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92
Chapter 5
Exercise 5B
1 Work out the gradients of these lines:
a y = −2
x + 5 b y = −x
+ 7 c y = 4
+ 3x
d y = 1 _ 3 x − 2 e y = − 2 _ 3 x f y = 5 _ 4 x + 2 _ 3
g 2x −
4y + 5 = 0 h 10x −
5y + 1 = 0 i −x +
2y − 4 = 0
j −3x
+ 6y + 7 = 0 k 4x +
2y − 9 = 0 l 9x +
6y + 2 = 0
2 These lin... | [
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0.002828... |
93Straight line graphs
5 The line 3x +
2y = 0 meets the x-axis at the point R. Work out the coordinates of R.
6 The line 5x −
4y + 20 = 0 meets the y-axis at the point A and the x-axis at the point B.
Work out the coordinates of A and B.
7 A line l passes thr
ough the points with coordinates (0, 5) and (6, 7).
a Fi... | [
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0.0... |
94
Chapter 5
Example 7
Find the equation of the line that passes through the points (5, 7) and (3, −1).
m = y2 − y1 _______ x2 − x1 = 7 − (−1) ________ 5 − 3 = 8 __ 2 = 4
S
o y −
y1 = m(x − x1)
y + 1 = 4( x – 3)
y + 1 = 4 x − 12
y = 4 x – 13First find the slope of the line.
Here (x1, y1) = (3, −1) an... | [
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0.0069... |
95Straight line graphs
5 The straight line l
passes through (a, 4) and
(3a, 3). An equation of l is x + 6y + c = 0.
Find the value of a and the value of c. (3 marks)
6 The straight line l
passes through (7a, 5) and (3a, 3).
An equation of l is x + by − 12 = 0.
Find the value of a and the value of b. (3 mark... | [
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96
Chapter 5
1 The line y =
4x − 8 meets the x-axis at the point A. Find the equation of the line with gradient
3 that passes through the point A.
2 The line y = −2 x + 8 meets the
y-axis at the point B . Find the equation of the line with gradient
2 that passes through the point B.
3 The line y =
1 _ 2 x + ... | [
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97Straight line graphs
Example 10
A line is parallel to the line 6x + 3y − 2 = 0 and it passes through the point (0, 3).
Work out the equation of the line.
6x + 3 y − 2 = 0
3y − 2 = − 6x
3y = −6x + 2
y = −2x + 2 __ 3
The g
radient of this line is − 2.
The equation of the line is y = − 2x + 3.Rearrange the equat... | [
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98
Chapter 5
Example 11
Work out whether these pairs of lines are parallel, perpendicular or neither:
a 3x − y − 2 = 0 b y = 1 _ 2 x
x + 3y − 6 = 0 2x − y + 4 = 0
a 3x − y − 2 = 0
3x − 2 = y
So y =
3x − 2
The gradient of this line is 3.
x + 3 y − 6 = 0
3y − 6 = − x
3y = − x + 6
y = − 1 __ 3 x + 2
The gr... | [
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-0.0... |
99Straight line graphs
Example 12
A line is perpendicular to the line 2y − x − 8 = 0 and passes through the point (5, −7).
Find the equation of the line.
y = 1 __ 2 x + 4
Grad
ient of y = 1 __ 2 x + 4 is 1 __ 2
So th
e gradient of the perpendicular line is − 2.
y − y 1 = m(x − x 1 )
y +
7 ... | [
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100
Chapter 5
8 Find an equation of the line tha
t passes through the point (−2, −3) and is perpendicular to the
line y = − 4 _ 7 x + 5. W rite your answer in the form ax + by + c = 0, where a, b and c are integers.
9 The line l passes thr
ough the points (−3, 0) and (3, −2)
and the line n passes through the po... | [
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101Straight line graphs
Example 13
Find the distance between (2, 3) and (5, 7).
y
x Od
(2, 3)/four.ss01
3(5, 7)
d 2 = (5 − 2)2 + (7 − 3)2
d 2 = 32 + 42
d = √ __________ (32 + 42)
= √ ___ 25
= 5Draw a sketch.
Let the distance between the points be d.
The difference in the y-coordinates is 7 − 3 = 4.
The differ... | [
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102
Chapter 5
1 Find the distance between these pairs of points:
a (0, 1), (6, 9) b (4, −6), (9, 6) c (3, 1), (−1, 4)
d (3, 5), (4, 7) e (0, −4), (5, 5) f (−
2, −7), (5, 1)
2 Consider the points A(
−3, 5), B(−2, −2) and C (3, −7).
Deter
mine whether the line joining the points A and B is
congruent to the line... | [
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-0... |
103Straight line graphs
10 The straight line l 1 has equation 4x − 5y − 10 = 0 and intersects the x-axis at point A.
The straight line l 2 has equation 4x − 2y + 20 = 0 and intersects the x-axis at the point B.
a Work out the coor
dinates of A.
b Work out the coor
dinates of B.
The straight lines l 1 ... | [
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104
Chapter 5
Example 15
The graph shows the extension, E, of a spring
when different masses, m, are attached to the end of the spring.
a
Calculate the gr
adient, k, of the line.
b Write an equation linking
E and m.
c Explain what the v
alue of k represents in this
situation.E
m100 0 200
Mass on spring (grams)Exten... | [
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-0.001... |
105Straight line graphs
a
20 0/four.ss010
Time (seconds)Depth of w/a.ss01ter
d
tDepth of w/a.ss01ter (cm)
60801001202
0/four.ss016810121/four.ss01161820
The points form a straight line, therefore a
linear model is appropriate.
b m = 6.1 − 19.1 ___________ 100 − 0
= − 13 ____ 100 = − 0.13
Th
e d-intercept is ... | [
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0.04... |
106
Chapter 5
a 1991 is the first year, so t = 0.
Wh
en t = 0, the population is 18 500
.
18 500 i
s the p -intercept.
The population is expected to increase
by 350 each year.
350 represents the gradient of the line.p = at + b
p = 350 t + 18
50
0
b The n
umber of people living in Bradley
Stoke would probably not i... | [
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-0.... |
107Straight line graphs
a Draw a graph of the data. (3 marks)
b Explain how you kno
w a linear model would be appropriate. (1 mark)
c Deduce an equation in the for
m E = ah + b. (2 marks)
d Interpret the meaning of the coefficients
a and b. (2 marks)
e Use the model to find the cost of 65 kilo watt hours. (1 mark... | [
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-0.0... |
108
Chapter 5
1 The straight line passing through the point
P(2, 1) and the point Q (k, 11) has gradient − 5 __ 12
a Find the equation of the line in ter
ms of x and y only. (2 marks)
b Determine the value of k. (2 marks)
2 The points A and
B have coordinates (k, 1) and (8, 2k − 1) respectively, where k is a... | [
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-0.... |
109Straight line graphs
9 The line l has equa tion 2x − y − 1 = 0.
The line m passes through the point A(0, 4) and is perpendicular to the line l.
a Find an equation of
m. (2 marks)
b Show that the lines
l and m intersect at the point P(2, 3). (2 marks)
The line n passes thr
ough the point B(3, 0) and is parallel to... | [
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... |
110
Chapter 5
15 The points A and
B have coordinates (4, 6) and (12, 2) respectively.
The straight line l1 passes through A and B.
a Find an equation for
l1 in the form ax + by + c = 0, where a, b and c are integers. (3 marks)
The straight line l2 passes through the origin and has gradient − 2 _ 3
b Write down... | [
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-0.0111... |
111Straight line graphs
1 Find the area of the triangle with vertices A(−2, −2), B(13, 8) and C(−4, 14).
2 A tria
ngle has vertices A(3, 8), B(9, 9) and C(5, 2) as shown in
the diagram.
The l
ine l 1 is perpendicular to AB and passes through C.
The l
ine l 2 is perpendicular to BC and passes through A.... | [
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-... |
112
Chapter 5
1 The gradient m of the line joining the point with
coor
dinates (x1, y1) to the point with coordinates
(x2, y2) can be calculated using the formula
m = y2 – y1 ______ x2 – x1
2 ● The equation of a str
aight line can be written in the form
y = mx + c,
where m is the gradient and (0, c) is the y-i... | [
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113
Circles
After completing this unit you should be able to:
● Find the mid point of a line segment → pages 114 – 115
● Find the equation of the perpendicular bisector t o a line segment
→ pages 116 – 117
● Know how to find the equation of a circle → pages 117 – 120
● Solve geometric problems involving straight l... | [
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0... |
114
Chapter 6
6.1 Midpoints and perpendicular bisectors
You can find the midpoint of a line segment by averaging the
x- and y-coordinates of its endpoints.
■ The midpoint of a line segment with endpoints ( x1, y1) and ( x2, y2)
is ( x 1 + x 2 _______ 2 , y 1 + y 2 _______ 2 ) .
Example 1
E... | [
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0.... |
115Circles
1 Find the midpoint of the line segment joining each pair of points:
a (4, 2), (6, 8) b (0, 6), (12, 2) c (2, 2), (−4, 6)
d (−
6, 4), (6, −4) e (7, −4), (
−3, 6) f (−
5, −5), (−11, 8)
g (6a, 4
b), (2a, −4b) h (−4
u, 0), (3u, −2v) i (a
+ b, 2a − b), (3a − b, −b)
j (4 √ __
2 , 1) (2 √ __
2 , 7)... | [
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0.041... |
116
Chapter 6
■ The perpendicular bisector of a line segment AB is the straight line
that is perpendicular to AB and passes through the midpoint of AB.lB
Amidpoint
If the gradient of AB is m then the gradient
of its perpendicular bisector, l , will be − 1 __ m
OB(5, 2)A(–1, /four.ss01)
xy
l
C
The centre of th... | [
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0.015... |
117Circles
6.2 Equation of a circle
A circle is the set of points that are equidistant from a fixed point. You can use Pythagoras’ theorem
to derive equations of circles on a coordinate grid.
For any point (x, y) on the circumference of a circle, you can use Pythagoras’ theorem to show
the relationship between x, y a... | [
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118
Chapter 6
Example 6
The line segment AB is a diameter of a circle, where A and B are (4, 7) and (−8, 3) respectively.
Find the equation of the circle.
Length of AB = √ _____________________ (4 − ( −8))2 + (7 − 3)2
= √ ________ 122 + 42
= √ ____ 160
= √ ___ 16 × √ ___ 10
= 4 √ ___ 10 ... | [
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119Circles
Example 7
Find the centre and the radius of the circle with the equation x2 + y2 − 14x + 16y − 12 = 0.
Rearrange into the form ( x − a)2 + (y − b)2 = r2.
x2 + y2 − 14 x + 16 y − 12 = 0
x2 − 14 x + y2 + 16 y − 12 = 0 (1)
Co
mpleting the square for x terms and y terms.
x2 − 14 x = ( x − 7)2 − 49
y2 + 16 y =... | [
-0.013587024062871933,
-0.02658182941377163,
-0.03625372424721718,
-0.04968440160155296,
-0.0037310225889086723,
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-0.09757407009601593,
0.00029939107480458915,
-0.013461329974234104,
0.05191003158688545,
... |
120
Chapter 6
5 The line PQ is the diameter of
the circle, where P and Q are (5, 6) and (−2, 2) respectively.
Find the equation of the circle. (5 marks)
6 The point (1, −3) lies on the cir
cle (x − 3)2 + (y + 4)2 = r2. Find the value of r. (3 marks)
7 The points P(2, 2),
Q(2 + √ __
3 , 5) and R(2 − √ __
3 ... | [
-0.008733023889362812,
0.056478578597307205,
-0.02962067723274231,
0.008690333925187588,
0.006081746891140938,
0.06273674219846725,
0.04862157627940178,
0.05178618058562279,
-0.051096342504024506,
0.017094779759645462,
0.0747215673327446,
-0.03312540426850319,
0.08164791017770767,
0.008157... |
121Circles
Solve the equations simultaneously, so substitute
y = x + 5 into the equation of the circle.
← Sec tion 3.26.3 Intersections of straight lines and circles
You can use algebra to find the coordinates of intersection
of a straight line and a circle.
■ A straight line can intersect a cir
cle once, by just... | [
-0.04096598923206329,
0.03999536111950874,
0.07947146147489548,
0.027968011796474457,
-0.009988892823457718,
-0.029627539217472076,
0.06490939110517502,
-0.03191050887107849,
-0.046449147164821625,
0.05377505347132683,
0.024482250213623047,
-0.050643354654312134,
0.10618838667869568,
0.056... |
122
Chapter 6
1 Find the coordinates of
the points where the circle
(x − 1)2 + (y − 3)2 = 45 meets the x-axis. Substitute y = 0 in to the equation. Hint
2 Find the coordinates of
the points where the circle (x − 2)2 + (y + 3)2 = 29 meets the y-axis.
3 The line y =
x + 4 meets the circle (x − 3)2 + (y − 5)2 = 34... | [
0.016685517504811287,
0.06314210593700409,
-0.017824463546276093,
0.01060181949287653,
-0.010282652452588081,
0.036660611629486084,
0.014778590761125088,
-0.00008823576354188845,
-0.05943148583173752,
0.04931158572435379,
0.01077586691826582,
-0.05165594816207886,
0.03964177146553993,
-0.0... |
123Circles
6.4 Use tangent and chord pr operties
You can use the properties of tangents and chords within circles to solve
geometric problems. A tangent to a circle is a straight line that intersects the circle at only one point.
■
A tangent to a cir
cle is perpendicular to the radius of the circle at
the point of... | [
0.025313593447208405,
0.06657038629055023,
0.010548954829573631,
0.026717407628893852,
-0.05663568526506424,
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0.005795586854219437,
0.07968275249004364,
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-0.011962978169322014,
0.04334903880953789,
-0.015203365124762058,
0.052395936101675034,
0.067... |
124
Chapter 6
Draw a sketch showing the circle and the two
possible tangents with gradient −3. If you are solving a problem involving tangents and circles there is a good chance you will need to use the radius at the point of intersection, so draw this on your sketch.Problem-solvingExample 11
A circle C has equation (... | [
0.028359798714518547,
0.05905742943286896,
0.10354050248861313,
0.039763111621141434,
0.0002901677798945457,
-0.005296540446579456,
0.04931645467877388,
0.0069785164669156075,
-0.0714481845498085,
0.0428650937974453,
0.06807560473680496,
-0.046834055334329605,
0.06221175566315651,
0.088796... |
125Circles
This is the other possible equation for the tangent. So the tangents will intersect the circle at
(8, −2) and (2, − 4)
y − y1 = m(x − x1)
y + 2 = − 3(x − 8)
y = −3
x + 22
y −
y1 = m(x − x1)
y + 4 = − 3(x − 2)
y = −3
x + 2 This is one possible equation for the tangent.
Substitute ( x 1 , y ... | [
0.015823811292648315,
0.049730852246284485,
0.04562961310148239,
0.009676025249063969,
-0.0387934073805809,
0.00036698620533570647,
0.06945742666721344,
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0.006467076949775219,
0.1134432926774025,
-0.05865982547402382,
0.07266496121883392,
0.04201... |
126
Chapter 6
c The centre of the circle is ( −4, −8).
To find the radius of the circle:
CQ =
√ ___________________ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
= √ _______________________ (3 − (− 4)) 2 + (−5 − (−8)) 2
= √ _______ 49 + 9 = √ ___ 58
So th
e circle has a ra... | [
0.04560043662786484,
0.025704780593514442,
-0.07717897742986679,
0.02039719559252262,
-0.05920351296663284,
0.01323859952390194,
0.030377237126231194,
0.09132060408592224,
-0.03447676822543144,
-0.010607442818582058,
0.00454510934650898,
-0.08545887470245361,
0.0945052057504654,
0.02040170... |
127Circles
6 The tangent to the circle (x + 4)2 + (y − 1)2 = 242 at (7, − 10) meets the y -axis at S and the x -axis at T .
a Find the coordinates of
S and T. (5 marks)
b Hence, find the area of
△OST, where O is the origin. (3 marks)
7 The circle C
has equation (x + 5)2 + (y + 3)2 = 80.
The line l is a tangent to... | [
-0.002312507014721632,
0.06603080779314041,
0.10306563228368759,
0.012434727512300014,
-0.04758135601878166,
-0.007778261322528124,
0.0177980437874794,
0.018519733101129532,
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0.002974879229441285,
0.04767107218503952,
-0.08302076905965805,
0.015795031562447548,
0.050... |
128
Chapter 6
11 The circle C
has equation x2 − 4x + y2 − 6y = 7.
The line l with equation x − 3y + 17 = 0 intersects the circle
at the points P and Q.
a Find the coordinates of
the point P and the
point Q. (4 marks)
b Find the equation of the tangent a
t the point P
and the point Q. (4 marks)
c Find the equa... | [
-0.01624411903321743,
0.11154145002365112,
0.0008701924234628677,
0.0179268941283226,
0.00026461234665475786,
0.036531656980514526,
0.023736771196126938,
0.06228247284889221,
-0.09701704233884811,
-0.010518662631511688,
0.06177204102277756,
-0.055247578769922256,
0.014230675995349884,
-0.0... |
129Circles
Example 13
The points A(−8, 1), B(4, 5) and C(−4, 9) lie on the
circle, as shown in the diagram.
a Show that
AB is a diameter of the circle.
b Find an equation of the cir
cle.y
xC(–4, 9)
OA(–8, 1)B(4, 5)
a Test triangle ABC to s ee if it is a right-
angled triangle.
AB2 = (4−(−8))2 + (5 − 1)2
= 122 + 42... | [
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0.046089231967926025,
-0.03474186360836029,
-0.00586772570386529,
-0.0380689837038517,
-0.04171242192387581,
-0.08220583200454712,
0.038890521973371506,
-0.07385935634374619,
-0.021444974467158318,
-0.01847250945866108,
-0.11232104897499084,
0.08084414154291153,
0.048... |
130
Chapter 6
Example 14
The points P(3, 16), Q(11, 12) and R(−7, 6) lie on the circumference of a circle. The equation of the
perpendicular bisector of PQ is y = 2x.
a Find the equation of the perpendicular bisector of
PR.
b Find the centre of the cir
cle.
c Work out the equa
tion of the circle.
a The midpoint of P... | [
-0.04358202591538429,
0.05861734598875046,
-0.08464455604553223,
-0.026515820994973183,
0.015193612314760685,
0.012081916444003582,
-0.02277006022632122,
0.06666318327188492,
-0.08945479989051819,
-0.07585961371660233,
0.0377173013985157,
-0.05794407054781914,
0.04596494883298874,
-0.00793... |
131Circles
Exercise 6F
1 The points U( −2, 8), V(7, 7) and W(−3, −1) lie on a circle.
a Show that triangle
UVW has a right angle.
b Find the coordinates of
the centre of the circle.
c Write down an equa
tion for the circle.
2 The points A(2, 6),
B(5, 7) and C(8, −2) lie on a circle.
a Show that
AC is a diameter of ... | [
-0.031715285032987595,
0.0014280967880040407,
-0.02272462099790573,
-0.030236242339015007,
-0.019994722679257393,
0.09717120975255966,
-0.08287722617387772,
0.021702421829104424,
-0.08336213231086731,
-0.06233981251716614,
-0.018144069239497185,
-0.028934268280863762,
0.040562573820352554,
... |
132
Chapter 6
9 A circle has equation
x2 + 2x + y2 − 24y − 24 = 0
a Find the centre and radius of
the circle. (3 marks)
b The points A(
−13, 17) and B(11, 7) both lie on the circumference of the circle.
Show that AB is a diameter of the circle. (3 marks)
c The point C lies on the nega
tive x-axis and the angle ACB ... | [
-0.0008427576394751668,
0.07502005994319916,
-0.076657734811306,
0.025541359558701515,
-0.010760791599750519,
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0.0004134928167331964,
0.09094370901584625,
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-0.05441748723387718,
-0.006311298813670874,
-0.04712012782692909,
0.038331080228090286,
-0.... |
133Circles
10 The circle (x − 5)2 + y2 = 36 meets the x-axis at P and Q. Find the coordinates of P and Q.
11 The circle (x
+ 4)2 + (y − 7)2 = 121 meets the y-axis at (0, m) and (0, n).
Find the values of m and n.
12 The circle C
with equation (x + 5)2 + (y + 2)2 = 125 meets the positive coordinate axes at
A(a, 0)... | [
0.00959866214543581,
0.047889988869428635,
0.01650942489504814,
-0.013221293687820435,
-0.05408952385187149,
0.03192116320133209,
-0.016806112602353096,
0.003864490194246173,
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-0.013276496902108192,
-0.008744859136641026,
-0.07513685524463654,
0.09029419720172882,
0.017... |
134
Chapter 6
18 The circle C
has equation (x − 3)2 + (y + 3)2 = 52.
The baselines l1 and l2 are tangents to the circle and
have gradient 3 _ 2
a Find the points of intersection, P and
Q, of the
tangents and the circle. (6 marks)
b Find the equations of lines
l1 and l2, giving your
answers in the form ax... | [
0.016976263374090195,
0.11281070858240128,
0.024972954764962196,
0.005246651358902454,
-0.007000241428613663,
0.04625209420919418,
0.022948138415813446,
0.02941630221903324,
-0.07348848879337311,
0.015629224479198456,
0.08200100064277649,
-0.03776732459664345,
0.07963304966688156,
-0.00531... |
135Circles
23 The points R(
−4, 3), S(7, 4) and T(8, −7) lie on the circumference of a circle.
a Show that
RT is the diameter of the circle. (4 marks)
b Find the equation of the cir
cle. (4 marks)
24 The points A(
−4, 0), B(4, 8) and C(6, 0) lie on the circumference of circle C.
Find the equation of the circle.
25 Th... | [
0.018732460215687752,
0.03563275560736656,
-0.022175513207912445,
-0.028656279668211937,
-0.0064820945262908936,
0.01670677587389946,
-0.06564196199178696,
0.09161341935396194,
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-0.08370407670736313,
0.015605954453349113,
-0.11840322613716125,
0.06099521368741989,
0.015... |
136
Chapter 6
3 The equation of a circl e with centre (0, 0) and radius r is x2 + y2 = r2.
4 The equation of the circl
e with centre (a, b) and radius r is (x − a)2 + (y − b)2 = r2.
5 The equation of a circl
e can be given in the form: x2 + y2 + 2fx + 2gy + c = 0
This circle has centre (−f, −g) and radius √ _______... | [
0.020493673160672188,
0.06028351932764053,
-0.04143655300140381,
0.01961212418973446,
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0.05639573559165001,
0.0579652339220047,
-0.05415625125169754,
-0.011039352975785732,
0.03205914795398712,
-0.03907279670238495,
0.05740852653980255,
0.0814548... |
137
Algebraic methods
After completing this unit you should be able to:
● Cancel fact
ors in algebraic fractions → pages 138–139
● Divide a polynomial by a linear expr ession → pages 139–142
● Use the factor theorem t o factorise a cubic expression
→ pages 143–146
● Construct mathematical proofs using algebra → p... | [
-0.09974712133407593,
0.02420107275247574,
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-0.0339592769742012,
-0.011171494610607624,
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0.02679169736802578,
-0.05967917665839195,
-0.16203299164772034,
-0.052605170756578445,
-0.... |
138
Chapter 7
7.1 Algebraic fractions
You can simplify algebraic fractions using division.
■ When simplifying an algebr
aic fraction, where
possible factorise the numerator and denominator
and then cancel common factors.5x2 – 245
2x2 – 15x + 7=5(x + 7)(x – 7)
(2x – 1)(x – 7)=5(x + 7)
2x – 1Factorise
Cancel common f... | [
-0.04866039380431175,
0.05268817022442818,
0.06990083307027817,
-0.00004713485759566538,
0.027884824201464653,
0.0016997139900922775,
-0.04377235844731331,
-0.03354059159755707,
-0.004628953989595175,
-0.0067831976339221,
-0.030997853726148605,
-0.09467411786317825,
-0.039517682045698166,
... |
139
Algebraic methods
d 7x5 − x3 − 4 ___________ x e 8x4 − 4x3 + 6x _____________ 2x f 9x2 − 12x3 − 3x ______________ 3x
g 7x3 − x4 − 2 ___________ 5x h −4x2 + 6x4 − 2x ______________ −2x i −x8 + 9x4 − 4x3 + 6 _________________ −2x
j −9x9 − 6x6 + 4x4 − 2 _______________... | [
-0.06037140637636185,
0.05892029404640198,
-0.021282775327563286,
-0.07843983173370361,
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-0.019666574895381927,
-0.00037175443139858544,
-0.06668762117624283,
-0.013132475316524506,
... |
140
Chapter 7
2 x2 + 5 x
x 3 + 2 x 2 − 17 x + 6 ____ x − 3)
x3 − 3 x2
5x2 − 17 x
5x2 − 15 x
−2x + 6
3 x2 + 5 x − 2
x 3 + 2 x 2 − 17 x + 6 ____ x − 3)
x3 − 3 x2
5x2 − 17 x
5x2 − 15 x
−2x + 6
−2x + 6
0
So x3 + 2x2 − 17 x + 6 __________________ x − 3 = x2 + 5 x − 2Repeat th... | [
-0.02724640816450119,
0.06433182954788208,
0.001313717570155859,
-0.08469488471746445,
0.03986818715929985,
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0.013774454593658447,
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0.03222329169511795,
-0.030557826161384583,
-0.06304746866226196,
0.06823636591434479,
-0.003229... |
141
Algebraic methods
Example 4
Find the remainder when 2x3 − 5x2 − 16x + 10 is divided by (x − 4).
2x2 + 3 x − 4
2 x 3 − 5 x2 − 16 x + 10 _____ x − 4)
2x3 − 8 x2
3x2 − 16 x
3x2 − 12 x
−4x + 10
−4x + 16
−6
So the remainder is − 6. (x − 4) is not a factor of 2x3 − 5x2 − 16x + 10 as
the remainder ≠ 0.
Th... | [
0.0009149094112217426,
0.06321510672569275,
-0.01567232795059681,
-0.025018446147441864,
0.06755169481039047,
0.04036771506071091,
-0.056090645492076874,
0.014738323166966438,
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0.021105730906128883,
-0.031639695167541504,
-0.06983289122581482,
0.029158184304833412,
-0... |
142
Chapter 7
7 Show that
x3 + 2x2 − 5x − 10 = (x + 2)(x2 − 5)
8 Find the remainder when:
a x3 + 4x2 − 3x + 2 is divided by (x + 5) b 3x3 − 20x2 + 10x + 5 is divided by (x − 6)
c −2x3 + 3x2 + 12x + 20 is divided by (x − 4)
9 Show that w
hen 3x3 − 2x2 + 4 is divided by (x − 1) the remainder is 5.
10 Show that w
hen... | [
-0.049399103969335556,
0.1187511682510376,
0.05388299748301506,
-0.08434280753135681,
0.034740712493658066,
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0.05151218920946121,
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0.011235466226935387,
0.0021801914554089308,
-0.11436519026756287,
0.034329041838645935,
-0.009... |
143
Algebraic methods
7.3 The factor theorem
The factor theorem is a quick way of finding simple linear factors of a polynomial.
■ The factor theorem stat
es that if f( x) is a
polynomial then:
• If f( p) = 0, then ( x – p) is a factor of f( x).
• If (x – p) is a factor of f( x), then f( p) = 0.
You can use the f... | [
-0.07705669850111008,
0.1121988594532013,
0.007891807705163956,
-0.029118744656443596,
0.06790745258331299,
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0.0013954584719613194,
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0.04041450098156929,
-0.016008930280804634,
-0.056914739310741425,
-0.0785837396979332,
-0.... |
144
Chapter 7
Example 6
a Fully factorise 2x3 + x2 – 18x – 9 b Hence sketch the gra ph of y = 2x3 + x2 − 18x − 9
a f(x) = 2 x3 + x2 − 18x − 9
f(−1) = 2(− 1)3 + (−1)2 − 18( −1) − 9 = 8
f(1) = 2(1)3 + (1)2 − 18(1) − 9 = − 24
f(2) = 2(2)3 + (2)2 − 18(2) − 9 = − 25
f(3) = 2(3)3 + (3)2 − 18(3) − 9 = 0
So (x − 3) is a facto... | [
-0.08419398218393326,
0.09223900735378265,
-0.007840902544558048,
-0.03670160099864006,
0.07662847638130188,
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0.007072285283356905,
-0.019650926813483238,
-0.048866722732782364,
0.033475182950496674,
0.025462443009018898,
-0.03100656345486641,
-0.028940755873918533,
-0... |
145
Algebraic methods
Example 7
Given that (x + 1) is a factor of 4x4 − 3x2 + a, find the value of a.
f(x) = 4 x 4 − 3 x2 + a
f(−1) = 0
4(−1)4 − 3(−1)2 + a = 0
4 − 3 + a = 0
a = −1Write the polynomial as a function.
Use statement 2 from the factor theorem.
(x − p) is a factor of f(x), so f( p) =
0
Here p = −1.
Substi... | [
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0.11629907786846161,
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-0.010297943837940693,
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0.00011987598554696888,
0.03444385901093483,
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0.06619545072317123,
0.01616172306239605,
-0.024183865636587143,
-0.010451280511915684,
-0... |
146
Chapter 7
7.4 Mathematical proof
A proof is a logical and structured argument to show that a
mathematical statement (or conjecture) is always true. A mathematical proof usually starts with previously established mathematical facts (or theorems) and then works through a series of logical steps. The final step i... | [
-0.0343727245926857,
0.06819811463356018,
0.012907715514302254,
-0.03855685517191887,
-0.07136751711368561,
0.07106637209653854,
-0.046169769018888474,
0.057932280004024506,
-0.01614036038517952,
-0.05966667830944061,
-0.019604487344622612,
-0.07655564695596695,
0.058796558529138565,
-0.01... |
147
Algebraic methods
Example 8
Prove that (3x + 2)(x − 5)(x + 7) ≡ 3x3 + 8x2 − 101x − 70
(3x + 2)( x – 5)( x + 7)
= (3x + 2)( x2 + 2x – 35)
= 3x3 + 6 x2 – 105 x + 2x2 + 4 x – 70
= 3x3 + 8 x2 – 101 x – 70
So
(3x + 2)(x – 5)(x + 7) ≡ 3 x3 + 8x2 – 101x – 70
Example 9
Prove that if (x − p) is a factor of f(x) then f( ... | [
-0.03660247474908829,
0.062075499445199966,
0.06445693224668503,
-0.08078251034021378,
-0.004609762690961361,
0.11321792751550674,
-0.04842440038919449,
-0.025646187365055084,
-0.09064748138189316,
-0.04906437173485756,
-0.037314046174287796,
-0.019065121188759804,
0.05698249116539955,
0.0... |
148
Chapter 7
Example 11
The equation kx2 + 3kx + 2 = 0, where k is a constant, has no real roots.
Prove that k satisfies the inequality 0 < k < 8 _ 9
kx2 + 3 kx + 2 = 0 has no real roots,
so b2 – 4 ac < 0
(3k)2 – 4k(2) < 0
9k2 – 8k < 0
k(9k – 8) < 0
y
k Oy = k(9k – 8)
8
9State which assumption or information y... | [
-0.05249692127108574,
0.1068231463432312,
-0.009880072437226772,
0.052256595343351364,
0.036311712116003036,
-0.02999960072338581,
-0.13059838116168976,
0.01835399493575096,
-0.10197977721691132,
0.014287246391177177,
-0.01072283647954464,
-0.03025582619011402,
0.029890671372413635,
-0.024... |
149
Algebraic methods
1 Prov
e that n2 − n is an even number for all values of n.
2 Prov
e that x ______ 1 + √ __
2 ≡ x √ __
2 − x.
3 Prov
e that (x + √ __
y )(x − √ __
y ) ≡ x2 − y.
4 Prov
e that (2x − 1)(x + 6)(x − 5) ≡ 2x3 + x2 − 61x + 30.
5 Prov
e that x2 + bx ≡ (x + b __ 2 ) ... | [
0.01814783178269863,
0.07034771889448166,
-0.0409371517598629,
0.0005870932945981622,
-0.01719832234084606,
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-0.0016862436896190047,
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-0.08396506309509277,
-0.06617584079504013,
-0.03272349014878273,
0.058537907898426056,
-0.01... |
150
Chapter 7
7.5 Methods of proo f
A mathematical statement can be proved by exhaustion. For example, you can prove that the sum
of two consecutive square numbers between 100 and 200 is an odd number. The square numbers between 100 and 200 are 121, 144, 169, 196.
121 + 144 = 265 which is odd
144 + 169
= 313 which i... | [
-0.04302125424146652,
0.07979383319616318,
-0.002527705393731594,
-0.06782735139131546,
-0.05627875030040741,
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0.003986333962529898,
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-0.07953400164842606,
-0.042587872594594955,
0.01874990016222,
0.12457770109176636,
-0.00196... |
151
Algebraic methods
A mathematical statement can be disproved using a counter-example. For example, to prove that
the statement ‘3n + 3 is a multiple of 6 for all values of n’ is not true you can use the counter-example when n = 2, as 3 × 2 + 3 = 9 and 9 is not a multiple of 6.
■
You can pro
ve a mathematical state... | [
-0.05582578107714653,
0.07603441178798676,
-0.04730967432260513,
0.060064006596803665,
-0.052604012191295624,
0.0041335998103022575,
-0.021039674058556557,
0.017513761296868324,
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-0.06549783051013947,
-0.0552486814558506,
-0.06890080869197845,
0.11996257305145264,
0.0... |
152
Chapter 7
1 Prov
e that when n is an integer and 1 < n < 6, then m = n + 2 is
not divisible by 10.
2 Prov
e that every odd integer between 2 and 26 is either prime or the product of two primes.
3 Prov
e that the sum of two consecutive square numbers between 12 to 82 is an odd number.
4 Prov
e that all cube number... | [
0.006219703704118729,
0.10198097676038742,
-0.07117259502410889,
-0.051657725125551224,
0.06919070333242416,
0.06610190123319626,
-0.04015722498297691,
-0.02121402509510517,
-0.07574445754289627,
-0.02532501146197319,
0.01159086637198925,
-0.03887370601296425,
0.036645565181970596,
-0.0318... |
153
Algebraic methods
5 Find a counter-example to dispr
ove each of the following statements:
a If n
is a positive integer then n4 − n is divisible by 4.
b Integers alw
ays have an even number of factors.
c 2n2 − 6n + 1 is positive for all values of n.
d 2n2 − 2n − 4 is a multiple of 3 for all integer values of n.
6 A... | [
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0.03423655405640602,
-0.027140285819768906,
0.014332447201013565,
0.015399723313748837,
0.020896488800644875,
-0.04856323078274727,
-0.03773583099246025,
-0.08187869936227798,
-0.05466211959719658,
-0.05380309373140335,
-0.024358797818422318,
0.11335387825965881,
0.056... |
154
Chapter 7
1 Simplify these fractions as far as possible:
a 3x 4 − 21x _________ 3x
b x2 − 2x − 24 ___________ x2 − 7x + 6
c 2x2 + 7x − 4 ___________ 2x2 + 9x + 4
2 Divide 3x3 + 12x2 + 5x + 20 by (x + 4).
3 Simplify 2x3 + 3x + 5 ___________ x + 1
4 a Show that (
x − 3) is a factor o... | [
-0.05644458904862404,
0.09656356275081635,
0.007675596512854099,
-0.0676240548491478,
0.04358283802866936,
0.033244673162698746,
-0.07572420686483383,
0.056084588170051575,
-0.041605930775403976,
0.04919491708278656,
-0.04918566346168518,
-0.09128958731889725,
-0.02010395936667919,
-0.0553... |
155
Algebraic methods
11 a Show that (
x − 2) is a factor of f(x ) = x3 + x2 − 5x − 2. (2 marks)
b Hence, or otherwise, find the e
xact solutions of the equation f(x ) = 0. (4 marks)
12 Given tha
t −1 is a root of the equation 2x3 − 5x2 − 4x + 3, find the two positive roots. (4 marks)
13 f(x
) = x3 – 2x2 – 19x + 20
... | [
-0.06272023171186447,
0.1377633661031723,
0.052612971514463425,
-0.0005350247956812382,
0.048151809722185135,
0.0730699896812439,
-0.034129802137613297,
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-0.09220712631940842,
0.012505541555583477,
-0.055549606680870056,
-0.10861926525831223,
-0.019552648067474365,
-0... |
156
Chapter 7
24 a Prov
e that the difference of the squares of two consecutive even numbers is always divisible
by 4.
b Is this statement true for od
d numbers? Give a reason for your answer.
25 A student is trying to prov
e that 1 + x2 < (1 + x)2.
The student writes:
(1 + x)2 = 1 + 2x + x2.
So 1 + x2 < 1 + 2x + x... | [
-0.003695450024679303,
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-0.042559944093227386,
0.03686699643731117,
-0.0034858451690524817,
0.07366257160902023,
0.0... |
157
Algebraic methods
1 When simplifying an algebr aic fraction, factorise the numerator and denominator where
possible and then cancel common factors.
2 You can use long division t
o divide a polynomial by (x ± p), where p is a constant.
3 The factor theor
em states that if f(x) is a polynomial then:
• If f( p) = 0,... | [
-0.0664624273777008,
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-0.08957108110189438,
-0.06604062765836716,
-0.02705... |
1588
The binomial
expansion
After completing this chapter you should be able to:
● Use Pascal
’s triangle to identify binomial coefficients and use them
to expand simple binomial expressions → pages 159–161
● Use combinations and factorial notation → pages 161–163
● Use the binomial expansion t o expand brackets ... | [
-0.02065727300941944,
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0.008290720172226429,
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-0.008641866035759449,
-0.03199578449130058,
-0.0256... |
159
The binomial expansion
8.1 Pascal’s triangle
You can use Pascal’s triangle to quickly expand expressions such as (x + 2y)3.
Consider the expansions of (a + b)n for n = 0, 1, 2, 3 and 4:
(a + b)0 =
(a + b)1 =
(a + b)2 =
(a + b)3 =
(a + b)4 = + 1a44a3b +6a2b2+4ab3+ 1a33a2b+ 3ab2+ 1a22ab+ 1a1
1b
+ 1b2
+ 1b3
+1b4
Every... | [
-0.04643777757883072,
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0.0027603162452578545,
0.03440758213400841,
-0.07236023992300034,
0.0491715669631958,
0.026116393506526947,
-0.05736623704433441,
0.0641961470246315,
-0.026922060176730156,
-0.0770965963602066,
-0.024626780301332474,
0.03135383501648903,
0.016054... |
160
Chapter 8
The coefficients are 1, 3, 3, 1:
The term in x2 is 3 × 2(− cx)2 = 6 c2x2
So 6 c2 = 294
c2 = 49
c = ±7Example 2Index = 4 so look at the 5th row of Pascal’s
triangle.
Index = 3 so use the 4th row of Pascal’s triangle.The coefficient of x2 in the expansion of (2 − cx)3 is 294.
Find the possible value(s) o... | [
-0.01225487794727087,
0.07012869417667389,
0.01835969276726246,
0.043423302471637726,
-0.019825639203190804,
0.08076386153697968,
0.030163127928972244,
0.015287816524505615,
-0.050505124032497406,
0.037520043551921844,
-0.06696594506502151,
-0.08791688829660416,
-0.02164718136191368,
-0.08... |
161
The binomial expansion
7 The coefficient of x3 in the expansion of (2 − x)(3 + bx)3 is 45. Find possible values of the
constant b.
8 Work out the coefficient of
x2 in the expansion of ( p − 2x)3. Give your answer in terms of p.
9 After 5 years, the v
alue of an investment of £500 at an interest rate of X % p... | [
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0.09867081791162491,
0.06837796419858932,
0.05488542839884758,
0.007322384510189295,
0.02441088855266571,
0.03789100795984268,
-0.03269188106060028,
0.009206158109009266,
0.07240523397922516,
-0.044152725487947464,
-0.03033100813627243,
-0.05070309340953827,
-0.03601... |
162
Chapter 8
1 Work out:
a 4! b 9! c 10 ! ___ 7 ! d 15 ! ___ 13 !
2 Without using a calcula
tor, work out:
a ( 4 2 ) b ( 6 4 ) c 6C3 d ( 5 4 ) e 10C8 f ( 9 5 )
3 Use a calculator to w
ork out:
a ( 15 6 ) b 10C7 c ( 20 10 ) d ( 20 17 ) e 14C9 f 18C5
4 Write each value a t... | [
-0.0511852502822876,
0.046651288866996765,
-0.0623679980635643,
-0.06599415838718414,
-0.04061164706945419,
0.010850428603589535,
-0.057984527200460434,
-0.014040939509868622,
-0.01931731216609478,
-0.016012495383620262,
-0.020446602255105972,
-0.0636293962597847,
0.0800246000289917,
-0.00... |
163
The binomial expansion
8.3 The binomial expansion
The binomial expansion is a rule that allows you to expand brackets. You can use ( n r ) to work out the
coefficients in the binomial expansion. For example,
in the expansion of (a + b)5 = (a + b)(a + b)(a + b)(a + b)(a + b), to find the b3 term you can choos... | [
-0.036077115684747696,
0.04503415524959564,
0.017005715519189835,
-0.0028763648588210344,
-0.00499691953882575,
0.08642075955867767,
0.01841404289007187,
-0.06600549072027206,
0.06678467243909836,
0.02784477174282074,
-0.07292522490024567,
-0.012013699859380722,
-0.007492745760828257,
0.00... |
164
Chapter 8
a (1 + 2x)10
= 1 10 + ( 10 1 ) 1 9 (2x) + ( 10 2 ) 1 8 (2x) 2
+ ( 10 3 ) 1 7 (2x) 3 + . . .
= 1 + 20 x
+ 180 x2 + 960 x3 + . . .
b (10 − 1 _ 2 x) 6
= 10 6 + ( 6 1 ) 10 5 (− 1 _ 2 x) + ( 6 2 ) 10 4 (− 1 _ 2 x) 2
+ ( 6 3 ) 10 ... | [
-0.08072605729103088,
0.12385401874780655,
-0.0138058140873909,
-0.0320022776722908,
-0.06396426260471344,
0.07894974946975708,
-0.019808301702141762,
-0.02408558875322342,
-0.0033614833373576403,
0.014333093538880348,
-0.06799570471048355,
-0.07576864212751389,
0.06798627972602844,
0.0054... |
165
The binomial expansion
8.4 Solving binomial problems
You can use the general term of the binomial expansion to find individual coefficients in a binomial
expansion.
■ In the expansion of ( a + b)n the general term is given by ( n r ) an − rbr.
x3 term = ( 10 3 ) 17(kx)3 = 15 x3
120k3x3 = 15 x3
k = 1 _... | [
-0.013950631022453308,
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0.004857539664953947,
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0.03785979002714157,
-0.010921908542513847,
-0.09430468082427979,
0.0039608050137758255,
0.022344816476106644,
-0.04... |
166
Chapter 8
a (1 + qx)8
= 18 + ( 8 1 ) 17(qx)1 + ( 8 2 ) 16(qx)2 + …
= 1 + 8 qx + 28 q2x2 + …
b 8q
= −r and 28 q2 = 7r
8q = −4 q2
4q2 + 8 q = 0
4q(q + 2) = 0
q = −2, r = 16Example 8
Using 28q2 = 7r, r = 4q2 and −r = −4q2.
q is non-zero so q = −2.a Write down the first three terms, in ascending powers of x,... | [
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0.009178265929222107,
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0.014039441011846066,
-0.09158772230148315,
0.07132166624069214,
-0.027... |
167
The binomial expansion
7 a Find the first four terms
, in ascending powers of x, of the binomial expansion
of (1 + qx)10, where q is a non-zero constant. (2 marks)
b Given tha
t in the expansion of (1 + qx)10 the coefficient of x3 is 108 times the
coefficient of x, work out the value of q. (4 marks)
8 a Find ... | [
-0.025759749114513397,
0.09196252375841141,
0.07181905210018158,
0.07854732125997543,
-0.03226485848426819,
0.1105085089802742,
0.010900059714913368,
-0.00690256804227829,
0.018600016832351685,
0.013503264635801315,
-0.06862495094537735,
-0.07312136888504028,
-0.018519625067710876,
-0.0477... |
168
Chapter 8
b We want (1 − x __ 4 ) = 0.975
x __ 4 = 0.025
x = 0.1
Substitute x = 0.1 into the expansion
for (1 − x __ 4 ) 10
from part a :
0.97510 ≈ 1 − 0.25 + 0.028 12 5
− 0.001 87
5
= 0.77625
0.97510 ≈ 0.7763 to 4 d.p.Substitute x = 0.1 into your expansion.
Using a calculator, 0.975... | [
0.008765924721956253,
-0.011639543808996677,
0.022916704416275024,
-0.0006057939608581364,
-0.025600438937544823,
-0.0006767971790395677,
0.07708311825990677,
0.0309405867010355,
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0.08436866104602814,
0.047871723771095276,
-0.11025359481573105,
-0.03355992212891579,
-0... |
169
The binomial expansion
8 a Find the first 4 terms
, in ascending powers of x, of the binomial expansion of (1 − 3x)5.
Give each term in its simplest form. (4 marks)
b If x
is small, so that x2 and higher powers can be ignored, show that
(1 + x)(1 −3x)5 ≈ 1 − 14x. (2 marks)
9 A microchip company mode
ls the p... | [
-0.07660514861345291,
0.05329667404294014,
0.07196085900068283,
0.026488035917282104,
-0.01815871335566044,
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0.0421762615442276,
-0.01595969870686531,
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-0.0344008207321167,
-0.003853442147374153,
-0.06901440024375916,
0.0038525089621543884,
-0.060... |
170
Chapter 8
6 a Expand (1 − 2
x)10 in ascending powers of x up to and including the term in x3,
simplifying each coefficient in the expansion. (4 marks)
b Use your expansion to f
ind an approximation of 0.9810, stating clearly the
substitution which you have used for x. (3 marks)
7 a Use the binomial series to ... | [
-0.056510768830776215,
0.055605363100767136,
0.06694052368402481,
0.03768892586231232,
0.009463922120630741,
0.08327744901180267,
-0.01654724031686783,
0.010141218081116676,
-0.022108003497123718,
0.036009930074214935,
-0.0964670404791832,
-0.0851970911026001,
-0.06180466711521149,
-0.0533... |
171
The binomial expansion
14 a Expand (2 +
x)6 as a binomial series in ascending powers of x, giving each coefficient
as an integer. (4 marks)
b By making suitab
le substitutions for x in your answer to part a, show that
(2 + √ _
3 )6 − (2 − √ _
3 )6 can be simplified to the form k √ __
3 , stati... | [
0.009102705866098404,
0.09899607300758362,
0.04818892478942871,
0.05441189557313919,
0.02476748637855053,
0.1191992536187172,
0.041159894317388535,
-0.03736047074198723,
-0.04609894007444382,
-0.006878565065562725,
-0.060902513563632965,
-0.03646443411707878,
-0.029024144634604454,
-0.0587... |
172
Chapter 8
1 Pascal’s triangle is formed by adding adjacent pairs of numbers to find the numbers on the
next r
ow.
2 The (n
+ 1)th row of Pascal’s triangle gives the coefficients in the expansion of (a + b)n.
3 n! =
n × (n − 1) × (n − 2) × … × 3 × 2 × 1.
4 You can use factorial notation and y
our calculator to f... | [
-0.0173038337379694,
0.03280153498053551,
-0.008434999734163284,
0.04708942025899887,
-0.04391319677233696,
0.06805361062288284,
0.08287862688302994,
0.0017450993182137609,
0.022945143282413483,
-0.04006945341825485,
-0.11081137508153915,
0.026746641844511032,
0.03904439136385918,
-0.00323... |
173
Trigonometric ratios
After completing this unit you should be able to:
● Use the cosine rul
e to find a missing side or angle → pages 174–179
● Use the sine rule to find a missing side or angl e → pages 179–185
● Find the area of a t riangle using an appropriate
formula → pages 185–187
● Solve problems involvi... | [
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0.00798764917999506,
0.010939967818558216,
0.014507255516946316,
-0.04700862243771553,
0.0025593077298253775,
-0.07241015881299973,
0.054889481514692307,
-0.05998857319355011,
-0.07092821598052979,
-0.01032259315252304,
-0.0766172781586647,
-0.07576487958431244,
0.0281... |
174
Chapter 9
9.1 The cosine rule
The cosine rule can be used to work out missing sides or angles in triangles.
■ This version of the c
osine rule is used to find a missing side if you know two sides and the
angle between them:
a2 = b2 + c2 − 2bc cos A
A
CB
ac
b
You can use the standard trigonometric ratios for righ... | [
-0.030881453305482864,
0.05293969810009003,
-0.019996879622340202,
-0.02770770527422428,
-0.02847610041499138,
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0.035933997482061386,
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-0.06520150601863861,
0.004998387303203344,
-0.016855919733643532,
0.03401757776737213,
0.054... |
175
Trigonometric ratios
Example 1
Calculate the length of the side AB of the triangle ABC in which AC = 6.5 cm, BC = 8.7 cm and
∠ACB
= 100°.
A
B Cb = 6.5 cm
a = 8.7 cm100°c
c2 = a2 + b2 − 2 ab cos C
c2 = 8.72 + 6.52 − 2 × 8.7 × 6.5 × cos 100°
= 75.69 + 42.25 − ( −19.639…)
= 137. 57…
So c = 11.729…
So AB = 1... | [
0.024313611909747124,
0.02689305879175663,
0.023522738367319107,
0.009592472575604916,
-0.021677449345588684,
0.07126600295305252,
-0.03237256780266762,
-0.012850377708673477,
-0.04908846318721771,
-0.06954643875360489,
0.016098368912935257,
-0.10914734750986099,
0.027289224788546562,
0.00... |
176
Chapter 9
Example 3
Coastguard station B is 8 km, on a bearing of 060°, from coastguar d station A. A ship C is 4.8 km,
on a bearing of 018°, aw
ay from A. Calculate how far C is from B.
ABCa km
4.8 km
8 km18°
60°N
a2 = b2 + c2 − 2 bc cos A
a2 = 4.82 + 82 − 2 × 4.8 × 8 × cos 42°
= 29.966…
a = 5.47 (3 s.f.)
C is ... | [
0.07628364115953445,
0.033617716282606125,
0.054287005215883255,
0.01908985711634159,
0.0415218323469162,
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-0.07136378437280655,
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0.00795094296336174,
0.0309038907289505,
-0.11102370172739029,
0.025155682116746902,
-0.0081918044... |
177
Trigonometric ratios
Give answers to 3 significant figures, where appropriate.
1 In each of the follo
wing triangles calculate the length of the missing side.
a
20°
AB
C
8.4 cm6.5 cm b
60°AB
C2 cm
1 cm c
160°A
BC4.5 cm
5.5 cm
d
45°AB
C6 cm5 cm e
40°
AB
C10 cm10 cm f
108°AB
C5.6 cm6.5 cm
2 In the following t... | [
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0.08955269306898117,
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-0.0863887295126915,
0.016454726457595825,
-0.07029444724321365,
-0.007285166531801224,
-0... |
178
Chapter 9
4 From a point
A a boat sails due north for 7 km to B. The boa
t leaves B and moves on a bearing
of 100° for 10 km until it reaches C
. Calculate the distance of C from A.
5 A helicopter flies on a bearing of 080° fr
om A to B, where AB = 50 km.
It then flies for 60 km to a point C.
Gi
ven tha... | [
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0.009534664452075958,
-0.0321684405207634,
-0.023525215685367584,
0.022138... |
179
Trigonometric ratios
15 In △ABC
, AB = x cm, BC
= 5 cm, AC =
(10 − x) cm.
a Show that cos
∠ABC = 4x
− 15 _______ 2x
b Given tha
t cos ∠ABC = − 1 _ 7 , work out the value of x.
16 A farmer has a field in the sha
pe of a quadrilateral as shown.
D
ABC
75 m60 m
120 m135 m
The angle between fen... | [
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0.06548936665058136,
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-0.13522402942180634,
-0.0043253591284155846,
... |
180
Chapter 9
and sin A = h __ b ⇒ h = b sin A
So a sin
B = b
sin A
So a _____ sin A = b _____ sin B
In a si
milar way, by drawing the perpendicular
from B to the side AC , you can show that:
a _____ sin A = c _____ sin C
So a _____ sin A = b _____ sin B = c _____ sin C ... | [
-0.023754067718982697,
0.11423072218894958,
-0.043841514736413956,
-0.04010525718331337,
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0.019673967733979225,
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-0.10357405990362167,
0.007706068456172943,
-0.04360411688685417,
0.0012479191645979881,
... |
181
Trigonometric ratios
sin C _____ c = sin
A _____ a
sin
C _____ 3.8 = sin
35° _______ 5.2
So si
n C = 3.8
sin 35° ___________ 5.2
C = 24
.781…
So B = 12
0° (3 s.f.)Use sin C _____ c = sin A _____ a
Write the f
ormula you are going to use as the first
line of workin... | [
-0.0004767813952639699,
0.010192066431045532,
0.04222476854920387,
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0.010401089675724506,
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-0.10781503468751907,
0.04026538133621216,
-0.11539500206708908,
-0.01398495677858591,
0.... |
182
Chapter 9
3 In each of the follo
wing sets of data for a triangle ABC,
AB
ca
bC
find the value of x.
a AB =
6 cm, BC
= 9 cm, ∠BA
C = 117°, ∠ACB = x
b AC =
11 cm, BC
= 10 cm, ∠ABC
= 40°, ∠CAB = x
c AB =
6 cm, BC
= 8 cm, ∠BA
C = 60°, ∠ACB = x
d AB =
8.7 cm, AC =
10.8 cm, ∠ABC
= 28°, ∠BAC = x
4 In e... | [
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0.06442134827375412,
-0.11544070392847061,
0.02136838808655739,
-0.01... |
183
Trigonometric ratios
9 In the diagram
AD = DB = 5 cm, ∠ABC
= 43°
43°5 cm
5 cmA
CDB72°
and ∠ACB = 72°.
Calculate:
a AB
b CD
10 A zookeeper is building an enclosur
e for some llamas. 66°
76 m
80 m98°A B
CD
The enclosure is in the shape of a quadrilateral as shown.
If the length of the diagonal BD is 136 ... | [
0.033465828746557236,
0.03453167527914047,
0.03738616034388542,
-0.012726512737572193,
-0.010675025172531605,
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-0.016391610726714134,
0.021302592009305954,
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-0.018596375361084938,
0.06177680939435959,
-0.08579233288764954,
-0.0033843887504190207,
0... |
184
Chapter 9
Example 7
In △ABC, AB = 4 cm, AC = 3 cm and ∠ABC = 44°. Work out the two possible values of ∠ACB .
Give answers to 3 significant figures, where appropriate.
1 In △ABC
, BC = 6 cm, AC =
4.5 cm and ∠ABC
= 45°.
a Calculate the tw
o possible values of ∠BAC .
b Draw a dia
gram to illustrate your answers.
... | [
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0.04728871211409569,
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0.011561471968889236,
0.014969308860599995,
-0.1470191925764084,
0.010553406551480293,
0.016... |
185
Trigonometric ratios
4 Triangle ABC
is such that AB = 4 cm, BC
= 6 cm and ∠AC
B = 36°. Show that one of the possible
values of ∠ABC is 25.8° (to 3 s.f.). Using this value, calculate the length of AC .
5 Two triangles
ABC are such that AB = 4.5 cm, BC
= 6.8 cm and ∠AC
B = 30°. Work out the value
of the larg... | [
0.011469509452581406,
-0.0000733290144125931,
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0.06713338941335678,
-0.08613119274377823,
-0.008283047005534172,
0.... |
186
Chapter 9
Example 8
Work out the area of the triangle shown below.
B
C A75°
6.9 cm4.2 cm
Area = 1 __ 2 bc sin A
A
rea of △ ABC = 1 __ 2 × 6.9 × 4.2 × sin 75° cm2
= 14.0 cm2 (3 s.f.)Here b = 6.9 cm, c = 4.2 cm and angle A = 75°,
so use:
Area = 1 _ 2 bc sin A .
1 Calculate the area of each tri... | [
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0.05277446284890175,
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-0.02890520542860031,
-0.11818168312311172,
-0.010393870063126087,
-0... |
187
Trigonometric ratios
2 Work out the possib
le sizes of x in the following triangles.
x
x30 cm
6.5 cm8.5 cm
8 cm6 cm
x40 cm
Area =
400 cm2Area =
12.4 cm2Area =
12 3 cm2ab c B
ACBBC
A
AC
3 A fenced triangular plot of ground has ar ea 1200 m2. The fences along the two smaller sides are
60 m and 80 m respectiv
... | [
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0.049043599516153336,
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-0.10504314303398132,
-0.008102426305413246,
-0.... |
188
Chapter 9
For questions involving area, check first whether you can use Area = 1 _ 2 × b ase × height, before using
the formula involving sine.
• to find an unknown angl
e given two sides and one opposite angle, use the sine rule
• to find an unknown side giv
en two angles and one opposite side, use the sine... | [
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-0.02523803897202015,
-0.0464274100959301,
0.0... |
189
Trigonometric ratios
AB2 = AD2 + BD2 − 2(AD)(BD)cos(∠BDA)
AB2 = 702 + 71.708…2
− 2(70)(71.708…)cos (81.045…)
AB2 = 8479.55…
AB = √ ___________ 8479.55… = 92 .084…
= 92.1 m (3 s.
f.)
b sin(∠B
AD) __________ BD = sin(∠B
DA) __________ AB
sin(∠B
AD) = sin(81.045…) × 71.708 _____________... | [
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-0.0034067430533468723,
... |
190
Chapter 930°x
x
84°y
xa
4.2 cm
5.7 cm
ABB
AC
A CB
C z cm56°48°b
14.6 cm20 cmy cm
y cmz cmc
z cm
120°
yxd
12.8 cm
6 cmz cm
130°A
AACCC BB
B
yxy
x
zze3 cm
5 cm6 cm8 cm
8 cm12 cmf
y
x
z cm40°10.5 cm9.5 cmg
z45°
yz x
4.8 cm9.6 cm
x cmh
12.3 cm
15.6 cmy cmi
20.4 cm
C
C
CAB
D
A
AB B
2 In △ABC, ca lculate the size ... | [
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0.05679132789373398,
... |
191
Trigonometric ratios
9 In △ABC
, AB = √ __
2 cm, BC = √ __
3 cm and ∠BAC = 60°. Show that ∠ACB = 45° and find AC .
10 In △ABC
, AB = (2 − x) cm, BC
= (x + 1) cm
and ∠
ABC = 120°.
a Show that
AC 2 = x2 − x + 7.
b Find the value of
x for which AC has a
minimum value.
11 Triangle ABC
is suc... | [
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0.03450021147727966,
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0.014467095956206322,
-0.0119... |
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