + 14 = 0
3y − 4x – 15 = 0, (p − 5)x − 3y − 18 = 0 px + 6y − 12 = 0 1
2x − 3y + 14 = 0 2
Solution
3y − 4x − 15 = 0 1 From 1 , 6y = –px + 12
(p − 5)x − 3y − 18 = 0 2
y = – p6 x + 12
6
From/Daripada 1 , y = – 6px + 2
p
3y = 4x + 15 y = mx + c m1 = – 6
4
From 2 , 3y = 2x + 14
y = 3 x + 5
4 2 14
3 3
m1 = 3
From/Daripada 2 , y = x +
2
3y = (p − 5)x – 18
m2 = 3
p − 5 x
y = 3 − 6 y = mx + c m1 = m2
p−5 – p6 = 2 x2
m2 = 3 3 x2
m1 = m2 – 6p = 4
6
4 p−5
3= 3 p = –4
4=p−5
p=9
2 4y + 3x + 24 = 0, (2p + 1)x + 3y + 6 = 0 3 8x + 2py = 5, 2x − (p + 1)y − 8 = 0
4y + 3x + 24 = 0 1 8x + 2py = 5 1
(2p + 1)x + 3y + 6 = 0 2 2x − (p + 1)y − 8 = 0 2
From 1 , 4y = –3x − 24 From 1 , 2py = –8x + 5
y = – 43 x − 6 y = – 28p x + 5
2p
m1 = – 3 m1 = – 8 = – 4
4 2p p
From 2 , 3y = –(2p + 1)x − 6 From 2 , (p + 1)y = 2x − 8
y = – (2p + 1) x − 2 y = 2 x − p 8 1
p+1 +
– ( 2p + 1) 3 2
3 +
m2 = m2 = p 1
m1 = m2 m1 = m2
– 3 = – (2p + 1) – 4 = p 2 1
4 3 p +
9 = 2p + 1 – 4(p + 1) = 2p
4
–2(p + 1) = p
5 –2p − 2 = p
4
= 2p –2p − p = 2
p = 85 –3p = 2
p = – 2
3
108
Exercise 5 Determine whether each of the following pairs of lines are perpendicular or not.
Tentukan sama ada setiap pasangan garis lurus berikut adalah berserenjang atau tidak.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 5 1 2x + y = 1, –x + 2y = 6
3y − x = 6, 3x + y = 12 2x + y = 1 1
–x + 2y = 6 2
Solution
3y − x = 6 1 Smart Tip From 1 ,
3x + y = 12 2 y = –2x + 1
m1 = –2
From/Daripada 1 , Perpendicular lines are
3y = x + 6 lines which intersect at a From 2 ,
right angle (90°). 2y = x + 6
x Garis serenjang ialah garis
y=3+2 yang bersilang pada sudut x
tegak (90°). y=2+3
1 1
m1 = 3 Perpendicular lines m2 = 2 1
Garis berserenjang 2
From/Daripada 2 , m1 m2 = –1 m1 m2 = (–2) = –1
y = –3x + 12
∴ The pair of straight lines are perpendicular.
m2 = –3 1
m1 m2 = 3 (–3) = –1
∴ The pair of straight lines are perpendicular.
Pasangan garis lurus adalah berserenjang.
2 3y − x = 12, 3x – 2y = 8 3 8y + 6x = 24, –4x + 3y = 12
3y − x = 12 1 From 2 , 8y + 6x = 24 1 From 2 , 3y = 4x + 12
3x − 2y = 8 2 –4x + 3y = 12 2
2y = 3x − 8 4
From 1 , 3
3y = x + 12 3 From 1 , 8y = –6x + 24 y = x + 4
y = 2x − 4
1 y = – 68x + 3 m2 = 4
y = 3x + 4 3 3
m2 =2
1 m1 m2 = 1 3 = 1 ≠ –1 m1 = – 6 = – 3 m1m2 = – 3 4 = –1
m1 = 3 3 2 2 8 4 4 3
∴ The pair of straight lines ∴ The pair of straight
are not perpendicular. lines are perpendicular.
Exercise 6 It is given that each of the following pairs of straight lines are perpendicular. Find the value of n.
Diberi bahawa setiap pasangan garis lurus berikut adalah berserenjang. Cari nilai n.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 6 1 y − 2x + 6 = 0, nx − 6y − 12 = 0
2y − 3x + 14 = 0, (n – 2)x − 4y − 16 = 0 y – 2x + 6 = 0 1
nx − 6y − 12 = 0 2
Solution From 1 ,
2y − 3x + 14 = 0 1 y = 2x – 6
(n − 2)x − 4y − 16 = 0 2 m1 = 2
From 2 ,
From/Daripada 1 , m2 = n − 2 6 y = nx − 12
4
2y = 3 x − 14
3 3 n − 2 = –1 y = n6 x − 2
4
y = 2 x − 7 m3(1nm−2 2=) =2–8 n
m2 = 6
3 3n − 6 = –8
m1 = 2
3n = –2 m 1 n
6
From/Daripada 2 , 2 m2 = (2) = –1
4y = (n − 2)x − 16 n = – 3
n3 = –1
y = n − 2 x − 4
4 n = –3
109
2 2y − x + 8 = 0, (n + 1)x + 4y − 20 = 0 3 4y – 3x + 8 = 0, (2n – 1)x + 2y – 6 = 0
2y − x + 8 = 0 1
(n + 1)x + 4y − 20 = 0 2 4y − 3x + 8 = 0 1
From 1 , From 2 , (2n − 1)x + 2y − 6 = 0 2
From 1 ,
2y = x − 8 4y = 3x − 8
y = 1 x − 4 4y = –(n + 1)x + 20 y = 3 x − 2
2 4
y = – (n + 1) x + 5 3
m1 = 1 4 m1 = 4
2
m2 = – ( n + 1) From 2 , 3 – (2n − 1)
4 4 2
m1 m2 = = –1
m1 m2 = 1 – (n + 1) = –1 2y = –(2n − 1)x + 6 3 (2n − 1) = 8
2 4 y = – (2n − 1) x + 3
6n − 3 = 8
n + 1 = 8 2
6n = 11
n = 7
m2 = – (2n − 1) n = 11
2 6
Exercise 7 Find the equation of the straight line that passes through point P and parallel to the given
straight line in general form.
Cari persamaan garis lurus yang melalui titik P dan selari dengan garis lurus yang diberi dalam bentuk am.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 7 1 2y − 6x + 5 = 0, P(4, 1)
3y − 2x − 12 = 0, P(2, –5) 2y − 6x + 5 = 0
Solution 2y = 6x − 5
3y − 2x − 12 = 0 y = 3x − 5
2
3y = 2x + 12
2 y = mx + c m=3
y = 3 x + 4 The equation of the straight line,
2 y – y1 = m(x − x1)
y − 1 = 3(x − 4)
m= 3 y − 1 = 3x − 12
y = 3x − 11
The equation of the straight line, 3x − y − 11 = 0
Persamaan garis lurus,
y – y1 = m(x – x1)
2
y − (–5) = 3 (x − 2)
3(y + 5) = 2(x − 2)
3y + 15 = 2x − 4
2x − 3y − 19 = 0
2 4y + 3x + 6 = 0, P(1, 8) 3 2y + 3x + 18 = 0, P(4, –2)
4 y + 3x + 6 = 0 2y + 3x + 18 = 0
4y = –3x − 6 2y = –3x − 18
y = –43 x − 3 ,m=– 3 y = – 3 x − 9
2 4 2
The equation of the straight line, m = – 23
y − y1 = m(x − x1)
y − 8 = – 3 (x − 1) The equation of the straight line,
4
y − y1 = m(x − x1)
4y − 32 = –3x + 3 y − (–2) = – 3 (x − 4)
2
4 y + 3x − 35 = 0
3x + 4y − 35 = 0 2(y + 2) = –3(x − 4)
2y + 4 = –3x + 12
3x + 2y − 8 = 0
110
4 x + y = 1, P(–3, –2) 5 x − y = 1, P(–7, 6)
9 2 5 3
x + y =1 Intercept form: x + y =1 5x − y =1 Intercept form: x – y =1
9 2 a b 3 a b
y – intercept b 2 ∴m = – b –3 3
∴m = – x – intercept =– a = – 9 y – intercept = – a =– 5 = 5
x – intercept
The equation of the straight line,
The equation of the straight line,
y – y1 = m(x – x1)
2 y − y1 = m(x – x1)
9 3
y − (–2) = – [x − (–3)] y − 6 = 5 [x − (–7)]
y + 2 = – 2 x − 6 y − 6 = 3 21
9 9 5x+ 5
2 24
y = – 9 x − 9 y = 3 51
5x+ 5
9y = –2x − 24
5y = 3x + 51
2 x + 9y + 24 = 0
3x − 5y + 51 = 0
Exercise 8 Find the equation of the straight line that passes through point P and perpendicular to the
given straight line in general form.
Cari persamaan garis lurus yang melalui titik P dan berserenjang dengan garis lurus yang diberi dalam bentuk am.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 8 1 y – 2x + 7 = 0, P(2, 6)
4y − 3x − 8 = 0, P(1, –4) y – 2x + 7 = 0
y = 2x − 7
Solution m1 = 2
m2 = – 12
4 y − 3x − 8 = 0 m1 m2 = –1
4y = 3x + 8 3
3 4
y = 4 x + 2 × m2 = –1 The equation of the straight line,
4
3 – 34 m2 = – 3 y – y1 = m2(x − x1)
4,
m1 = m2 = y − 6 = – 1 (x − 2)
2
The equation of the straight line,
2(y − 6) = –(x − 2)
Persamaan garis lurus,
2y − 12 = –x + 2
y − y1 = m2(x − x1)
x + 2y – 14 = 0
4
y − (–4) = – 3 (x − 1)
3(y + 4) = –4(x − 1)
3y + 12 = –4x + 4
4x + 3y + 8 = 0
2 3x + y − 7 = 0, P(–1, 5) 3 4x + 3y + 24 = 0, P(–2, –6)
3 x + y − 7 = 0 4x + 3y + 24 = 0
y = –3x + 7 3y = –4x − 24
m1 = –3 y = – 4 x − 8
1 3
m2 = 3
4
m1 = – 3
The equation of the straight line, 3
4
y − y1 = m2(x − x1) m2 =
y − 5 = 1 (x − (–1)) The equation of the straight line,
3
y – y1 = m2(x − x1)
3(y − 5) = (x + 1)
3y − 15 = x + 1 y − (–6) = 3 (x − (–2))
x − 3y + 16 = 0 4
4(y + 6) = 3(x + 2)
4y + 24 = 3x + 6
3x − 4y − 18 = 0
111
4 x − y = 3, P(0, –11) 5 2x + y = 2, P(–3, 8)
2 3 7 21
x − y = 3 (÷ 3) 6x2+1 y = 2
2 3
x − y =1 Intercept form: x − y =1 6x + y = 42
6 9 a b
y = –6x + 42
m1 = – y – intercept = –(–69) m1 = –6, m2 = 1
x – intercept 6
m1 = 3 , m2 = –2 The equation of the straight line,
2 3 y − y1 = m2(x − x1)
The equation of the straight line,
y − 8 = 1(x + 3)
y − y1 = m2(x − x1) 6
y − (–11) = – 2(x – 0) 6y − 48 = x + 3
3 x − 6y + 51 = 0
y + 11 = – 2x
3
3y + 33 = –2x
2x + 3y + 33 = 0
Exercise 9 Solve each of the following.
Selesaikan setiap yang berikut.
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang geometri koordinat dalam konteks penyelesaian masalah rutin yang mudah.
1 A developer is required to build a new straight 2 The diagram below shows the positions of four
road which passes through the point (–2, 5). It is rest huts, E, F, G and H in a park. It is given that
given that a straight road that was already built EF is parallel to GH.
can be expressed as 2y = 3x − 4. The developer is
required to build the new road such that it does Rajah di bawah menunjukkan kedudukan empat buah
not intersect with the existing road. Determine pondok rehat, E, F, G dan H di sebuah taman. Diberi
the equation of the new road. bahawa EF adalah selari dengan GH.
Pemaju dikehendaki untuk membina sebatang jalan y
raya lurus baharu yang melalui titik (–2, 5). Diberi
bahawa sebatang jalan raya lurus yang telah dibina E(3, 6) F(7, 4)
dapat diungkapkan sebagai 2y = 3x − 4. Pemaju tersebut
dikehendaki untuk membina jalan raya baharu supaya O x
tidak bersilang dengan jalan raya yang sedia ada. Tentukan G(2, –3)
persamaan bagi jalan raya baharu itu.
Gradient of the existing road H(12, k)
2y = 3x – 4 Find/Cari
(a) the equation of the straight line GH,
y = 3x – 2 persamaan garis lurus GH,
2
3 (b) the value of k.
So, m1 = 2 nilai k.
The new road must be parallel to the existing 4 – 6 –2 –1
7 – 3 4 2
road, so that both roads do not intersect. mEF= = =
So, m2 = 3 Given EF//GH
2
The equation of the new road: 1
So, mGH= mEF = – 2
y = m2x + c
(a) The equation of the straight line GH,
(5) = 1 3 2 (–2) + c
2 1 2 y – (–3) = – 1 (x – 2)
2
5 = –3 + c
c=8 y + 3 = – 1x + 1
2
So, y = 3x + 8 or 2y = 3x + 16
2 y = – 1x – 2
2
(b) When x = 12, y = k
k = – 1(12) – 2 = –8
2
112
3 In the diagram below, E is the point of 4 It is given that a straight line has equation
intersection of two perpendicular lines. If the y = 2x + 6 and a point P(12, 10). Find the coordinates
equation of straight line OE is y = 3x, find the of Q such that Q is on the straight line y = 2x + 6
values of h and k.
and PQ is the shortest distance. HOTS Applying
Dalam rajah di bawah, E ialah titik persilangan bagi dua Diberi bahawa satu garis lurus mempunyai persamaan
garis yang berserenjang. Jika persamaan garis lurus OE
ialah y = 3x, cari nilai h dan k. y = 2x + 6 dan titik P(12, 10). Cari koordinat Q dengan
keadaan Q berada pada garis lurus y = 2x + 6 dan PQ ialah
y jarak terdekat.
E(4, h) • PQ is the shortest distance, hence PQ is
• G(7, k) perpendicular to the straight line y = 2x + 6.
The gradient of the straight line y = 2x + 6 is 2.
Ox Let the gradient of PQ = m2. PAK-21
Hence, (2)(m2) = –1 perpendicular
The equation of OE is y = 3x. m2 = – 1 QR CODE
2
The gradient of straight line OE is m1.
m1 = 3 The equation of PQ,
m1 = h–0 = h y − y1 = m2(x − x1), P(12, 10)
4–0 4
y − 10 = – 1 (x – 12)
2
Hence, h = 3
4 2 y − 20 = –x + 12
h = 12 x + 2y = 32
The gradient of EG is m2. Solve
y = 2x + 6 1
m2 = k–h = k–h x + 2y = 32 2
7–4 3 From 2 , x = 32 − 2y
Substitute x into 1 ,
m1m2 = –1 y = 2(32 − 2y) + 6
y = 64 – 4y + 6
(3) k – h = –1 5y = 70
3 y = 14
From 1 ,
k – h = –1
14 = 2x + 6
k – (12) = –1 2x = 8
x=4
k = 11 Hence, Q = (4, 14)
∴ h = 12, k = 11
7.3 Areas of Polygons / Luas Poligon
Smart Tip QR CODE
Area of a triangle/Luas segi tiga Scan or visit
https://www.
1 x1 x2 x3 x1 y mathopenref.com/
y2 A(x1, y1) coordtrianglearea.
2 y1 y3 y1 html to learn
the alternative
=1 (x1y2 + x2y3 + x3y1) − (y1x2 + y2x3 + y3x1) methods in
2 determining
area of a triangle
or/atau with given
coordinates.
1 x1 x2 x3 x1 B(x2, y2) C(x3, y3)
y2 O
2 y1 y3 y1
=1 (x1y2 + x2y3 + x3y1) − (x1y3 + x3y2 + x2y1) x
2
113
Exercise 10 Find the area of each of the following triangles with the given vertices.
Cari luas bagi setiap segi tiga berikut dengan bucu-bucu yang diberi.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 9 1 E(0, 1), F(2, 3), G(4, –1)
E(2, 3), F(5, 4), G(3, –6) Area12
0 2 4 0
= 1 3 –1 1
Solution 12[(0)(3)
− [(1)(2)
Area/Luas 12 = + (2)(–1) + (4)(1)]
= 2 5 3 2 +
3 4 –6 3 (3)(4) + (–1)(0)]
= 12[(2)(4) + (5)(–6) + (3)(3)] = 21 (2) − (14)
− [(3)(5) + (4)(3) + (–6)(2)] = 12(–12)
= 12 [8 + (–30) + 9] − [15 + 12 + (–12)] = 6
= 12 (–13) − (15)
= 12(–28) The area of triangle EFG is 6 units2.
= –14
2 E(–2, 0), F(–1, 3), G(4, –2)
= 14
Area21
The area of triangle EFG is 14 units2. = –2 –1 4 –2
Luas segi tiga EFG ialah 14 unit2. 0 3 –2 0
= 21 [(–2)(3) + (–1)(–2) + (4)(0)]
Smart Tip − [(0)(–1) + (3)(4) + (–2)(–2)]
= 21 (–4) − (16)
The notation of modulus (or absolute value),
= 21 (–20)
, is the non-negative value of a number
= 10
regardless of its sign.
Tatatanda modulus (atau nilai mutlak), , ialah nilai The area of triangle EFG is 10 units2.
bukan negatif bagi suatu nombor tanpa mengambil kira
tandanya.
∴ -x = x
Smart Tip
Area of a quadrilateral/Luas sisi empat y
B(x2, y2)
1 x1 x2 x3 x4 x1
2 y1 y2 y3 y4 y1
= 1 (x1y2 + x2y3 + x3y4 + x4y1 − (y1x2 + y2x3 + y3x4 + y4x1) A(x1, y1)
2
1 If the vertices are arranged in the anticlockwise order, the
answer is positive.
Jika bucu-bucu disusun mengikut tertib arah lawan jam, jawapan yang x
diperoleh adalah positif. O
C(x3, y3)
2 If the vertices are arranged in the clockwise order, the answer is D(x4, y4)
negative.
Jika bucu-bucu disusun mengikut tertib arah jam, jawapan yang diperoleh
adalah negatif.
114
Exercise 11 Find the area of each of the following quadrilaterals with the given vertices.
Cari luas bagi setiap sisi empat berikut dengan bucu-bucu yang diberi.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 10
P(2, 0), Q(7, 1), R(5, 9), S(–1, 4)
Solution
Area/Luas = 12 2 7 5 –1 2
0 1 9 4 0
= 12[(2)(1) + (7)(9) + (5)(4) + (–1)(0)] − [(0)(7) + (1)(5) + (9)(–1) + (4)(2)]
= 12 [2 + 63 + 20 + 0] − [0 + 5 – 9 + 8]
= 12 (85) − (4)
= 1281
= 40.5
The area of quadrilateral PQRS is 40.5 units2.
Luas sisi empat PQRS ialah 40.5 unit2.
1 A(3, 6), B(7, 8), C(6, 0), D(1, 2) 2 A(–1, 4), B(5, 3), C(6, –1), D(0, 4)
Area = 12 Area = 21
3 7 6 1 3 –1 5 6 0 –1
6 8 0 2 6 4 3 –1 4 4
= 21[(3)(8) + (7)(0) + (6)(2) + (1)(6)] = 12[(–1)(3) + (5)(–1) + (6)(4) + (0)(4)]
− [(6)(7) + (8)(6) + (0)(1) + − [(4)(5) + (3)(6) +
(2)(3)] (–1)(0) + (4)(–1)]
= 12 [24 + 0 + 12 + 6] – [42 + 48 + 0 + 6] = 21 [–3 + (–5) + 24 + 0] – [20 + 18 + 0 – 4]
= 21 (42) – (96) = 12 (16) – (34)
= 12(–54) = 21(–18)
= 27 =9
The area of quadrilateral ABCD is 27 units2. The area of quadrilateral ABCD is 9 units2.
3 E(–1, 2), F(6, 3), G(3, 2), H(1, –4) 4 E(–2, 1), F(5, 4), G(4, –1), H(2, –3)
Area Area = 21
= 12 –1 6 3 1 –1 –2 5 4 2 –2
2 3 2 –4 2 1 4 –1 –3 1
= 21 [(–1)(3) + (6)(2) + (3)(–4) + (1)(2)] = 21 [(–2)(4) + (5)(–1) + (4)(–3) + (2)(1)]
− [(2)(6) + (3)(3) + (2)(1) + (–4)(–1)] – [(1)(5) + (4)(4) + (–1)(2) + (–3)(–2)]
= 12 (–1) − (27)
= 12 (–28) = 21 (–23) − (25)
= 12 (–48)
= 14
= 24
The area of quadrilateral EFGH is 14 units2. The area of quadrilateral EFGH is 24 units2.
115
Smart Tip
Area of a polygon/Luas poligon
1x1 x2 x3 x4 xn x1
y1 y2 y3 y4
=2 yn y1
1 If the vertices are arranged in the anticlockwise order, the answer is positive.
Jika bucu-bucu disusun mengikut tertib arah lawan jam, jawapan yang diperoleh adalah positif.
2 If the vertices are arranged in the clockwise order, the answer is negative.
Jika bucu-bucu disusun mengikut tertib arah jam, jawapan yang diperoleh adalah negatif.
Exercise 12 Find the area of the following polygons.
Cari luas bagi poligon berikut.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 11
A(1, 0), B(2, –1), C(3, 6), D(0, 3), E(–3, 4)
Solution
Area/Luas = 21
1 23 0 –3 1
0 –1 6 3 4 0
= 21[(1)(–1) + (2)(6) + (3)(3) + (0)(4) + (–3)(0)] – [(0)(2) + (–1)(3) + (6)(0) + (3)(–3) + (4)(1)]
= 21 [–1 + 12 + 9 + 0 + 0] – [0 – 3 + 0 – 9 + 4]
= 21 (20) − (–8)
= 2128
= 14
The area of polygon ABCDE is 14 units2.
Luas poligon ABCDE ialah 14 unit2.
1 P(–2, –1), Q(1, 0), R(2, 5), S(1, 4), T(0, 6) 2 P(0, 6), Q(2, 7), R(3, 5), S(4, 0), T(3, –1)
Area = –2 1 2 Area =
12 –1 0 5 1 0 –2 21 0 2 3 4 3 0
4 6 –1 6 7 5 0 –1 6
= 21 [(–2)(0) + (1)(5) + (2)(4) + (1)(6) + (0)(–1)] = 21 [(0)(7) + (2)(5) + (3)(0) + (4)(–1) + (3)(6)]
− [(–1)(1) + (0)(2) + (5)(1) + (4)(0) + (6)(–2)] – [(6)(2) + (7)(3) + (5)(4) + (0)(3) +(–1)(0)]
= 12 [0 + 5 + 8 + 6 + 0] – [–1 + 0 + 5 + 0 – 12] = 21 [0 + 10 + 0 + (–4) + 18] – [12 + 21 + 20 + 0 + 0]
= 12 (19) – (–8) = 21 (24) – (53)
= 21 (27) = 21 (–29)
= 13.5 = 14.5
The area of polygon PQRST is 13.5 units2 The area of polygon PQRST is 14.5 units2
116
Exercise 13 Solve each of the following.
Selesaikan setiap yang berikut.
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang geometri koordinat dalam konteks penyelesaian masalah rutin yang mudah.
Example 12
In the diagram below, the shaded region represents a newly cleared campsite.
Dalam rajah di bawah, rantau berlorek mewakili kawasan tapak perkhemahan yang baru dibersihkan.
N(–7, 6) y
(5 m)
P(3, 4)
M(–5, 2)
O x
(5 m)
Calculate the area, in m2, of the campsite.
Hitung luas, dalam m2, kawasan tapak perkhemahan itu.
Solution
Area/Luas = 21 0
0 3 –7 –5 0
4 6 2 0
= 12[(0)(4) + (3)(6) + (–7)(2) + (–5)(0)] – [(0)(3) + (4)(–7) + (6)(–5) + (2)(0)]
= 21 [0 + 18 – 14 + 0] – [0 – 28 – 30 + 0]
= 21 (4) – (–58)
= 1262
= 31
The area of quadrilateral OMPN is (31 × 5 × 5) m2 = 775 m2.
Luas sisi empat OMPN ialah (31 × 5 × 5) m2 = 775 m2.
1 The diagram below shows the region of an orchard that belongs to Lokman.
Rajah di bawah menunjukkan kawasan dusun milik Lokman.
y
A(4, 8)
B(–3, 6)
O x
D(8, 0)
C(0, –4)
Calculate the area, in units2, of the orchard.
Hitung luas, dalam unit2, dusun itu.
Area =
12 4 –3 0 8 4
8 6 –4 0 8
= 21 [(4)(6) + (–3)(–4) + (0)(0) + (8)(8)] – [(8)(–3) + (6)(0) + (–4)(8) + (0)(4)]
= 12 [24 + 12 + 0 + 64] – [–24 + 0 – 32 + 0]
= 12 (100) – (–56)
= 12 156
= 78
The area of the orchard is 78 units2.
117
2 Determine whether the points (–2, 7), (1, 4), (7, –2) are collinear or not.
Tentukan sama ada titik-titik (–2, 7), (1, 4), (7, –2) adalah segaris atau tidak.
Area = 21
–2 1 7 –2
7 4 –2 7
= 12 [(–2)(4) + (1)(–2) + (7)(7)] – [(7)(1) + (4)(7) + (–2)(–2)]
= 12 [–8 + (–2) + 49] – [7 + 28 + 4]
= 12 (39) – (39)
= 0
The points (–2, 7), (1, 4), (7, –2) are collinear.
3 Show that the points (–1, –7), (0, –6), (8, 2), (9, 3), (2, –4) are collinear.
Tunjukkan bahawa titik-titik (–1, –7), (0, –6), (8, 2), (9, 3), (2, –4) adalah segaris.
Area = 21
–1 0 8 9 2 –1
–7 –6 2 3 –4 –7
= 12 [(–1)(–6) + (0)(2) + (8)(3) + (9)(–4) + (2)(–7)] – [(–7)(0) + (–6)(8) + (2)(9) + (3)(2) +(–4)(–1)]
= 12 [6 + 0 + 24 + (–36) + (-14)] – [0 + (–48) + 18 + 6 + 4]
= 12 (–20) – (–20)
= 12 –20 + 20
= 0
The points (–1, –7), (0, –6), (8, 2), (9, 3), (2, –4) are collinear.
7.4 Equations of Loci / Persamaan Lokus A(x, y)
d
Smart Tip •
B(x1, y1)
The equation of the locus of point A(x , y) which moves such that its distance is d units from a
fixed point B(x1 , y1):
Persamaan lokus bagi titik A(x , y) yang bergerak dengan keadaan jaraknya ialah d unit dari satu titik tetap B(x1 , y1):
(x – x1)2 + (y − y1)2 = d2
Exercise 14 Find the equation of the locus of point L(x, y) which moves such that its distance is d units
from a fixed point P.
Cari persamaan lokus bagi titik L(x, y) yang bergerak dengan keadaan jaraknya ialah d unit dari satu titik
tetap P.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 13 1 P(0, 2); d = 3 units/unit
P(3, –4); d = 7 units/unit L(x, y) PL = 3
•
Solution (x − 0)2 + (y − 2)2 = 32
PL = 7 7 units x2 + (y − 2)2 = 32
(x – x1)2 + (y – y1)2 = d2 • x 2 + y2 − 4y + 4 − 9 = 0
(x – 3)2 + (y – (–4))2 = 72 P(3, –4) x2 + y2 − 4y − 5 = 0
(x – 3)2 + (y + 4)2 = 49
x2 – 6x + 9 + y2 + 8y + 16 = 49
x2 + y2 – 6x + 8y – 24 = 0
118
2 P(–5, 0); d = 2 units/unit 3 P(–3, 6); d = 5 units/unit
PL = 5
PL = 2 [x – (–3)]2 + (y – 6)2 = 52
[x – (–5)]2 + (y – 0)2 = 22 (x + 3)2 + (y – 6)2 = 52
(x + 5)2 + (y)2 = 22 x 2 + 6x + 9 + y2 – 12y + 36 = 25
x 2 + 10x + 25 + y2 – 4 = 0 x2 + y2 + 6x – 12y + 20 = 0
x2 + y2 + 10x + 21 = 0
4 L (x, y) PL = 6 5 PL = 4
• [x – (–2)]2 + [y – (–8)]2 = 62
(x – 3)2 + (y – 7)2 = 42
6 units (x + 2)2 + (y + 8)2 = 62
• x2 + 4x + 4 + y2 + 16y + 64 = 36 4 un itsP( 3•, 7) x2 –x62x++y92 + y2 – 14y + 49 = 16
– 6x – 14y + 42 = 0
P(–2, –8) x2 + y2 + 4x + 16y + 32 = 0
•
L(x, y)
Smart Tip
The equation of the locus of point A(x, y) which moves such that its distance from point A(x, y)
P(x1 , y1) and point Q(x2 , y2) are in the ratio of m : n
Persamaan lokus bagi titik A(x , y) yang bergerak dengan keadaan jaraknya dari titik P(x1 , y1) dan titik Q(x2 , y2)
adalah dalam nisbah m : n
(x – x1)2 + (y – y1)2 = mn or/atau n(x – x1)2 + (y – y1)2 = m(x – x2)2 + (y – y2)2 P(x1, y1) Q(x2, y2)
(x – x2)2 + (y – y2)2
Exercise 15 Find the equation of the following loci.
Cari persamaan lokus berikut.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 14
Given A(2, –3) and B(–1, 2). P(x, y) is a moving point Smart Tip
such that AP : PB = 2 : 3. Find the equation of locus P.
Diberi A(2, –3) dan B(–1, 2). P(x, y) ialah satu titik yang If the locus of point P is equidistant (m : n = 1 : 1)
bergerak dengan keadaan AP : PB = 2 : 3. Cari persamaan lokus P.
from two fixed points, A(x1, y1) and B(x2, y2), AB, the
Solution locus is the perpendicular bisector of the line AB.
AP 2
PB = 3 Jika lokus titik P adalah sama jarak (m : n = 1 : 1) dari dua titik
3AP = 2PB
tetap, A(x1, y1) dan B(x2, y2), AB, lokus itu ialah pembahagi dua
3(x − 2)2 + [y − (–3)]2 = 2[x − (–1)]2 + (y – 2)2
sama serenjang bagi garis AB. Locus P
9[(x − 2)2 + (y + 3)2] = 4[(x + 1)2 + (y − 2)2]
9(x2 − 4x + 4 + y2 + 6y + 9) = 4(x2 + 2x + 1 + y2 − 4y + 4) AP = PB Lokus P
9(x2 − 4x + y2 + 6y + 13) = 4(x2 + 2x + y2 − 4y + 5)
9x2 − 36x + 9y2 + 54y + 117 = 4x2 + 8x + 4y2 − 16y + 20 (x – x1)2 + (y – y1)2 = (x – x2)2 + (y – y2)2 A B
5x2 + 5y2 − 44x + 70y + 97 = 0
(x – x1)2 + (y – y1)2 = (x – x2)2 + (y – y2)2
119
1 Given A(–1, 2) and B(3, 4). P(x, y) is a moving 2 Given A(2, –1) and B(–1, 3). P(x, y) is a moving
point such that AP : PB = 1 : 2. Find the equation point such that AP : PB = 2 : 1. Find the equation
of locus P. of locus P.
Diberi A(–1, 2) dan B(3, 4). P(x, y) ialah satu titik yang bergerak DiberiA(2,–1)danB(–1,3).P(x,y)ialahsatutitikyangbergerak
dengan keadaan AP : PB = 1 : 2. Cari persamaan lokus P. dengan keadaan AP : PB = 2 : 1. Cari persamaan lokus P.
AP 1 AP 2
PB = 2 PB = 1
2AP = PB AP = 2PB
2A(x − (–1))2 + (y − 2)2 = A(x – 3)2 + (y − 4)2 A(x − 2)2 + [y − (–1)]2 = 2A[x − (–1)]2 + (y − 3)2
4[(x + 1)2 + (y − 2)2] = (x − 3)2 + (y − 4)2 (x − 2)2 + (y + 1)2 = 4[(x + 1)2 + (y − 3)2]
4(x2 + 2x + 1 + y2 − 4y + 4) = x2 − 6x + 9 + y2 – 8y + 16 x2 − 4x + 4 + y2 + 2y + 1 = 4(x2 + 2x + 1 + y2 − 6y + 9)
4(x2 + 2x + y2 − 4y + 5) = x2 − 6x + y2 − 8y + 25 x2 − 4x + y2 + 2y + 5 = 4(x2 + 2x + y2 − 6y + 10)
4x2 + 8x + 4y2 − 16y + 20 = x2 – 6x + y2 − 8y + 25 x2 − 4x + y2 + 2y + 5 = 4x2 + 8x + 4y2 − 24y + 40
3x2 + 3y2 + 14x − 8y − 5 = 0 0 = 3x2 + 12x + 3y2 − 26y + 35
3x2 + 3y2 + 12x − 26y + 35 = 0
3 Given A(–2, –1) and B(3, 2). P(x, y) is a moving 4 A point P(x, y) moves such that it is always
point such that AP : PB = 2 : 3. Find the equation equidistant from two fixed points, A(7, 7) and
of locus P. B(–4, 0). Find the equation of locus P.
DiberiA(–2,–1)danB(3,2).P(x,y)ialahsatutitikyangbergerak Titik P(x, y) bergerak dengan keadaan sentiasa berjarak
dengan keadaan AP : PB = 2 : 3. Cari persamaan lokus P. sama dari dua titik tetap, A(7, 7) dan B(–4, 0). Cari
AP 2 persamaan lokus P.
PB = 3
3AP = 2PB AP = PB
(x − x1)2 + (y − y1)2 = (x − x2)2 + (y − y2)2
3A[x − (–2)]2 + [y − (–1)]2 = 2A(x − 3)2 + (y − 2)2 (x − 7)2 + (y − 7)2 = [x − (–4)]2 + (y − 0)2
x2 − 14x + 49 + y2 − 14y + 49 = x2 + 8x + 16 + y2
9[(x + 2)2 + (y + 1)2] = 4[(x − 3)2 + (y − 2)2] 22x + 14y − 82 = 0 (÷2)
9(x2 + 4x + 4 + y2 + 2y + 1) = 4(x2 − 6x + 9 + y2 − 4y + 4) 11x + 7y − 41 = 0
9(x2 + 4x + y2 + 2y + 5) = 4(x2 − 6x + y2 − 4y + 13)
9x2 + 36x + 9y2 + 18y + 45 = 4x2 − 24x + 4y2 − 16y + 52
5x2 + 5y2 + 60x + 34y − 7 = 0
Exercise 16 Solve each of the following.
Selesaikan setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang geometri koordinat untuk melaksanakan tugasan mudah.
Example 15 1 Determine whether the locus x2 + y2 − 3x + 4y − 9
= 0 passes through the x-axis or not.
Determine whether the locus x2 + y2 + 2x + 3y – 12
= 0 passes through the x-axis or not. Tentukan sama ada lokus x2 + y2 – 3x + 4y – 9 = 0 melalui
Tentukan sama ada lokus x2 + y2 + 2x + 3y – 12 = 0 paksi-x atau tidak.
melalui paksi-x atau tidak.
On the x–axis, y = 0
Solution x2 + (0)2 – 3x + 4(0) – 9 = 0
On the x-axis, y = 0 x2 – 3x – 9 = 0
Pada paksi-x, y = 0 b2 – 4ac = (–3)2 – 4(1)(–9)
x2 + (0)2 + 2x + 3(0) – 12 = 0 = 45
x2 + 2x – 12 = 0 45 > 0
a = 1, b = 2, c = –12
b2 – 4ac = (2)2 – 4(1)(–12) = 52 > 0 The locus passes through the x-axis.
The locus passes through the x–axis.
Lokus itu melalui paksi–x.
Smart Tip
b2 – 4ac > 0
(locus passes through the x-axis/y-axis)
b2 – 4ac > 0
(lokus melalui paksi-x/paksi-y)
b2 – 4ac < 0
(locus does not pass through the x-axis/y-axis)
b2 – 4ac < 0
(lokus tidak melalui paksi-x/paksi-y)
120
2 Determine whether the locus 2x2 + 2y2 – 5x + 6y 3 Determine whether the locus x2 + y2 + 4x – 7y + 12
+ 13 = 0 cuts the y-axis or not. = 0 intersects with the straight line y = –x + 1 or not.
Tentukan sama ada lokus 2x2 + 2y2 – 5x + 6y + 13 = 0 Tentukan sama ada lokus x2 + y2 + 4x – 7y + 12 = 0
memotong paksi-y atau tidak.. bersilang dengan garis lurus y = –x + 1 atau tidak.
On the y-axis, x = 0 On the line y = –x + 1,
2(0)2 + 2y2 – 5(0) + 6y + 13 = 0 x2 + (–x + 1)2 + 4x – 7(–x + 1) + 12 = 0
2y2 + 6y + 13 = 0 x2 + x2 – 2x + 1 + 4x + 7x – 7 + 12 = 0
b2 – 4ac = (6)2 – 4(2)(13) 2x2 + 9x + 6 = 0
= –68 b2 – 4ac = (9)2 – 4(2)(6)
= 33
–68 < 0
33 > 0
The locus does not cut the y-axis.
The locus intersects with the straight line
y = –x + 1.
Exercise 17 Solve each of the following.
Selesaikan setiap yang berikut.
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang geometri koordinat dalam konteks penyelesaian masalah rutin yang mudah.
1 The diagram shows part of the plan of a park drawn on a Cartesian y
plane. H is a point on PQ such that PH : HQ = 4 : 1. A straight
walkway perpendicular to PQ will be built through point H. Find the Hut P/Pondok P
(–120, 30) •
equation of the walkway. Hence, determine whether the flag pole R
should be moved to another place or not. HOTS Applying O x
Rajah menunjukkan sebahagian daripada pelan kawasan sebuah taman yang dilukis Hut Q/Pondok Q • Flag pole R
pada satah Cartes. H ialah satu titik pada PQ dengan keadaan PH : HQ = 4 : 1. Satu (–20, –100) • Tiang bendera R
laluan pejalan kaki lurus yang berserenjang dengan PQ akan dibina melalui titik H.
(25, –24)
Cari persamaan laluan pejalan kaki itu. Seterusnya, tentukan sama ada tiang bendera
R harus dipindahkan ke tempat lain atau tidak.
nx1 + mx2 ny1 + my2 The equation of the If x = 25, R(25, –24)
m+n m+n walkway,
H = , 13y = 10(25) − 562
10
1(–120) + 4(–20) 1(30) + 4(–100) y – (–74) = 13 [x – (–40)] 13y = –312
13(y + 74) = 10(x + 40)
= 4+1 , 4+1 13y + 962 = 10x + 400 y = –24
= (–40, –74) 13y = 10x – 562 The flag pole R is on the
walkway and should be
Gradient of PQ, moved to another place.
mPQ = –100 – 30 =– 13
–20 – (–120) 10
Let the gradient of walkway = m2
mPQ × m2 = –1
m2 = 10
13
121
2 The diagram shows a part of camp sites, A, B and C drawn on a y Camp B/Kem B
Cartesian plane. A watchtower will be built such that equidistant (300, 500)
from both Camp A and Camp B and nearest to Camp C. Find the Camp A/Kem A •
coordinates of the watchtower. HOTS Applying (–600, 200) •
x
Rajah menunjukkan sebahagian kawasan perkhemahan, A, B dan C yang dilukis 0
pada satah Cartes. Sebuah menara pemerhati akan dibina dengan keadaan berjarak • Camp C/Kem C
sama dari Kem A dan Kem B serta paling dekat dengan Kem C. Cari koordinat (500, –600)
menara pemerhati itu.
Midpoint AB = –600 + 300 , 200 + 500 y − 350 = –3x − 450 Solve
2 2 3x + y = –100 3x + y = –100 1
x − 3y = 2 300 2
= (–150, 350) Let Q = the watchtower and 9x + 3y = –300 1 × 3
perpendicular to x − 3y = 2 300 2 (+)
500 − 200 3x + y = –100.
Gradient AB = 300 − (–600) 10x = 2 000
Hence, the gradient of x = 200
300 1 1 From 1 , y = –100 − 3(200)
= 900 = 3
QC = 3 = –700
Hence, the gradient that perpendicular Equation of QC is Q = (200, –700)
∴The coordinates of the
to AB is –3. The equation of the locus 1 watchtower are (200, –700)
y − (–600) = 3(x − 500)
that is equidistant from A and B: 3(y + 600) = 1(x − 500)
3y + 1 800 = x − 500
y − y1 = m2(x − x1)
y − 350 = –3[x − (–150)] x − 3y = 2 300
Review 7
Paper 1 Questions
1 The following information refers to the equation pada satah Cartes, dengan keadaan E dan F terletak di
sebelah tebing sungai yang sama.
SPM of two straight lines, PQ and RS.
CLONE y
`18 Maklumat berikut adalah merujuk kepada persamaan dua
P1Q10 garis lurus, PQ dan RS. F(8, 6)
PQ: y = 4ax – 6 E(–1, 3)
RS: x + y = 1
Ox
6b 8
G(4, –2)
where a and b are constants.
dengan keadaan a dan b ialah pemalar. Diagram 1/ Rajah 1
Khalid wants to cross the river from tent G to
Given the straight lines PQ and RS are
perpendicular to each other. Express a in terms of the opposite riverbank where the tents E and F
b. are located. Find the shortest distance, in m, that
Diberi garis lurus PQ dan RS adalah berserenjang antara he can take to cross the river. Give your answer
correct to three decimal places.
satu sama lain. Ungkapkan a dalam sebutan b. Khalid hendak menyeberangi sungai dari khemah G ke tebing
[3 marks/markah] sungai bertentangan di mana khemah E dan F terletak.
Cari jarak terpendek, dalam m, yang dia boleh lalui untuk
2 Diagram 1 shows the position of three tents, E, menyeberangi sungai itu. Beri jawapan anda betul kepada
tiga tempat perpuluhan.
SPM F and G near a riverbank drawn on a Cartesian
CLONE [4 marks/markah]
`18 plane, such that E and F lie on the same side of
P1Q23
the riverbank.
Rajah 1 menunjukkan kedudukan bagi tiga buah khemah, E,
F dan G berhampiran tebing sebatang sungai yang dilukis
122
3 A straight line passes through K(5, 3) and L(15, 8). (a) State the value of p.
Nyatakan nilai p.
SPM The point N divides the line segment KL such that
CLONE (b) Find the coordinates of H.
`17 2KL = 5NL. Find the coordinates of N. Cari koordinat H.
P1Q18 Satu garis lurus melalui K(5, 3) dan L(15, 8). Titik N
membahagi tembereng garis KL dengan keadaan 2KL = 5NL. [3 marks/markah]
Cari koordinat N.
[3 marks/markah] 6 The straight line y = –3x + 12 is parallel to the
4 The straight line 4y = 2x + p + 8 intersects the SPM straight line y = (h + 4)x + 9, where h is a constant.
CLONE
SPM y-axis at (0, 3q), where p and q are constants. `14 Determine the value of h.
P1Q13 Garis lurus y = –3x + 12 adalah selari dengan garis lurus
CLONE Express p in terms of q.
`16 y = (h + 4)x + 9, dengan keadaan h ialah pemalar.
P1Q8 Garis lurus 4y = 2x + p + 8 bersilang dengan paksi-y pada
Tentukan nilai h.
(0, 3q), dengan keadaan p dan q ialah pemalar. Ungkapkan p
[2 marks/markah]
dalam sebutan q.
[2 marks/markah] 7 Diagram 3 shows a straight line FH on a Cartesian
plane.
5 Diagram 2 shows two straight lines on a Cartesian Rajah 3 menunjukkan garis lurus FH pada satah Cartes.
SPM plane. y
CLONE
`16 Rajah 2 menunjukkan dua garis lurus pada satah Cartes.
P1Q9
y
y = 2x + 6
F(p, 5)
G(6, q)
x O x
O H(8, 0)
H Diagram 3/ Rajah 3
y = px − 4
Point G lies on the straight line FH such that FG :
Diagram 2/ Rajah 2 GH = 4 : 1. Find the values of p and q.
Both straight lines are perpendicular to each other. Titik G terletak pada garis lurus FH dengan keadaan FG :
GH = 4 : 1. Cari nilai p dan q.
Kedua-dua garis lurus itu adalah berserenjang antara satu [3 marks/markah]
sama lain.
Paper 2 Questions y
1 Diagram 1 shows a triangle OFG.
SPM Rajah 1 menunjukkan sebuah segi tiga OFG.
CLONE
`18
P2Q3
F(–7, 9)
G(p, 3) [2 marks/markah]
Ox [4 marks/markah]
Diagram 1/ Rajah 1
(a) Given the area of the triangle OFG is 33 units2, find the value of p.
Diberi luas segi tiga OFG ialah 33 unit2, cari nilai p.
(b) Point H(1, 5) lies on the straight line FG.
Titik H(1, 5) terletak pada garis lurus FG.
(i) Find FH : HG.
Cari FH : HG.
(ii) Point L moves such that LG = 2LH.
Find the equation of locus L.
Titik L bergerak dengan keadaan LG = 2LH.
Cari persamaan lokus L.
123
2 Diagram 2 shows the location of town G and town H on a Cartesian plane.
SPM Rajah 2 menunjukkan kedudukan bandar G dan bandar H pada satah Cartes.
CLONE
`17
P2Q5 y
K
Town G
Bandar G
(–5, 2)
Ox
Town H
Bandar H
(7, –4)
L
Diagram 2/ Rajah 2
KL is a straight road such that the distance from town G and town H to any point on the road is always equal.
KL ialah sebuah jalan raya lurus dengan keadaan jarak dari bandar G dan bandar H ke mana−mana titik pada jalan raya adalah
sentiasa sama.
(a) Find the equation of KL.
Cari persamaan bagi KL.
[3 marks/markah]
(b) Another straight road, PQ with the equation y = 3x − 7 is to be built.
Sebuah jalan raya lurus yang lain, PQ dengan persamaan y = 3x − 7 akan dibina.
(i) A traffic light is to be installed at the crossroads of the two roads. Find the coordinates of the traffic
light.
Sebuah lampu isyarat akan dipasang di persimpangan kedua-dua jalan raya itu. Cari koordinat bagi lampu isyarat itu.
1 2(ii) Which road passes through town T 2 , −5 ?
3
1 2Jalan yang manakah melalui bandar T 2
3 , −5 ?
[4 marks/markah]
3 Diagram 3 shows a quadrilateral DEFG. The equation of the straight line DE is y = 2x + 5.
SPM Rajah 3 menunjukkan sebuah sisi empat DEFG. Persamaan bagi garis lurus DE ialah y = 2x + 5.
CLONE
`13
P2Q9 yE
y = 2x + 5 F(7, 7)
D(–1, 1) x
O
G
Diagram 3/ Rajah 3
Find/Cari
(a) the equation of the straight line GF,
persamaan garis lurus GF,
(b) the equation of the straight line DG, [2 marks/markah]
persamaan garis lurus DG, [3 marks/markah]
[2 marks/markah]
(c) the coordinates of G, [3 marks/markah]
koordinat G,
(d) the area, in unit2, of the quadrilateral DEFG.
luas, dalam unit2, sisi empat DEFG.
124
H OTS Zone
1 Diagram 1 shows the position of two squirrels, F and G.
SPM Rajah 1 menunjukkan kedudukan dua ekor tupai, F dan G. y
CLONE
`15
P1Q12
Squirrel F Squirrel G
Tupai F Tupai G
x
O
Diagram 1/ Rajah 1
The coordinates of squirrel F and squirrel G are (–5, 4) and (11, 12) respectively. Both squirrels move towards
each other on a straight line at different speeds. The speed of squirrel F is three times the speed of squirrel G.
Find the distance of squirrel F from its initial position when it meets squirrel G. HOTS Applying
Koordinat bagi tupai F dan tupai G masing-masing ialah (–5, 4) dan (11, 12). Kedua-dua tupai itu bergerak ke arah satu sama
lain pada suatu garis lurus dengan laju yang berbeza. Laju bagi tupai F ialah tiga kali ganda laju tupai G. Cari jarak tupai F dari
kedudukan awalnya apabila tupai F bertemu dengan tupai G.
2 Diagram 2 shows a campsite OMNP that had been cleared by a group of scouts. Points F, G and H are centre
SPM of the tents, M, N and P, respectively. F, G and H are collinear.
CLONE
`15 Rajah 2 menunjukkan sebuah tapak perkhemahan OMNP yang telah dibersihkan oleh sekumpulan pengakap. Titik-titik F, G dan H
P2Q3 masing-masing ialah pusat khemah M, N dan P. F, G dan H adalah segaris.
y
N(10, 14)
H(11, 8)
M(–5, 3) G(2, 5) P(15, 1)
F
x
O
Diagram 2/ Rajah 2
Given the distance of point H is three times from point G and four times from point F.
Diberi jarak titik H ialah tiga kali ganda dari titik G dan empat kali ganda dari titik F.
(a) Find/Cari,
(i) the area, in units2, of the campsite OMNP,
luas, dalam unit2, tapak perkhemahan OMNP,
(ii) the coordinates of F.
koordinat bagi F.
(b) A scout spread sulphur powder around tent H such that the distance of the sulphur powder track from
the centre of H is always 3 m. Find the equation of the track of the sulphur powder. HOTS Applying
Seorang pengakap menabur serbuk belerang mengelilingi khemah H dengan keadaan jarak serbuk belerang dari pusat khemah
H ialah sentiasa 3 m. Cari persamaan laluan serbuk belerang itu.
125
Chapter Learning Area: Geometry
8 Vectors
Vektor
8.1 Vectors / Vektor
Smart Tip
Scalar quantity is a quantity that has magnitude only.
Kuantiti skalar ialah suatu kuantiti yang hanya mempunyai magnitud.
Vector quantity is a quantity that has magnitude and direction.
Kuantiti vektor ialah kuantiti yang mempunyai magnitud dan arah.
Exercise 1 Identify whether the quantity given is a scalar quantity or a vector quantity.
Kenal pasti sama ada kuantiti yang diberi ialah kuantiti skalar atau kuantiti vektor.
TP 1 Mempamerkan pengetahuan asas tentang vektor.
Quantity Scalar quantity Vector quantity
Kuantiti Kuantiti skalar Kuantiti vektor
Example 1 Solution Displacement
Sesaran
Distance and displacement Distance
Jarak dan sesaran Jarak
1 Velocity and speed Speed Velocity
Halaju dan laju Mass Force
Volume
2 Force and mass Speed Acceleration
Daya dan jisim Work Momentum
3 Acceleration and volume Weight
Pecutan dan isi padu
4 Speed and momentum
Laju and momentum
5 Work and weight
Kerja dan berat
Smart Tip
1 A vector can be represented by a directed line segment. An arrow shows the direction of the vector from the initial
point to the terminal point and the length of the line represents the magnitude of the vector.
Suatu vektor boleh diwakilkan menggunakan tembereng garis berarah. Anak panah menunjukkan arah vektor dari titik awal ke titik terminal
dan panjang garis mewakili magnitud vektor.
Terminal point/Titik terminal
Initial point/Titik awal
→
2 Vector can be denoted by ~a , AB, a, or AB.
→
VTheketomr baoglenhidtuitadnedaokfanthseebvageacit~oa,rAcBa,nab, aetadueAno→Bt.ed
3 by |~a |, | → | a| , or | AB| .
AB|,
Magnitud vektor boleh ditandakan sebagai |~a |, | AB|, | a|, atau | AB|.
126
Exercise 2 Find the magnitude of each of the following vectors.
Cari magnitud bagi setiap vektor yang berikut.
TP 2 Mempamerkan kefahaman tentang vektor.
Example 2 → →
1 |PQ| 2 |PQ|
→
|AD| 1 unit P 1 unit 1 unit
D 1 unit Q 1 unit
1 unit
Q
→ P
|PQ| = 4 units
→
A |PQ| = 3 units
Solution
→
|AD| = 62 + 32
= 36 + 9
= 45
= 6.708 units/unit
→ 4 |a| 5 |v| 1 unit
3 |AB| 1 unit
1 unit 1 unit ~v
A 1 unit 1 unit
~a
B
→ 32 + 42 |a| = 62 + 22 |v| = 122 + 52
|AB| = = 36 + 4 = 144 + 25
= 9 + 16 = 40 = 169
= 25
= 6.325 units = 13 units
= 5 units
Exercise 3 Draw each of the following vectors.
Lukiskan setiap vektor yang berikut.
TP 2 Mempamerkan kefahaman tentang vektor.
Example 3 1 RT(aah)j eah2d~umia,e gnuranmjukskhanowv(ebsk)tt ohr−e~u2.v~uLe,u cktiosrka~un. Draw
(c) 3~u .
Smart Tip
The diagram shows the
1 If a is a vector
vReajcathomr ~aen. uDnrjuakwkan vektor ~a . and k is a scalar,
Lukiskan then magnitude
oJikfaka~aiailsahksutiamtuevsek~ato.r
(a) −~a , (b) 2~a , (c) 23 ~a . dan k ialah skalar, ~u
maka magnitud ka
Solution (a) (b)
2 Iiiasflakinh>kth0ka,eltish~aae.mn ek~a 2u~ –2~u
Jdaidkiaraleakch>tsioa0m,nmawadkeiatnhgk~aa~an. (c)
~a 3 Iiasfrakihn<~at.h0,etohpenpoks~aite
dJaidkiaraleakch<tbio0er,ntmeonaftka~aan.kg~aan
–~a dengan arah ~a. 3u~
(a)
(b) 2~a
(c) 3 ~a
2
127
Exercise 4 Solve the following problems. 1 F ~u G
Selesaikan masalah yang berikut. J
TP 2 Mempamerkan kefahaman tentang vektor. M
P
Example 4 S
~a B C K
Q
AF N
R
D
Q
E
P
→ ~u. State each the
Given FG = of following in
→ each following
Given AB = ~a . State of the in
terms of ~u.
terms of ~a . →
→ Diberi FG = ~u. Nyatakan setiap yang berikut dalam sebutan ~u.
Diberi AB = ~a . Nyatakan setiap yang berikut dalam → −→
sebuta→n ~a. → → (a) JK, (b) MN,
→→
(c) PQ, (d) RS.
(a) CD, (b) EF, (c) PQ.
Solution (a) → 1 ~u
→ JK = 2
(a) CD = −~a −→
→
(b) EF = 3~a (b) MN = 2~u
→ (c) → 3
(c) PQ = 4~a PQ = 2 ~u
(d) → = − 5 ~u
RS 2
2 Q 3 P
w~ ~x
P Q
BCE BC
A D
H
E
F
AD
→ F
Given PQ = w~. State each of the following in
terms
of w~. G
→
Diberi PQ = w~. Nyatakan setiap yang berikut dalam →
sebuta→n w~. → → Given PQ = ~x. State each of the following in
terms
(a) AB, (b) CD, (c) EF. of ~x.
→
Diberi PQ = ~x. Nyatakan setiap yang berikut dalam sebutan ~x.
→ →
→ 5 (a) AB, (b) CD,
(a) AB = 2 w~ →→
→ 5 (c) EF, (d) GH.
CD 2
(b) → = − w~ →
(c) EF = −2w~ (a) AB = −~x
→
((cb)) EC→→FD==322~x~x
(d) GH = −3~x
128
Smart Tip
If vector ~u is parallel to the vector ~v, then ~u = λ~v.
Note: if h~u = k~v, where h and k are constants, then λ =
1 k
h
3 4
Jika vektor ~u selari dengan vektor ~v, maka ~u = λ~v. k
3 4Nota: jika hu = kv, dengan keadaan h dan k ialah pemalar, maka λ = h
2 If vector ~u~u daanndvveketcorto~vrt~vidaarkesneloatripdaarnahll~ue=l ak~nv,dmha~uka=h k=~v0, then h = 0 and k = 0.
Jika vektor dan k = 0.
Exercise 5 Show that each pair of the following vectors are parallel.
Tunjukkan bahawa setiap pasangan vektor berikut adalah selari.
TP 3 Mengaplikasikan kefahaman tentang vektor untuk melaksanakan tugasan mudah.
Example 5 →→ →→
1 AB = 10~x and/dan CD = 2~x. 2 BC = 3~x and/dan EF = −5~x.
→
1 3 → EF = −−51~xE→F
~u = 2 ~x + 2 ~y and/dan ~v = ~x + 3~y. AB = 10~x →~x = 5
Solution = 5(2→~x)
= 5CD
~u = 1 ~x + 3 ~y →→ 1 2 BC = 3~x 1 →
2 2 AB = 5CD = 3− EF
1 1 →→
= 2 (~x + 3~y) = 2 ~v ∴ AB and CD are parallel. 5
3 →
= − EF
1 5
~u = 2 ~v → 3 →
B C = − EF
∴ ~~uu daannd~v~vadaarleahpsaerlaarlil.el. →5 →
∴ BC and EF are parallel.
3 ~u = 8~x + 2~y and/dan ~v = 2~x + 12~y. 4 ~x = 4~a – 1 ~b and/dan y = 24~a – 3~b .
2 ~
1 2~u = 1 2~y =
= =
8~x + +2~y1 24~a – 3~b
4 2~x 2 1
~y 6 4~a − 2 ~b
~ u∴ == ~u44a~~vvnd ~v are parallel. ~y∴ == ~y66a~~xxnd ~x are parallel.
Exercise 6 GDiibveerindtuwa ovenktoonr -ypanagratlildealkasneldarindoann-zbuekraonvseifcatro, r~usd, a~una~vn. dCa~vr.i Find the values of h and k.
nilai h dan k.
TP 3 Mengaplikasikan kefahaman tentang vektor untuk melaksanakan tugasan mudah.
Example 6 1 h~u = (k + 3)~v
h=0
(h + 1)~u = (3k – 5)~v , k + 3 = 0
k = –3
Solution
h + 1 = 0 , 3k – 5 = 0 ∴ h = 0 and k = –3
3k = 5
h = −1 k = 5
3
∴ h = −1 and/dan k = 5
3
129
2 (3h – 2)~x = (2k + 8)~y 3 (2h + 1)~u = (k2 + 2k – 3)~v
3h – 2 = 0 , 2k + 8 = 0 2h + 1 = 0 , k2 + 2k – 3 = 0
h = 2 2k = −8 2h = –1 (k + 3)(k −1) = 0
3 k = −4
h = – 1 k = −3, 1
2
∴ h = 2 and k = −4
3 ∴ h = − 1 and k = −3, k = 1
2
8.2 Addition and Subtraction of Vectors / Penambahan dan Penolakan Vektor
Smart Tip
Triangle Law Parallelogram Law Polygon Law
Hukum Segi Tiga Hukum Segi Empat Selari Hukum Poligon
C DC D
~d ~c
~a + ~b ~b ~a + ~b
EC
~b A ~a B
~e ~b
A ~a B →→ →
AC = AB + AD A ~a B
→ →→ = ~a + ~b → →→ → →
AC = AB + BC AE = AB + BC + CD + DE
= ~a + ~b ~e = ~a + ~b + ~c + ~d
Exercise 7 Determine the single vector for each of the following.
Tentukan vektor tunggal bagi setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang vektor untuk melaksanakan tugasan mudah.
Example 7 1 Given PQRS is a parallelogram. 2 Given PSR is a triangle.
Diberi PQRS ialah sebuah segi empat Diberi PSR ialah sebuah segi tiga.
Given ABCDE is a regular
hexagon. selari.
Diberi ABCDE ialah sebuah
heksagon sekata. SR
CD SR
BE P Q P
AF Find/Cari →→
→→ Find SR + RP.
Find/Cari
→→ (a) PS + PQ, →→
→→ Cari SR + RP.
(a) AE + ED,
→→ (b) QP + QR. →
→ SP
(b) AC − BC.
Solution (a) PR
→
→
(a) AD (b) QS
→→
(b) AC − BC
→→
= AC – (–CB)
→→
= AC + CB
→
= AB
130
3 Given ABCDEF is a regular 4 Given PQR is a triangle. 5 Given PQRST is a regular
hexagon. Diberi PQR ialah sebuah segi tiga. pentagon.
Diberi ABCDEF ialah sebuah Q Diberi PQRST ialah sebuah
heksagon sekata. pentagon sekata.
CD R
RQ S
BO E P
A F Find/Cari PT
→→
Find/Cari Find/Cari
→→ (a) PQ – RQ, →→
→→
(a) AB + AF, (a) PT – ST,
→→ (b) QP – QR. →→
→→
(b) OD + OF. (b) PR – QR.
→ (a) PQ − RQ →→
→→
(a) AO (a) PT − ST
→ = PQ – (–QR) →→
→→
(b) OE = PT – (–TS)
= PQ + QR →→
→
= PT + TS
= PR →
→→
= PS
(b) QP − QR →→
→→ (b) PR − QR
= QP – (–RQ) →→
→→ = PR – (–RQ )
= QP + RQ →→
→ = PR + RQ
= RP →
= PQ
Exercise 8 Simplify each of the following.
Permudahkan setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang vektor untuk melaksanakan tugasan mudah.
Example 8 1 2~a – 1 ~a 2 ~u – 4 ~u + 1 ~u
4 5 5
(4~p + 3~q) – (~p – 2~q)
= 8 ~a – 1 ~a = 5 ~u – 4 ~u + 1 ~u
Solution 4 4 5 5 5
2
= 34~~pp + 53~~qq – ~p + 2~q = 7 ~a = 5 ~u
= + 4
3 =(=5~a53~~aa++–~b2~b)~b––(22~a~a –+33~b~b) 4 (==~x~x−+3+~x4~y4+~y) –5–~y(44~x~x – ~~yy) 5 =(=2~u~u2~u−+–~v2~v~v) – (~u~u+– 33~v~v)
+ –
6 ==(3~x34~~xx+++2~y24~~yy) ++(~x~x + 22~y~y) 1 27 12~a – 5~b + (2~a + 3~b) 1 28 3~u – 1 ~v + (~u – 2~v)
+ 3
= 12~a – 5~b + 2~a + 3~b
= 25~a – 2~b = 3~u – 1 ~v + ~u – 2~v
3
131
= 4~u – 7 ~v
3
Exercise 9 Solve the following problems.
Selesaikan masalah yang berikut.
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang vektor dalam konteks penyelesaian masalah rutin yang mudah.
Example 9 1 The d→iagram shows→a quadrilater→al EFGH such
that kEHare=c(ohn–st2a)~nxt,sH. G = k~y and FG = k~x , where h
The diagram shows a triangle EFH and G is a and
point on the line FH.
Rajah menunjukkan sebuah segi tiga EFH dan G ialah satu Rajah menunjukkan sebuah sisi H
titik pada garis FH.
empat EFGH dengan keadaan
F →→
EH = (h – 2)~x , HG = ky dan E
→ ~
FG = k~x, dengan keadaan h dan k
ialah pemalar. FG
→ ~x ~y.
(a) Express EF in terms of h, k, and
EG →
Ungkapkan EF dalam sebutan h, k, ~x dan y.
→ = h– ~
2 ~y and 1
Find the values of h if EF 6 3
H 1 1 2 2(b) k = .
→→
Given EF = 5p, EH = 10q and FG = 4GH, find
→ ~→ ~ → h –2 ~y dan k = 1.
Cari nilai-nilai h jika EF = 6 3
Diberi EF = 5p, EH = 10q dan FG = 4GH, cari
→~ ~ → → → →→
(a) EF = EH + HG + GF
(a) FH, (b) EG.
= h((hh~x –––2k2)–~x~x 2+–+)~xk13k~y~y+–––k~2ykk~x~x~x
Solution =
→→ → =
(a) EF + FH = EH h
→ 1 2 1 2(b) –2 ~y = h + 1 ~y
5~p + FH = 10~q 6 3
→
FH = 10~q – 5~p
(b) 1h – 2 2~y = 1h – 7 2~x + 1 ~y
6 3 3
F →→→ Compare scalar for ~x, Compare scalar for y,
EG = EF + FG h – 2 = 1 ~
5~p h – 7 = 0
(4) 10~q – 5~p = 5~p + 4 (10~q – 5~p) 3 63
5 h – 2 = 2
G = 5p + 8q – 4p h = 7
E (1) = ~p~+ 8~q~
10~q H
3 h = 4
2 Gcoinvsetnan~pt=. F3i~ni d– 4t~hj eanpdos~qsi=blke~i v–a5l~uj,ews ohferkeskucishathat 3 The diagram shows a triangle OAC and B is a
= 170. point on the line AC.
D|3~ipbe–ri~qp|
pemala~r. = 3~i – 4j dan q = ki – 5j, dengan keadaan k ialah Rajah menunjukkan sebuah segi tiga OAC dan B ialah satu
Cari nil~ai-nila~i yang m~ungkin bagi k dengan titik pada garis AC.
170 A
keadaan |3p – q| = .
~ ~ ~a B
|3p – q| = 170 O ~c
|3(3|~9i~i––(49~|1j(–)92~j–k−–)(2kkk~+)i~~ii~–(+––57~75j~~~))jj2||| ==== 170 C
170 →→
170 Given OA = ~a ,→OC = ~c and AB : BC = 1 : 3, find
170 →
Diberi OA = ~a , OC = ~c dan AB : BC = 1 : 3, cari
→ →
(a) AC, (b) OB.
(9 – k)2 + 49 = 170 →→→
(a) OA + AC = OC
81 – 18k + k2 + 49 = 170 →
~a + AC = ~c
k2 – 18k – 40 = 0 →
(k + 2)(k – 20) = 0 → AC =→~c – ~a
→
k = −2, k = 20 (b) OB = OA + AB
= ~a + 1 (~c – ~a) A
4
41 ~c 1 B ~c – ~a
= ~a + – 4 ~a ~a (1) (3)
= 3 ~a + 1 ~c O ~c
4 4
C
132
8.3 Vectors in a Cartesian Plane / Vektor dalam Satah Cartes
Smart Tip
1 2It yj x , therefore the magnitude of the vector, |~r | = x2 + y2 vector, ~^r = ~r .
is given that ~r = x~i + ~ = y and the unit |~r |
1 2 Diberi
bahawa ~r = x~i + yj = x x2 + y2 dan vektor unit, ~^r = ~r .
~ y , maka magnitud vektor, |~r | = |~r |
Exercise 10 Express each of the following vectors in the form of
Ungkapkan setiap vektor yang berikut dalam bentuk
(a) x~i + y~j, (b) 1 xy 2.
TP 2 Mempamerkan kefahaman tentang vektor.
Example 10 1 y 2 y
y8 8
6
8 6 4 p
m~ 2 ~
6
4 ~a 4
2
2 x x
246 8 8
O O 24 6
O 246 8 x 1 2(a) 7~i + 3~j 1 2(a) 5~i + 7~j
Solution (b) 7 (b) 5
3 7
1 2(a) 5~i − 6~j (b) –56
3 y 4 y 5 y
8 4
6
4 q 2 ~v 6
2 ~ 2
–2 O 4 w~
–2 4 x
2
–2 O 24 x
O 24 6 x –4
8
1 2(a) –7~i – 6~j
1 2(a) 6~i – 7~j 1 2(a) 9~i + 7~j (b) –7
–6
(b) 6 (b) 9
–7 7
Exercise 11 Find the unit vector in the direction of the vector given.
Cari vektor unit dalam arah vektor yang diberi.
TP 2 Mempamerkan kefahaman tentang vektor.
Example 11 1 ~u = −~i + 2j 2 ~u = 4~i – 3j
~ ~
~u = 2~i − 5j |~u| = (–1)2 + (2)2 |~u| = (4)2 + (–3)2
~ = 1 + 4 = 16 + 9
Solution
(2)2
|~u| = 4 + + (–5)2 = 5 = 25
= 25
= 29 ^~u = 1 (–~i + 2j ) =5
~ 1
1 5 ^~u = 5 (4~i – 3j )
~
^~u = 29 (2~i − 5j )
~
133
3 ~u = −5~i – 4~j 4 ~u = −5~i + 12~j 5 ~u = 4~i – 6~j
(–5)2 (–5)2 (4)2 + (–6)2
|~u| = 25 + + (–4)2 |~u| = 25 + + (12)2 |~u| = 16 + 36
= 16 = 144 =
= 41 = 169 = 52
^~u = 141(–5~i – 4~j) = 13 ^~u = 1 (4~i – 6~j )
1
= –141(5~i + 4~j) ^~u = 13 (–5~i + 12~j ) 52
Exercise 12 Solve the following.
Selesaikan yang berikut.
TP 3 Mengaplikasikan kefahaman tentang vektor untuk melaksanakan tugasan mudah.
. Example 12 1 GFCD===iaiin225br~~viii(ed~revi++ietn~+ehk67t~~~e=eo3jj~r~j=vi+)2e+~~ei+3c~3i++t~(oj~3+f~3rdi.~~ajj2+n~ae~~fnj+)=d~f3~.~fi =+~3j~.i +~j. 2 FGiinvdenth~ae=v4e~ic−to~jra3n~ad+~b =~i + 2~j .
2~b .
FGiinvdenth~ae=v4e~ict–o2r~j~aa−nd3~b~b.= 2~i +~j. DCaibriervie~akt=or4~3i~a−+~j dan ~b =~i + 2~j .
2~b .
Diberi ~a = 4~i – 23~~jb .dan ~b = 2~i +~j . = 311(42~~4ii~i+−−~j3~~jj) + 22~i(~i+ +4~j2~j )
Cari vektor ~a − = +
=
Solution
= –4(~4i2~~ii–––2~2j5~~jj–) –6~i3−(2~i3~j+~j )
=
=
3 ~q=G==DCai=i(37brv~~7iiie~2eir~vi++ine+~p−k53~pt~~~3=ojjj~j.r=−7)F~~pi7−i4~n−+i~i2d+23+(~~qj2t3~h.id2~~jeaj−nav~nj~qe)dc=to2~ri ~p−~−j.2~q. 4 FGiinvdenth~pe=v~eic+to2r~j3a~pn−d2~q~q=. 2~i +~j. 5 Given ~u = 4~i +~j and 12~v~v=. 6~i −~j .
Find the vector 2~u –
CDaibriervie~pkt=or~i3+p 2j dan q = 2~i + j. =D==Cai528br~~iii(e4rv~++iie~uk+252t~=oj~~jjr−)42~i−~u3+~i–21~j+(21d6a~~12vin~.j−~v~j=) 6~i − j.
−~ 2q . ~ ~ ~
= 33−~~i(i~i+++64~~2jj~j−) ~−4~i 2−(2~~2i~j+~j )
=
=
Exercise 13 Solve the following problems.
Selesaikan masalah yang berikut.
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang vektor dalam konteks penyelesaian masalah rutin yang mudah.
Example 13
The diagram shows three points, P, Q and R drawn on a Cartesian plane. y
O
Rajah menunjukkan tiga titik, P, Q dan R dilukis pada satah Cartes. P(–6, 8)
Q(–2, 6)
It is given that OP p=o–in6~it + 8i~nj , tOheQy=-a–x2i~si a+n6d~j and OR = –P4,~iQ+a3n~jd. PSoainrte S
is the reflection of R the points
→→
collinear such that SQ = bQP. Find the the value of b.
bDaigbiertiitbikahRawpaadOa Ppa=ks–i-6y~i d+an8~jt,itOikQ-ti=tik–2P~i, + 6j dan aOdRala=h–s4e~iga+ri3s~jd. eTnitgiaknSkieaaldahaapnantulan R(–4, 3) S
Q ~dan S
→→ x
SQ = bQP. Cari nilai b. x
Solution →→ y
→ →→ Given/Diberi SQ = βQP, Q(–2, 6)
SQ = OQ – OS
= –(––→26~~2ii~i+++36~~6→jj~j–) – (4~i + 3~j ) –6~i ++~3j~j) == β2β(–(–4~2i~i++2~~jj )) P(–6, 8)
= 4~i – 3~j 3(–2~i
=
→ Compare/Bandingkan,
QP = OP – OQ 2β = 3 –6~i + 8~j –2~i + 6~j S(4, 3)
4~i + 3~j
= (–––64~~6ii~i+++28~~8jj~j+) –2~i(––2~i6~j+ 6~j ) β = 3
= 2 O
=
134
→→ →
1 The coordinates of F, G and H are (−2, 3), (8, −4) and (10, −6) respectively. It is given that OH = aOF + βOG,
such that O is the origin and a and β are constants. Find the values of a and β.→ → →
Koordinat bagi F, G dan H masing-masing ialah (−2, 3), (8, −4) dan (10, −6). Diberi bahawa OH = aOF + βOG, dengan
keadaan O ialah asalan dan a dan β ialah pemalar. Cari nilai a dan β.
→ →→ Subtitute β = 9 into 3 ,
OF =→–2~i + 3~→j, OG =→8~i – 4~j, OH = 10~i – 6~j 8
OH = aOF + βOG 41 9 2
a = 8 – 5
C 11 1 000o~~~iiim–––p666~~~ajjj r===e,–(a–2(2–aa~2i~i+++833βa~j~)~j)i +++8β(β(3~8ia~i–––444β~~βjj))~j
−2a + 8β = 10 ——— 1 =–1
3a – 4β = –6 ——— 2 2
From 1 : a = 4β – 5 ——— 3
Substitute 3 into 2 ,
3(4β – 5) – 4β = –6
12β – 15 – 4β = –6
8β = 9
β= 9
8
2 The vq eacrteocrson~vsatanndtsw~. are non-zero and non-parallel vectors. It is given that (p – 3)~v = (4q + 5 – p)w~, where p
and Find the values of p and q.
pVeemktaolrar~v. dan nw~ilaiai lpahdavnekqt.or bukan sifar dan tidak selari. Diberi bahawa (p – 3)~v = (4q + 5 – p)w~, dengan keadaan p dan q ialah
Cari
p – 3 = 0 , 4q + 5 – p = 0
p = 3 4q + 5 – (3) = 0
4q = −2
q = − 2
4
= − 1
2
3 The diagram shows an explosive exploded into four parts, E, F, G and H, F
where each part has a momentum and represented by a vector.
–2~i + 4j
Rajah menunjukkan satu bahan letupan telah meletup kepada empat bahagian, E, F, G dan H, ~
dengan keadaan setiap bahagian mempunyai momentum dan diwakili oleh suatu vektor.
4~i
–3~i –j ~w G
~
If the sum of all the momentum vectors is a zero vector, find
E
Jika hasil tambah vektor momentum ialah vektor sifar, cari
(a) wwt~~hdeinaulatnmeirtmsevbseucotatfon~ir~iaidnnadnth~~jje,, direction H
(b) vektor unit dalam arah w~.
of w~.
(a) (−2~i + 4~j) + (4~i ) + w~ + (−3~i −~w~j) ==~0i − 3~j
(b) w^~ = (–3)2
(1)2 1
+ (~i − 3~j)
= 1 (~i − 3~j)
10
135
Review 8
Paper 1 Questions
→
OC =
1 2 1 2 1 2CSL OP1MN E Gp aivnednqOa→Are=con6pst,aOn→Bts.=Ex27pr,eassndpq , where Given ~u = 1 4 2 and ~v = 1 h + 2 2, where h is a
–3 q if the 5 3
in terms of
`15 constant. Find the value of h.
P1Q15
points A, B and C lie on a straight line. 1 2 1 2Diberi ~u =4
5 dan ~v = h+2 , dengan keadaan h ialah
1 2 1 2 1 2→ 3
Diberi OA =
p → 2 → q , dengan pemalar. Cari nilai bagi h.
6 , OB = 7 , dan OC = –3
keadaan p dan q ialah pemalar. Ungkapkan p dalam sebutan [3 marks/markah]
q, jika titik−titik A, B dan C terletak pada satu garis lurus. →→ →
[3 marks/markah] 4 Diagram 3 shows the vectors AB, AD and AC
2 Diagram 1 shows a regular pentagon ABCDE SPM drawn on a square grid. → → →
CLONE Rajah 3 menunjukkan vektor AB, AD dan AC yang dilukis
`18
P1Q8 pada grid segi empat sama.
SPM with centre O.
CLONE
`16 Rajah 1 menunjukkan sebuah pentagon sekata ABCDE
P1Q10 berpusat di O.
~x A
D B y
~
C
E C
O
D
AB
Diagram 1/ Rajah 1
→→ →
(a) Express EB + BC − DC as a single vector.
→→ →
Ungkapkan EB + BC − DC sebagai satu vektor tunggal.
→→ Diagram 3/ Rajah 3
(b) Given OD =pe~dn, tOagEo=n~eisa7nudntihtes.lFeningdththoef each → + p
side of the unit (a) Express AD in the form of p~x q~y, where
and q are constants.
→ terms of →
vector in the direction of ED, in ~d
and ~e . → → Ungkapkan AD dalam bentuk p~x + qy, dengan keadaan
p dan q ialah pemalar. ~
Diberi OD = i~da,laOhE7 =un~eitd.aCnaprai nvjeakntogrbuangiitsdetailaapmsiasriah (b) In Diagram 3, mark and label the point W
pentagon itu −→ → →
→
ED, dalam sebutan ~d dan ~e . such that DW + AB = 2AC.
[3 marks/markah] Pada Rajah 3, tanda dan labelkan titik W dengan
−→ → →
keadaan DW + AB = 2AC.
3 Diagram 2 shows a trapezium EFGH. [3 marks/markah]
SPM Rajah 2 menunjukkan sebuah trapezium EFGH. 5 F(3, 5) and G(−3, 6) lie on a Cartesian plane. It is
CLONE
`17
P1Q4 G SPM → →→ Find
H CLONE given that 3OF = 2OG + OH.
~v `18 F(3, 5) dan G(−3, 6) terletak pada satah Cartes. Diberi
P1Q9 → →→
E ~u bahawa 3OF = 2OG + OH. Cari
(a) the coordinates of H,
F koordinat H,
Diagram 2/ Rajah 2 →
(b) |FH|.
[4 marks/markah]
136
Paper 2 Questions
1 Diagram 1 shows a triangle PQR. H
SPM Rajah 1 menunjukkan sebuah segi tiga PQR.
CLONE
`15
P2Q9
R
GQ
F
P
Diagram 1/ Rajah 1
: = 2: 1, →→
It is given that PF : FQ = 1 : 2, QH HR → PF = 8→~x and PR = 12~y.
Diberi bahawa PF : FQ = 1 : 2, QH : HR = 2 : 1, PF = 8~x dan PR = 12y.
~
(a) EUxnpgkr→aepsksanindtaelarmmssebouft~xana~xndda~yn
(i) RF, y
~
→
(ii) RH.
→ [3 marks/markah]
(b) Given ~x = 3~i and ~y = –~i + 2~j, find |RH |.
→
Diberi ~x = 3~i dan ~y = –~i + 2~j, cari |RH|.
→→ →→ [2 marks/markah]
(c) Given RG = aRF and GH = bPH, where a and b are constants, find the values of a and b.
→→ →→
Diberi RG = aRF dan GH = bPH, dengan keadaan a dan b ialah pemalar, cari nilai a dan b.
[5 marks/markah]
2 Diagram 2 shows triangles OCG and OFD. The straight lines CG and FD intersect at point H.
SPM Rajah 2 menunjukkan segi tiga OCG dan OFD. Garis lurus CG dan FD bersilang pada titik H.
CLONE
`18
P2Q8 C
F
H
O GD
Diagram 2/ Rajah 2
It is given that →→ 24~y, 1 3 : →→ →→ where
are constants. OC = 27~x, OD = OF : FC = : 2, OG : GD = 1, FH = aFD and GH = bGC, a and b
Diberi bahawa →→ OF : FC = 1 : OG →→ →→ dengan keadaan a dan b ialah
pemalar. OC = 27~x, OD = 24y, 2, : GD = 3 : 1, FH = aFD dan GH = bGC,
~
→
(a) Express OH in terms of
→
Ungkapkan OH dalam sebutan
(i) a, ~x~x aanndd//ddaann~~yy.,
(ii) b,
[4 marks/markah]
(b) Hence, find the the values of a and β.
Seterusnya, cari nilai a dan β.
(c) Given |~x | = 2 units, = unit and OC is perpendicular OD, calculate → [4 marks/markah]
|~y| 1 to → |FH |. [2 marks/markah]
Diberi |~x | = 2 unit, |y| = 1 unit dan OC adalah berserenjang dengan OD, hitung |FH |.
~
137
3 Diagram 3 shows a triangle EFG. The straight line EL intersects with the straight line FG at the point K. Point
SPM H lies on the straight line EL.
CLONE
`17 Rajah 3 menunjukkan sebuah segi tiga EFG. Garis lurus EL bersilang dengan garis lurus FG pada titik K. Titik H terletak pada
P2Q8 garis lurus EL.
FL
K
H G
E
Diagram 3/ Rajah 3
Given → = 1 → → 8~x and → 12~y.
FK = FG, EG = EF =
4
→ → → →
Diberi FK = = 1 FG, EG = 8~x dan EF = 12~y.
4
(a) UExnpgkra→epsksanindtaelarmmssebouft~xana~xndda/no/ rat~yau
(i) FG, y →
~ (ii) EK.
→→ → [3 marks/markah]
(b) It is given that EH = aEK and FH = b(~x – 8~y), where a and b are constants. Find the values of a dan b.
→ → →
Diberi bahawa EH = aEK dan FH = b(~x – 8~y), dengan keadaan a dan b ialah pemalar. Cari nilai a dan b.
→
[5 marks/markah]
(c) Given EL = w~x + 15~y, where w is a constant, find the the value of w.
→
Diberi EL = w~x + 15y, dengan keadaan w ialah pemalar, cari nilai w.
~
[2 marks/markah]
H OTS Zone
1 Diagram 1 shows the position and the direction of boats P, Q and R in a boat competition.
SPM Rajah 1 menunjukkan kedudukan dan arah bagi bot P, Q dan R dalam satu pertandingan bot.
CLONE
`16
P2Q5
R
Q
P
SGtaarritsianngpelirnmeulaan
Diagram 1/ Rajah 1
Both boat P and boat Q move in the direction of the water current. The velocity of the water current is given by
w~ = 2~i + 23~j m s−1. Given the velocity of boat P is ~p = 10~i + 72~j m s−1 and the v=e2l~io+cit32y~jomf bso−1a. tDQibeirsi~qha=la4j~ui +~j m s−1.
Halaju arus air diberi oleh w~ bot P ialah
Kedua-dua bot P dan bot Q bergerak mengikut arah arus air.
~p = 10~i + 7 j m s−1 dan halaju bot Q ialah ~q = 4~i +~j m s−1.
2
(a) How many times is the resultant velocity of boat P compared to the resultant velocity of boat Q?
Berapa kali gandakah halaju paduan bot P berbanding halaju paduan bot Q? HOTS Evaluating
(b) On the way to the finishing point, boat R faces a technical problem and goes off track. The velocity of
boat R is given by ~r = 4~i – 21~j m s−1.
Dalam perjalanan ke titik penamat, bot R mengalami satu masalah teknikal dan terpesong dari laluan. Halaju bot R diberi oleh
~r = 4~i – 1j m s−1.
2~
Find/Cari
(i) the resultant velocity of boat R, (ii) the unit vector in the direction of boat R. HOTS Applying
halaju paduan bot R, vektor unit dalam arah bot R.
138
Chapter Elective Package: Application of Science and Technology
9 Solution of Triangles
Penyelesaian Segi Tiga
9.1 Sine Rule / Petua Sinus
Smart Tip
Derivation of sine rule / Terbitan petua sinus A To find sides:
c Untuk mencari sisi:
sin A = t and/dan sin B = t
b a B
tb
b sin A = t a sin B = t a = b = c C
sin sin sin
t = b sin A = a sin B A B
∴ b = a a To find angles:
sin sin
B A Untuk mencari sudut:
Apply the same method by drawing a line between vertex A and side a to get: C sin A sin B sin C
a b c
Gunakan kaedah yang sama dengan melukiskan satu garis di antara bucu A dengan sisi a untuk mendapatkan: = =
b B = c C Therefore,/Maka, a = b = c
sin sin sin sin sin
A B C
Exercise 1 Prove the sine rule.
Buktikan petua sinus.
TP 2 Mempamerkan kefahaman tentang petua sinus dan petua kosinus.
Example 1
The diagram shows a triangle ABC inside a circle with centre O. Given the radius = R. A
Construct straight line COE and BE where COE = 2R. Hence, show that sin A = 21R. c
a O
b
Rajah menunjukkan sebuah segi tiga ABC yang terterap di dalam sebuah bulatan berpusat di O. Diberi a C
jejari = R. sin A 1 B
a 2R
Bina garis lurus COE dan BE dengan keadaan COE = 2R. Seterusnya, tunjukkan bahawa = .
Solution
A So/Maka, ∠CBE = 90° (semicircle/ BC
semi bulatan) COE
∠A = ∠E (angles sin A =
subtended by a common
E c b sin A = a
O 2R
B aC arc/sudut yang dicangkum oleh sin A = 1
lengkok sepunya) a 2R
1 The diagram shows a triangle ABC inside a circle with centre O. Given the radius = R. A
Construct straight line COG and AG where COG = 2R. Hence, show that sin B = 21R. c
b O
Rajah menunjukkan sebuah segi tiga ABC yang terterap di dalam sebuah bulatan berpusat di O. b
a C
Diberi jejari = R. sin B 1
b 2R
Bina garis lurus COG dan AG dengan keadaan COG = 2R. Seterusnya, tunjukkan bahawa = . B
A So, ∠CAG = 90° (semicircle)
∠B = ∠G (angles subtended by a common arc)
Gc b sin B = AC
GOC
O
B aC sin B = b
2R
sin B = 1
b 2R
139
2 The diagram shows a triangle ABC inside a circle with centre O. Given the radius = R. A
Construct straight line AOF and BF where AOF = 2R. Hence, show that sin C = 21R.
c
Rajah menunjukkan sebuah segi tiga ABC yang terterap di dalam sebuah bulatan berpusat di O. c b
O C
Diberi jejari = R. sin C 1
c 2R a
Bina garis lurus AOF dan BF dengan keadaan AOF = 2R. Seterusnya, tunjukkan bahawa = . B
A So, ∠ABF = 90° (semicircle)
cb ∠C = ∠F (angles subtended by a common arc)
O
sin C = AB
B aC AOF
F sin C = c
2R
sin C = 1
c 2R
Exercise 2 Find the value of x for each of the following.
Cari nilai x bagi setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang petua sinus, petua kosinus dan luas segi tiga untuk melaksanakan tugasan mudah.
Example 2 Smart Tip 1
Q Apply the sine rule P
when given 50º 16.5 cm
8.4 cm x cm Gunakan petua sinus Q
apabila diberi 126º
(a) 1 side and R x cm
73º 44º 2 angles sinx50° = 16.5
P R 1 sisi dan 2 sudut sin 126°
(b) 2 sides and
Solution x = 16.5 sin 50°
1 non-included sin 126°
sinx73° = 8.4 angle
sin 44° 2 sisi dan 1 sudut x = 15.624 cm
bukan kandung
x = 8.4 sin 73°
sin 44°
x = 11.564 cm
2 3
P x cm Q Q
80º 54º
15.4 cm x cm
14.6 cm 47º 56º 42’ R
P
R
∠PRQ = 180° − 80° − 54° = 46° sinx47° = 15.4 42′
sin 56°
sinx46° = 14.6
sin 80° x = 15.4 sin 47°
sin 56°42′
x = 14.6 sin 46°
sin 80° x = 13.475 cm
x = 10.664 cm
140
4 5
x cm Q Q
P 43º 24’ 57º 36’ 36º 35’
8 cm
16 cm x cm R
P 26º 14’
R
∠PRQ = 180° − 43° 24′ − 57° 36′ = 79° ∠PRQ = 180° − 26° 14′ − 36° 35′ = 117° 11′
sinx79° = sin 16 36′ sin x = 8
57° 117° 26°
11 sin 14′
16 sin 79°
x = sin 57° 36′ x = 8 sin 117° 11′
sin 26° 14′
x = 18.602 cm
x = 16.099 cm
Exercise 3 Find the value of θ for each of the following.
Cari nilai θ bagi setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang petua sinus, petua kosinus dan luas segi tiga untuk melaksanakan tugasan mudah.
Example 3 1 B
A
7 cm 17 cm 9 cm 17.5 cm
30º
u C
115º A
B
u
sin θ sin 30°
C 17.5 = 9
Solution
sin q = sin 115° sin θ = 17.5 sin 30°
7 17 9
sin θ = 7 sin 115° sin θ = 0.9722
17
θ = 76° 28’
sin θ = 0.3732
θ = 21° 55’
2 3
24 cm B B
u
Au
26 cm
13 cm
117º
C 130º
A 18 cm C
sin θ = sin 117° sin θ = sin 130°
13 24 18 26
sin θ = 13 sin 117° 18 sin 130°
24 26
sin θ =
= 0.4826
= 0.5303
θ = 28° 51’
θ = 32° 2’
141
4 A 14 cm 5
72º u
B B
15 cm 9 cm u
4 cm
C A 58º
C
sin ∠ACB = sin 72° sin ∠BAC = sin 58°
14 15 4 9
sin ∠ACB = 14 sin 72° sin ∠BAC = 4 sin 58°
15 9
sin ∠ACB = 0.8877 sin ∠BAC = 0.3769
∠ACB = 62° 35’ ∠BAC = 22° 9’
θ = 180° − 72° − 62° 35’ θ = 180° − 58° − 22° 9’
= 45° 25’ = 99° 51’
Smart Tip
The Ambiguous Case of the Sine Rule M p
Kes Berambiguiti bagi Petua Sinus n
An ambiguous case of the sine rule will occur P’ P N
when two sides (n and p) and a non-included
acute angle (∠N) are given, where the opposite A non-included angle is the angle which is not between the two
side of the given angle is shorter than the other sides given.
side (n < p). There are two different triangles that Sudut bukan kandung ialah sudut yang tidak terletak di antara dua sisi yang
can be drawn (MNP and MNP’), therefore two diberi.
sets of answers would be obtained.
Kes berambiguiti bagi petua sinus akan berlaku apabila dua ∠N and ∠P are non-included angles. M p
sisi (n dan p) dan satu sudut tirus bukan kandung (∠N) ∠N dan ∠P ialah sudut bukan kandung.
diberi, dengan keadaan sisi yang bertentangan dengan sudut
yang diberi itu lebih pendek daripada sisi yang satu lagi ∠M is an included angle. n
(n < p). Terdapat dua buah segi tiga yang berbeza yang ∠M ialah sudut kandung.
dapat dilukis (MNP dan MNP’), maka dua set jawapan akan
diperoleh.
Exercise 4 Solve the following problems.
Selesaikan masalah yang berikut.
TP 3 Mengaplikasikan kefahaman tentang petua sinus, petua kosinus dan luas segi tiga untuk melaksanakan tugasan mudah.
Example 4 1 The diagram shows a triangle PQR. Sketch
triangle QRP’ such that the straight line RP is
The diagram shows a triangle PQR. Sketch extended to P’, QP’ = QP and ∠QRP’ = ∠QRP.
triangle PQR’ such that ∠QPR’ = ∠QPR and
QR’= QR. Rajah menunjukkan sebuah segi tiga PQR. Lakarkan segi
Rajah menunjukkan sebuah segi tiga PQR. Lakarkan segi tiga QRP’ dengan keadaan garis lurus RP dipanjangkan ke
tiga PQR’ dengan keadaan ∠QPR’ = ∠QPR dan QR’= QR. P’, QP’ = QP dan ∠QRP’ = ∠QRP.
Q P’
P R Q
Q P
Solution
P R’ R R
142
2 The diagram shows a triangle LKM. Sketch 3 The diagram shows a triangle ABC. Sketch
triangle LMK’ such that ∠LMK’ = ∠LMK and triangle AB’C such that ∠ACB’ = ∠ACB and
LK’ = LK. AB’ = AB.
Rajah menunjukkan sebuah segi tiga LKM. Lakarkan segi Rajah menunjukkan sebuah segi tiga ABC. Lakarkan segi
tiga LMK’ dengan keadaan ∠LMK’ = ∠LMK dan LK’ = LK. tiga AB’C dengan keadaan ∠ACB’ = ∠ACB dan AB’ = AB.
K K’ M A
L B B’ C
Exercise 5 Sketch triangle PQR and find the possible values of ∠R’ in an ambiguous case.
Lakarkan segi tiga PQR dan cari nilai-nilai yang mungkin bagi ∠R’ dalam kes berambiguiti.
TP 3 Mengaplikasikan kefahaman tentang petua sinus, petua kosinus dan luas segi tiga untuk melaksanakan tugasan mudah.
Example 5 1 ∠P = 32°, p = 9 cm, r = 14 cm
∠P = 27°, p = 7 cm, r = 11 cm Q
9 cm
Solution Q 14 cm
11 cm 7 cm 32º
R R’
P R
P 27º R’
sin11∠R = sin 27° sin14∠R = sin 32°
7 9
11 sin 27° sin ∠R = 14 sin 32°
7 9
sin ∠R =
= 0.8243
= 0.7134
∠R = 55° 31’
∠R = 45° 31’
∠R’ = 180° – 55° 31’
∠R’ = 180° – 45° 31’
= 124° 29’
= 134° 29’
2 ∠P = 24°, p = 4 cm, r = 9 cm 3 ∠Q = 43°, q = 7 cm, r = 9.5 cm
Q P
4 cm
9 cm
R
9.5 cm 7 cm
24º R’ Q 43º
P
R’ R
sin9∠R = sin 24° sin9.∠5 R = sin 43°
4 7
sin ∠R = 9 sin 24°
4
sin ∠R = 9.5 sin 43°
7
= 0.9152
∠R = 66° 14’ = 0.9256
∠R’ = 180° – 66° 14’ ∠R = 67° 45’
= 113° 46’ ∠R’ = 180° – 67° 45’
= 112° 15’
143
Exercise 6 Solve each of the following.
Selesaikan setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang petua sinus, petua kosinus dan luas segi tiga untuk melaksanakan tugasan mudah.
Example 6
23 cm F
84º
28º 130º G
E 9 cm H
Given EHG is a straight line, find the length of
Diberi EHG ialah garis lurus, cari panjang
(a) FH, (b) HG, (c) FG.
Solution
(a) FH = 23 (b) ∠FGH = 130° − 84° = 46° (c) ∠FHG = 180° − 130° = 50°
sin 28° sin 130°
HG = 14.096 FG = 14.096
23 sin 28° sin 84° sin 46° sin 50° sin 46°
FH = sin 130°
14.096 sin 84° 14.096 sin 50°
HG = sin 46° FG = sin 46°
= 14.096 cm
= 19.488 cm = 15.011 cm
1 F 5 cm 2 F
32º
E G
120º
14 cm
19 cm 8 cm
78º
H 68º 115º G
E H
Given EFG is a straight line, find the length of Given EHG is a straight line, find
Diberi EHG ialah garis lurus, cari
Diberi EFG ialah garis lurus, cari panjang
(a) ∠FGH, (b) EF, (c) ∠EFG.
(a) FH, (b) EF, (c) EH.
(a) FH = 19 (a) sin ∠8FGH = sin 115°
sin 32° sin 120° 14
FH = 19 sin 32° sin ∠FGH = 8 sin 115°
sin 120° 14
= 11.626 cm = 0.5179
∠FGH = 31.19° (or 31° 11’)
(b) ∠FEH = 120° − 78° = 42° EF 8
sin(180° – 115°) sin 68°
EF = 11.626 (b) =
sin 78° sin 42°
8 sin 65°
EF = 11.626 sin 78° EF = sin 68°
sin 42°
= 7.820 cm
= 16.995 cm
(c) ∠EFH = 180° − 120° = 60° (c) ∠EFG = 180° − 68° − 31.19°
= 80.81° (or 80° 49‘)
EH = 11.626
sin 60° sin 42°
EH = 11.6s2in6 sin 60°
42°
= 15.047 cm
144
3
F
19 cm
10 cm 7 cm
E 34º 112º
HG
Given EHG is a straight line, find
Diberi EHG ialah garis lurus, cari
(a) ∠FHG, (b) HG, (c) EH.
(a) sin ∠FHG = sin 112° (b) ∠HFG = 180° − 112° − 40.47° = 27.53° (c) ∠EFH = ∠FHG − ∠FEH
7 10
HG = 10 = 40.47° − 34°
sin 27.53° sin 112°
sin ∠FHG = 7 sin 112° = 6.47°
10
HG = 10 sin 27.53° EH = 10
= 0.6490 sin 112° sin 6.47° sin 34°
∠FHG = 40.47° = 4.985 cm EH = 10 sin 6.47°
sin 34°
= 2.015 cm
9.2 Cosine Rule / Petua Kosinus
Exercise 7 Prove the cosine rule.
Buktikan petua kosinus.
TP 2 Mempamerkan kefahaman tentang petua sinus dan petua kosinus.
Example 7
The diagram shows a triangle ABC with sides a, b and c. B
Rajah menunjukkan sebuah segi tiga ABC dengan sisi-sisi a, b dan c. b
Construct straight line BP which perpendicular to AC and P is on the straight c a
line AC. Hence, show that a2 = b2 + c2 – 2bc cos A. A
Bina garis lurus BP yang berserenjang dengan AC dan P berada pada garis lurus AC.
Seterusnya, tunjukkan bahawa a2 = b2 + c2 – 2bc kos A.
Solution C
In ∆ABP/Dalam ∆ABP,
x = cos A/kos A B
c
x = c cos A/c kos A
Pythagoras’ Theorem/Teorem Pythagoras, ch a
h2 = c2 – x2 ——— 1
In ∆BPC/Dalam ∆BPC, A P C
h2 = a2 – (b – x)2 x
= a2 – (b2 – 2bx + x2) (b – x)
= a2 – b2 + 2bx – x2 ——— 2
Substitute 1 into 2 /Gantikan 1 ke dalam 2 ,
a 2 – b2 + 2bx – x2 = c2 – x2
a2 = b2 + c2 – 2bx
a2 = b2 + c2 – 2b(c cos A/c kos A)
a2 = b2 + c2 – 2bc cos A/2bc kos A
145
1 The diagram shows a triangle ABC with sides a, b 2 The diagram shows a triangle ABC with sides a, b
and c. and c.
Rajah menunjukkan sebuah segi tiga ABC dengan sisi-sisi Rajah menunjukkan sebuah segi tiga ABC dengan sisi-sisi
a, b dan c. a, b dan c.
B B
ca ca
Ab C Ab C
PAK-21 Construct straight line AQ which is perpendicular Construct straight line BR which is perpendicular
to BC and Q is on the straight line BC. Hence, to AC and R is on the straight line AC. Hence,
QR CODE show that c2 = a2 + b2 – 2ab cos C.
show that b2 = a2 + c2 – 2ac cos B. Bina garis lurus BR yang berserenjang dengan AC dan R
Bina garis lurus AQ yang berserenjang dengan BC dan Q berada pada garis lurus AC. Seterusnya, tunjukkan bahawa
c2 = a2 + b2 – 2ab kos C.
berada pada garis lurus BC. Seterusnya, tunjukkan bahawa
b2 = a2 + c2 – 2ac kos B.
B
B ch a
xa A b–x R x C
Q
c
(a – x)
h
Ab C b
In ∆ABQ, In ∆BRC,
x = cos B x = cos C
c a
x = a cos C
x = c cos B
Pythagoras’ Theorem,
Pythagoras’ Theorem, h2 = a2 – x2 ——— 1
h2 = c2 – x2 ——— 1
In ∆ARB,
In ∆ACQ,
h2 = c2 – (b – x)2
h2 = b2 – (a – x)2
= c2 – (b2 – 2bx + x2)
= b2 – (a2 – 2ax + x2) = c2 – b2 + 2bx – x2 ——— 2
= b2 – a2 + 2ax – x2 ——— 2
Substitute 1 into 2 ,
Substitute 1 into 2 ,
c 2 – b2 + 2bx – x2 = a2 – x2
b2 – a2 + 2ax – x2 = c2 – x2
c2 – b2 + 2bx = a2
b2 = a2 + c2 – 2ax
c2 = a2 + b2 – 2bx
b2 = a2 + c2 – 2a(c cos B)
c2 = a2 + b2 – 2b(a cos C)
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
Smart Tip To find unknown sides: QR CODE
Untuk mencari sisi yang tidak diketahui:
Apply the cosine rule when given a2 = b2 + c2 − 2bc cos A/ kos A Scan or visit
Gunakan petua kosinus apabila diberi b2 = a2 + c2 − 2ac cos B/kos B http://www.cimt.
c2 = a2 + b2 − 2ab cos C/kos C org.uk/projects/
(a) 2 sides and 1 included angle mepres/step-up/
2 sisi dan 1 sudut kandung To find unknown angles: sect4/index.htm
(b) 3 sides for additional
3 sisi Untuk mencari sudut yang tidak diketahui:
notes and
A cos A/kos A = b²+ c² – a² exercises of the
2bc
cb cosine rule.
cos B/kos B = a²+ c² – b²
B aC 2ac
cos C/kos C = a²+ b² – c²
2ab
146
Exercise 8 Find the value of x for each of the following.
Cari nilai x bagi setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang petua sinus, petua kosinus dan luas segi tiga untuk melaksanakan tugasan mudah.
Example 8 1 2
Q Q
P 13 cm Q 50º
48º 16.5 cm
7 cm x cm 13 cm
R
8 cm x cm P 52º P
x cm
14 cm R
R a2 = b2 + c2 − 2bc cos A a2 = b2 + c2 − 2bc cos A
x2 = 72 + 142 − 2(7)(14) cos 52° x2 = 132 + 16.52 − 2(13)(16.5)
Solution cos 50°
= 124.330
a2 = b2 + c2 − 2bc cos A/ kos A x = 11.150 = 165.494
x2 = 132 + 82 − 2(13)(8) cos x = 12.864
48°/ kos 48°
= 93.821
x = 9.686
3 4 5 Q 14 cm
126º 43’
Q Q 6.7 cm
x cm
R
x cm 9.4 cm 9.6 cm x cm
P 54º 26’ 118º
15 cm R P R
13 cm P
a2 = b2 + c2 − 2bc cos A a2 = b2 + c2 − 2bc cos A a2 = b2 + c2 − 2bc cos A
x2 = 152 + 9.42 − 2(15)(9.4) x2 = 9.62 + 132 − 2(9.6)(13) x2 = 6.72 + 142 − 2(6.7)(14)
cos 54°26’ cos 118° cos 126°43’
= 378.34
= 149.335 x = 19.451 = 353.05
x = 12.220 x = 18.79
Exercise 9 Find the value of θ for each of the following.
Cari nilai θ bagi setiap yang berikut.
TP 3 Mengaplikasikan kefahaman tentang petua sinus, petua kosinus dan luas segi tiga untuk melaksanakan tugasan mudah.
Example 9 1 2 Q
Q
Q 6 cm 10 cm u
8 cm u 12 cm
9 cm 9 cm
P
13 cm R P
Pu 6.5 cm
14.5 cm R
cos θ = 62 + 132 − 102
2(6)(13)
Solution R
cos θ/kos θ = b2 + c2 − a2 = 0.6731
2bc θ = 47° 42’
92 + 122 − 6.52
92 + 14.52 − 82 cos θ = 2(12)(9)
= 2(9)(14.5)
= 0.8707 = 0.8461
θ = 29° 28’ or/atau θ = 32° 13’
29.46°
147
3 Q 4 5 19 cm
26 cm
P u 7 cm Q P
12 cm Q
u
7.5 cm 16.4 cm
θ 12 cm 13 cm
16 cm R P
122 + 72 − 162 R R
2(12)(7)
cos θ=
= –0.375 cos θ = 7.52 + 122 − 16.42 cos θ = 192 + 132 − 262
θ = 112° 1’ 2(7.5)(12) 2(19)(13)
= −0.3817 = −0.2955
θ = 112° 26’ θ = 107° 11’
Exercise 10 Solve the following problems.
Selesaikan masalah yang berikut.
TP 4 Mengaplikasikan pengetahuan dan kemahiran yang sesuai tentang penyelesaian segi tiga dalam konteks penyelesaian masalah rutin yang mudah.
Example 10 1 There are three ships, P, Q and R in the sea. The
sailor of ship P can see both the ships Q and R.
There are three tents, F, G and H at a campsite. The angle between the line of sight to Q and
Given the distance between tent F and tent G is the line of sight to R from P is 50°. Given the
52 m, the distance between tent F and tent H is distance between ship P and ship Q is 3 km and
86 m and the distance between tent G and tent H the distance between ship P and ship R is 6 km.
is 64 m. Calculate the angle between FH and GH. Calculate the distance, in km, between ship Q
Terdapat tiga buah khemah, F, G dan H di sebuah tapak and ship R.
perkhemahan. Diberi jarak di antara khemah F dan khemah
G ialah 52 m, jarak di antara khemah F dan khemah H Terdapat tiga buah kapal, P, Q dan R di dalam laut. Kelasi
ialah 86 m dan jarak di antara khemah G dan khemah H kapal P boleh melihat kedua-dua buah kapal Q dan R. Sudut
ialah 64 m. Hitung sudut di antara FH dan GH. di antara garis penglihatan Q dan garis penglihatan R dari
P ialah 50°. Diberi jarak di antara kapal P dan kapal Q ialah
Solution 3 km dan jarak di antara kapal P dan kapal R ialah 6 km.
Let θ be the angle between FH and GH. Hitung jarak, dalam km, di antara kapal Q dan R.
Katakan θ ialah sudut di antara FH dan GH.
Q
F 52 m
86 m
G 3 km
50°
64 m R
T P
6 km
H QR2 = PQ2 + PR2 – 2(PQ)(PR) cos P
= 32 + 62 – 2(3)(6) cos 50°
c os θ/kos θ = FH2 + GH2 – FG2 = 21.86
2(FH)(GH)
QR = 21.86
= 862 + 642 – 522
2(86)(64) = 4.675 km
∴ The distance between ship Q and ship R is
8 788
11 008 4.675 km.
=
= 0.7983
θ = 37°2’
148
2 The diagram shows a triangle PQR. 3 The diagram shows a triangle ABC.
Rajah menunjukkan sebuah segi tiga PQR. Rajah menunjukkan sebuah segi tiga ABC.
Q B
x cm 6 cm 6 cm 100° x cm
30° 9 cm R A 12 cm C
P
Find the value of x./Cari nilai x.
Find the value of x./Cari nilai x.
x2 + 92 – 2(x)(9) cos 30° = 62 62 + x2 – 2(6)(x) cos 100° = 122
x2 + 81 – 15.59x = 36 36 + x2 + 2.084x = 144
x2 – 15.59x + 45 = 0 x2 + 2.084x – 108 = 0
x = –(–15.59) ± (–15.59)2 – 4(1)(45) –(2.084) ± (2.084)2 – 4(1)(–108)
2(1) x = 2(1)
= 11.765, 3.825 = 9.403, –11.487
PQ = 11.765 cm BC = 9.403 cm
9.3 Area of a Triangle / Luas Segi Tiga
Exercise 11 Solve the following problems.
Selesaikan masalah yang berikut.
TP 3 Mengaplikasikan kefahaman tentang petua sinus, petua kosinus dan luas segi tiga untuk melaksanakan tugasan mudah.
Example 11 1 The diagram shows a triangle PQR.
Rajah menunjukkan sebuah segi tiga PQR.
The diagram shows a triangle ABC.
Rajah menunjukkan sebuah segi tiga ABC. Q
A 10 cm C 8 cm
54°
4 cm 36° 12 cm R
P
B Find/Cari
(a) the height, in cm, of ∆PQR,
Find/Cari tinggi, dalam cm, bagi ∆PQR,
(a) the height, in cm, of ∆ABC,
tinggi, dalam cm, bagi ∆ABC, (b) the area, in cm2, of ∆PQR.
(b) the area, in cm2, of ∆ABC. luas, dalam cm2, bagi ∆PQR.
luas, dalam cm2, bagi ∆ABC.
Solution Let the height of ∆PQR = h cm
Let the height of ∆ABC = h cm
Katakan tinggi bagi ∆ABC = h cm Q
A 10 cm C 8 cm
54°
h cm
h cm 36°
4 cm P 12 cm R
h B (a) PhQ = sin 36°
AB
(a) = sin 54° h
8
h = sin 36°
4
= sin 54° h = 8 sin 36°
h = 3.236 cm = 4.702 cm
1
(b) Area ∆ABC = 2 × base × height (b) Area ∆PQR = 1 × base × height
2
1
Luas ∆ABC = 2 × tapak × tinggi = 1 × 12 cm × 4.702 cm
2
1
= 2 × 10 cm × 3.236 cm = 28.21 cm2
= 16.18 cm2
149
2 The diagram shows a triangle ABC. Given ACD 3 The diagram shows a triangle ABC.
is a straight line. Rajah menunjukkan sebuah segi tiga ABC.
Rajah menunjukkan sebuah segi tiga ABC. Diberi ACD A
ialah garis lurus.
B
b cm
14 cm T
C
h cm a cm B
D
A 50° C Show that the area of ∆ABC = 1 ab sin θ cm2.
9 cm 2
Find/Cari Tunjukkan bahawa luas ∆ABC = 1 ab sin θ cm2.
(a) the value of h, 2
nilai h,
Let the height of ∆ABC = h cm
(b) the area, in cm2, of ∆ABC.
luas, dalam cm2, bagi ∆ABC. A
(a) h = sin 50° b cm
AB h cm
h = sin 50° T B
14 C
a cm
h = 14 sin 50° h = sin θ
= 10.725 cm b
1 h = b sin θ
2
(b) Area ∆ABC = × base × height Area ∆ABC = 1 × base × height
2
1
= 2 × 9 cm × 10.725 cm = 1 × a cm × b sin θ cm
2
= 48.26 cm2
1
= 2 ab sin θ cm2
Smart Tip
Area of ∆/Luas bagi ∆: A
c
1 ab sin C or/atau Remember/ Ingat:
2 The area of ∆ is applied when 1 included
b angle and 2 sides are given.
1 Luas ∆ digunakan apabila 1 sudut kandung dan
2 bc sin A or/atau 2 sisi diberi.
1 ac sin B Ba C
2
Exercise 12 Find the area of each of the following triangles by using the formula, area = 1 ab sin θ.
Cari luas bagi setiap segi 2
1 θ.
tiga yang berikut menggunakan rumus, luas = 2 ab sin
TP 3 Mengaplikasikan kefahaman tentang petua sinus, petua kosinus dan luas segi tiga untuk melaksanakan tugasan mudah.
Example 12 1
7 cm
4 cm 34°
68° 9 cm
7 cm Area = 1 ab sin θ
2
Solution 1
2
Area/Luas = ab sin θ = 1 × 7 cm × 9 cm × sin 34°
2
1
= 2 × 4 cm × 7 cm × sin 68° = 17.615 cm2
= 12.981 cm2
150
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