Contents Introduction
iii
1 – Algebra and functions 1: Manipulating algebraic expressions 1
1.1 Manipulating polynomials algebraically 1.2 Expanding multiple binomials 1.3 The binomial expansion 1.4 Factorisation 1.5 Algebraic division 1.6 Laws of indices 1.7 Manipulating surds 1.8 Rationalising the denominator Summary of key points Practice questions 1
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3 6 11 14 16 22 24 26 29 30
2.1 Quadratic functions 2.2 The discriminant of a quadratic function 2.3 Completing the square 2.4 Solving quadratic equations 2.5 Solving simultaneous equations 2.6 Solving linear and quadratic simultaneous equations 2.7 Solving linear inequalities 2.8 Solving quadratic inequalities Summary of key points Practice questions 2
33 36 38 40 43 47 48 53 57 57
3.1 Sketching curves of quadratic functions 61 3.2 Sketching curves of cubic functions 65 3.3 Sketching curves of quartic functions 69 3.4 Sketching curves of reciprocal functions 74 3.5 Intersection points 81 3.6 Proportional relationships 84 3.7 Translations 88 3.8 Stretches 93 Summary of key points 99 Practice questions 3 100
4 – Coordinate geometry 1: Equations of straight lines 102
4.1 Writing the equation of a straight line in the form ax + by + c = 0
103
111 114 120 126 126
5.1 Equations of circles 5.2 Angle in a semicircle 5.3 Perpendicular from the centre to a chord 5.4 Radius perpendicular to the tangent Summary of key points Practice questions 5
129 136 140 146 153 153
6 – Trigonometry 156
6.1 Sine and cosine 6.2 The sine rule and the cosine rule 6.3 Trigonometric graphs 6.4 The tangent function 6.5 Solving trigonometric equations 6.6 A useful formula Summary of key points Practice questions 6
157 160 168 170 174 176 179 179
7 – Exponentials and logarithms
181
ax 182
7.1 The function 7.2 Logarithms 186 7.3 The equation ax = b 191 7.4 Logarithmic graphs 193 7.5 The number e 198 7.6 Natural logarithms 203 7.7 Exponential growth and decay 208 Summary of key points 211 Practice questions 7 211
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3 – Algebra and functions 3: Sketching curves 60
105
5 – Coordinate geometry 2: Circles 128
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2 – Algebra and functions 2: Equations and inequalities 32
4.2 Finding the equation of a straight line using the formula y – y1 = m(x – x1) 4.3 Finding the gradient of the straight line between two points 4.4 Finding the equation of a straight line using the formula y – y1 / y2 – y1 = x – x1 / x2 – x1 4.5 Parallel and perpendicular lines 4.6 Straight line models Summary of key points Practice questions 4
8 – Differentiation 214
8.1 8.2 8.3 8.4 8.5
The gradient of a curve The gradient of a quadratic curve Differentiation of x² and x3 Differentiation of a polynomial Differentiation of xn
215 219 224 227 230
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Contents
8.6 Stationary points and the second derivative 8.7 Tangents and normals Summary of key points Practice questions 8
234 239 242 242
9 – Integration 245 9.1 Indefinite integrals 9.2 The area under a curve Summary of key points Practice questions 9
245 251 260 260
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10.1 Definition of a vector 10.2 Adding vectors 10.3 Vector geometry 10.4 Position vectors Summary of key points Practice questions 10
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10 – Vectors 265 266 271 274 283 288 288
11.1 Proof by deduction 11.2 Proof by exhaustion 11.3 Disproof by counter example Summary of key points Practice questions 11
288 291 296 300 300
12 – Data presentation and interpretation 300
13.1 Calculating and representing probability 346 13.2 Discrete and continuous distributions 353 13.3 The binomial distribution 362 Summary of key points 368 Practice question 13 368
14 – Statistical sampling and hypothesis testing 371
15.1 The language of kinematics 395 15.2 Equations of constant acceleration 398 15.3 Vertical motion 405 15.4 Displacement-time and velocity-time graphs 411 15.5 Variable acceleration 419 Summary of key points 425 Practice questions 15 425
16 – Forces 428
16.1 Forces 429 16.2 Newton’s laws of motion 432 16.3 Vertical motion 441 16.4 Connected particles 444 16.5 Pulleys 452 Summary of key points 458 Practice questions 16 458
Worked solutions 463
1 Algebra and functions 1: Manipulating algebraic expressions 463 2 Algebra and functions 2: Equations and inequalities 470 3 Algebra and functions 3: Sketching curves 483 4 Coordinate geometry 1: Equations of straight lines 499 5 Coordinate geometry 2: Circles 510 6 Trigonometry 520 7 Exponentials and logarithms 524 8 Differentiation 532 9 Integration 541 10 Vectors 545 11 Proof 553 12 Data presentation and interpretation 557 13 Probability and statistical distributions 565 14 Statistical sampling and hypothesis testing 572 15 Kinematic 578 16 Forces 593
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12.1 Measures of central tendency and spread 303 12.2 Variance and standard deviation 310 12.3 Displaying and interpreting data 319 Summary of key points 342 Practice question 12 343
13 – Probability and statistical distributions 345
15 – Kinematics 394
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Summary of key points 392 Practice question 14 392
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11 – Proof 287
Glossary 606 Index
xx
14.1 Populations and samples 372 14.2 Hypothesis testing 377
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4
cooRDInATe geoMeTRy 1: eQUATIonS oF STRAIghT lIneS
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It’s not easy comparing mobile phone tariffs from different providers. The following graph provides a simple, visual representation of three tariffs, which can easily be used to make comparisons over time. From this graph you could identify important features like:
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the tariff with the lowest monthly charge.
LEArninG oBJECtiVES
Cost (£)
200
Option C
150
Option B
100 50
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You will learn how to:
Option A
250
the tariff with the highest upfront cost
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› ›
300
›
write the equation of a straight line in the form ax + by + c = 0
›
understand and use the gradient conditions for parallel lines
›
understand and use the gradient conditions for perpendicular lines
›
be able to use straight line models in a variety of contexts.
0
1
2
3
4
5 6 7 Time (months)
8
9
10
11
12
FPO
ed
toPiC LinKS
0
Prior KnoWLEdGE
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You will need your knowledge of the gradient of a straight line to help you understand and solve differentiation problems in Chapter 8 Differentiation. The ability to draw and interpret straight line graphs will help you to solve distance, speed and time problems in Chapter 15 Kinematics. You will also need to determine and use the equations of straight lines when working with regression lines in Chapter 16 Data presentation and interpretation in context.
You should already know how to:
› › › ›
work with coordinates in all four quadrants plot graphs of equations that correspond to straight line graphs in the coordinate plane use the form y = mx + c to identify parallel and perpendicular lines find the equation of the line through two given points or through one point with a given gradient
227878 A-Level Maths_Chapter 4.indd 102
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4.1
Writing the equation of a straight line in the form ax + by + c = 0
›
identify and interpret gradients and intercepts of linear functions graphically and algebraically
›
recognise, sketch and interpret graphs of linear functions.
You should be able to correctly complete the following questions. 1 Draw the graph of y = 2x + 3 for −3 ⩽ x ⩽ 3. What are the coordinates of the intercepts on the x and y axes?
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2 Here are the equations of six lines: B y = 2x + 3
E y=3
F 4x + 2y = 3
1 C y= − x +3 2
D y = 3 − 2x
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A y=2
a Write down the gradient and y intercept of each line.
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b Identify the lines that are parallel. c Identify the lines that are perpendicular. 3 Find the equation of the line through the points (1, 8) and (2, 0) by sketching the graph.
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4 Find the equation of the line with a gradient of 2 and passing through the point (3, 4) by sketching the graph.
4.1 Writing the equation of a straight line in the form ax + by + c = 0 Key InFoRMATIon The equation of a straight line may be written in the form ax + by + c = 0, where a, b and c are integers.
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You should be very familiar with writing the equation of a straight line in the form y = mx + c, for example y = 2x + 3. You also need to be able to write the equation of a straight line in the form ax + by + c = 0 where a, b and c are integers. To rewrite an equation that is currently in the form y = mx + c in the form ax + by + c = 0 you will need to rearrange the equation.
Write y = 2x + 3 in the form ax + by + c = 0, where a, b and c are integers.
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Solution
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Example 1
Subtract y from both sides of the equation. 0 = 2x + 3 − y Rearrange. 2x − y + 3 = 0
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Modelling
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a, b and c are integers, so this is the final answer.
Problem solving
PF
Proof
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Communicating mathematically 103
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4 Coordinate Geometry 1: Equations of Straight Lines
Stop and think
Compare the following example to the one above and work out what is the same and what is different. The equation is still in the form y = mx + c but this time the gradient is both negative and a fraction. How will this affect the method?
Example 2
Solution
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1 Write y = − x + 3 in the form ax + by + c = 0, where a, b and c are 2 integers.
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Add 1 x to both sides of the equation. 2 y + 1x =3 2
Subtract 3 from both sides of the equation. y + 1 x −3 = 0 2
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Multiply both sides of the equation by 2.
2y + x − 6 = 0
Reorder.
x + 2y − 6 = 0
Alternatively:
a, b and c are integers, so this is the final answer.
Multiply both sides of the equation by 2. 2y = − x + 6
Add x to both sides of the equation. 2y + x = 6
Subtract 6 from both sides of the equation.
2y + x − 6 = 0
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Stop and think As a mathematician, which format of the equation do you find most useful and why?
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4.2
Finding the equation of a straight line using the formula y – y1 = m(x – x1)
exercise 4.1A
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1 Which of these lines are not written in the form ax + by + c = 0, where a, b and c are integers. Give reasons for your answers.
2
a 2x + 3y + 4 = 0
b −x − 4y − 3 = 0
c 5x + 3y = 1
d 4x − 3y + 2 = 5
e x + y − 12 = 0
f
4x 3
− 53 y +
7 3
=0
Write these lines in the form ax + by + c = 0, where a, b and c are integers.
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Write down the values of a, b and c in each case.
a y = 4 + 5x
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b y = 3 − 2x
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3 Write these lines in the form ax + by + c = 0, where a, b and c are integers. Write down the values of a, b and c in each case.
a y = 13 x − 7
2 b y = −5 x + 6
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c y = 43 x + 72
Write 8x − 2y + 3 = 0 in the form y = mx + c.
Write down the gradient and the coordinates of the y intercept.
5
Review and correct this method to write the line y = 3 − 52 x in the form ax + by + c = 0, where a, b and c are integers. y = 2 − 52 x 2y = 2 − 5x
So 5x + 2y − 2 = 0
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5x + 2y = 2
3 5 Write down the values of a, b and c.
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6 Write y = x − 4 in the form ax + by + c = 0, where a, b and c are integers.
7 Work out the coordinates of the axes intercepts of the line −3x − 5y + 2 = 0.
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4.2 Finding the equation of a straight line using the formula y – y1 = m(x – x1) If you know the gradient m of a line and the coordinates (x1, y1) of a point on the line, then you can use the formula y − y1 = m ( x − x1 ) to work out the equation of the line.
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4 Coordinate Geometry 1: Equations of Straight Lines
In Kinematics in Mechanics if you know a point on a straight line distance−time graph (for example, at x hours the object will be y distance from the start) and you know the constant speed (the gradient on a distance−time graph), then you will be able to work out the equation that links the time and distance travelled for this part of the journey. This method is covered in Chapter 15.
Example 3
Solution
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Find the equation of the line with gradient 2 that goes through the point (3, 7). State the formula you are going to use.
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y − y1 = m ( x − x1 )
The values for substitution are m = 2 and (x1, y1) = (3, 7).
The value of m is the gradient stated in the question.
y − 7 = 2(x − 3) Expand the bracket.
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y − 7 = 2x − 6
Technology
Add 7 to both sides of the equation. y = 2x + 1
Example 4
You can also use a graphing software package to check the answer. Plot the graph of y = 2x + 1. Does the line have a gradient of 2? Does the line go through the point (3, 7)?
A line with a gradient of −1 and the line y = 2 meet the y axis at the same point.
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Find the equation of the line in the form ax + by + c = 0, where a, b and c are integers.
Solution State the formula you are going to use.
The values for substitution are m = −1 and (x1, y1) = (0, 2). y − 2 = −1(x − 0)
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y − y1 = m ( x − x1 )
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The line y = 2 meets the y axis at (0, 2).
Expand the bracket. y − 2 = −x Add x to both sides of the equation. x + y − 2 = 0
a, b and c are integers, so this is the final answer.
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4.3
Finding the gradient of the straight line between two points
exercise 4.2A 1
p xx
Find the equation of the line with the given gradient and passing through the given point.
a m = 2 and (x1, y1) = (3, 0) b m = 3 and (x1, y1) = (0, 3) c m = 2 and (x1, y1) = (3, 4) d m = −5 and (x1, y1) = (2, 3) Find the equation of the line with the given gradient and passing through the given point.
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2
a m = −4 and (x1, y1) = (−2, −5) A line with a gradient of −2 and the line y = 3 meet the y axis at the same point.
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3
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b m = −1 and (x1, y1) = (2, −2) Find the equation of the line in the form ax + by + c = 0.
4
A line with a gradient of 3 and the line x = −1 meet the x axis at the same point. Find the equation of the line in the form ax + by + c = 0.
5 The lines y = 2x + 4 and y = 7 − x intersect at the point P.
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Find the equation of the line with gradient 3 that passes through the point P. PS
6
A line with a gradient of 3 which passes through the point (1, 1) intersects with another line with a gradient −1 which passes through (4, 6). Work out the point of intersection of the two lines.
PS
7
A line with a gradient of 5 passes through the point (2, 3).
Does the line passing through the same point with a gradient half the size go through the origin?
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4.3 Finding the gradient of the straight line between two points
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Key InFoRMATIon
You can find the gradient m of the line joining two points with coordinates (x1, y1) and ( x 2, y 2) using the formula y −y m= 2 1. x 2 − x1
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In GCSE Mathematics, we know that to find the gradient of a line joining two points, we calculate rise divided by run. Another way of saying this is you divide the difference in the y coordinates by the difference in the x coordinates. In more formal terms, if you know or are given the coordinates of two points, (x1, y1) and y −y ( x 2, y 2), then using the formula m = 2 1 you can work out the x 2 − x1 gradient m of the line joining these points.
You need to remember the y −y formula m = 2 1 . x 2 − x1
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4 Coordinate Geometry 1: Equations of Straight Lines
y Technology You can also use a graphing software package to check the answer. Plot the graph of y = 2 − x (the equation in the form y = mx + c). Does the line have a gradient of −1? Does the line go through the point (0, 2)?
(x2, y2) 0
0
x
sa (x1, y1)
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m Example 5
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In Kinematics in Mechanics, if you know two points on a straight line speed−time graph (that is, if you know that at x hours the object will be travelling at y speed for two points) then you will be able to calculate the gradient between the two points. On a speed−time graph, the gradient is the acceleration of the object.
Work out the gradient of the line joining the points (−1, 2) and (3, 8).
Solution
State the formula you are going to use. y 2 − y1 x 2 − x1
Write down what you know. Let ( x1, y1) = (−1, 2) Substitute the values into the formula. m = 8−2 3 − −1
Do the calculations.
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Let ( x 2, y 2) = (3, 8)
You can also use graphing software to check the answer. Plot the points (−1, 2) and (3, 8). Then draw a line between the two points and determine the equation of the line. Is the gradient of the line (the coefficient of x) 32 or an equivalent value?
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m=
Technology
m = 64
Simplify.
m = 32
Alternatively:
m is a fraction in its simplest form.
State the formula you are going to use.
m=
y 2 − y1 x 2 − x1
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4.3
Finding the gradient of the straight line between two points
Write down what you know. Let ( x1, y1) = (3, 8) Let ( x 2, y 2) = (−1, 2) Substitute the values into the formula. m = 2−8 −1 − 3
Do the calculations.
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m = −6 4
Simplify.
m
m = 32
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m is a fraction in its simplest form.
Stop and think Compare the next example to the one above and work out what is the same and
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what is different. This time the gradient and the coordinates of one point are given. Some information is given about the second point. How will this affect the method?
Example 6
A line with a gradient of 2 passes through the point (5, 6). What are the coordinates of the x intercept of the line?
Solution m=2
x1 = ?, y1 = 0 (x intercept) x 2 = 5, y 2 = 6
2 = 6 −0 5−x
Using a graphing software package, investigate the location of the x-intercept as the gradient varies with both positive and negative and integer and fractional values.
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Substitute the values into the formula.
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Technology
State the formula you are going to use. y −y m= 2 1 x 2 − x1
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Write down what you know.
You might find if you make a quick sketch of this scenario it aids your understanding.
Multiply both sides of the equation by (5 − x).
2( 5 − x ) = 6
Expand the bracket. 10 − 2x = 6 109
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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES
Add 2x to both sides of the equation. 10 = 6 + 2x Subtract 6 from both sides of the equation. 4 = 2x Divide both sides of the equation by 2. 4 =x 2 Simplifying.
sa x=2
The coordinates of the x intercept are (2, 0).
1
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m exercise 4.3A
Find the gradient of the line passing through the given points.
b (5, 9) and (3, 3) c (1, −3) and (3, −9)
Find the gradient of the line passing through the given points.
a (8a, 5a) and (3a, 3a) b (a, a) and (3a, −5a) 3
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a (2, 3) and (7, 8)
2
p xx
Let ( x1, y1) = (1, 2)
2 So m = − 3
PS
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m = 1−5 8−2 m = −4 6
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Let ( x 2, y 2) = (5, 8)
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Review and correct this method to find the gradient of the line between the points (1, 2) and (5, −8). y −y m= 1 2 x1 − x 2
4 Which of the following pairs of points lie on a straight line with a gradient of −3? a (1, 1) and (4, −8) b (1, 3) and (4, 9) c (1, −7) and (4, −16) d (1, 7) and (3, 1)
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4.4
Finding the equation of a straight line using the formula y – y1 = x – x1 y 2 – y1 x 2 – x1
PS
5
Show that the points (2, 2), (5, 12 ) and (11, − 52 ) lie on a straight line.
6
A cyclist climbs to the top of a hill then descends the other side.
PF PS
The profile of his ascent and descent are plotted on a pair of axes.
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The cyclist starts the climb at (−10, 0) and reaches the summit at (0, 12 ). The cyclist then descends from the summit to finish at (2, 83 ).
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Which is steepest part of the ride − the ascent or the descent?
7
Find the gradient of the line passing through ( 12 , 13 ) and ( 34 , − 32 ).
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4.4 Finding the equation of a straight line using the formula
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If you know or are given the coordinates of two points, (x1, y1) and y − y1 x − x1 = (x 2, y 2), then using the formula you can work out y 2 − y1 x 2 − x1 the equation of the line joining these points.
(x2, y2) 0
0
x
You can find the equation of the line joining two points with coordinates (x1, y1) and (x 2, y 2) using the formula y – y1 x – x1 . = y 2 – y1 x 2 – x1 You need to remember the y – y1 x – x1 = . formula y 2 – y1 x 2 – x1
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(x1, y1)
Key InFORMATIOn
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y
y – y1 x – x1 = y 2 – y1 x 2 – x1
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If you have a straight line conversion graph between different units (for example, temperatures, currencies, units of measure), you can read off two pairs of coordinates to work out the equation of the line joining the points. This allows you to convert any values for the units to which the equation pertains. Proof Show that
y – y1 x – x1 = . y 2 – y1 x 2 – x1
State the formulae you are going to use. y − y1 = m ( x − x1 ) y −y m= 2 1 x 2 − x1 111
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4 Coordinate Geometry 1: Equations of Straight Lines
Replace m in the first equation by the second equation. y − y1 =
Rearrange.
y 2 − y1 ( x − x1 ) x 2 − x1
PROOF By direct proof you have y–y x–x shown that y – y1 = x – x1 . 2 1 2 1
y − y1 x − x1 = y 2 − y1 x 2 − x1
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Example 7
Find the equation of the line between the y intercept of y = 3x − 5 and the point (9, −8).
m
Write the equation of the line in the form ax + by + c = 0, where a, b and c are integers.
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Solution
You might find it aids your understanding to make a quick sketch of this scenario.
State the formula you are going to use.
Write down what you know. Let (x1, y1) = (0, − 5) Let (x 2, y 2) = (9, − 8)
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y − y1 x − x1 = y 2 − y1 x 2 − x1
Substitute the values into the formula. y − −5 x − 0 = −8 − −5 9 − 0
Simplify the left hand side. −3( y + 5) = x
Expand the bracket.
−3y − 15 = x
Add 3y and 15 to both sides of the equation.
Technology
You can also use a graphing software package to check the answer. Plot the points (0, −5) and (9, −8). Draw a line between the two points and determine the equation of the line.
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9( y + 5) =x −3
ed
Simplify the equation by manipulating the numeric values. y +5 x = −8 + 5 9 y +5 x = −3 9 Multiply both sides of the equation by 9.
0 = x + 3y + 15
Reorder.
x + 3y + 15 = 0
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4.4
Finding the equation of a straight line using the formula y – y1 = x – x1 y 2 – y1 x 2 – x1
exercise 4.4A 1
p xx
Find the equation of the line passing through the given points.
a (2, 3) and (7, 8) b (3, 3) and (5, 9) 2
Find the equation of the line passing through the given points.
a (1, −3) and (3, −9)
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b (−3, −4) and (−7, 6)
3
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m
Review and correct this method to find the equation of the line between the points (1, 2) and (5, −8). y − y1 x − x1 = y 2 − y1 x 2 − x1 Let ( x1, y1) = (1, 2)
Let ( x 2, y 2) = (5, −8)
y − 2 x −1 = 10 −4 4 ( y − 2) = 10 ( x − 1) 4y − 8 = 10x − 10 4y = 10x − 18 So y = 10x − 18 4
4
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y − 2 x −1 = 2 − −8 1 − 5
Find the equation of the line between the y intercept of y = 6 − 2x and the point (−1, −2).
ed
Write the equation of the line in the form ax + by + c = 0, where a, b and c are integers.
5
Find the equation of the line between the x intercept of 2x + y − 4 = 0 and the point (3, −7).
CM
6
Line A passes through (2, 7) and (5, 6).
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Line B passes through (5, −4) and (7, −6).
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Write the equation of the line in the form y = mx + c . What are the gradient and the coordinates of the y intercept of this line?
Find the equations of lines A and B in the form y = mx + c and state which line is steeper. Give a reason for your answer.
7
Find the equation of the line passing through the given points.
a ( 12 , 13 ) and ( 34 , − 32 ) b ( 72 ,− 15 ) and (− 13 , − 12 )
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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES
PS
8
Find the equation of the line between the y intercept of 2x − 5y + 7 = 0 and the x intercept of y = −3x + 5 in the form ax + by + c = 0, where a, b and c are integers.
PS
9
Line A passes through (3, 5) and (4, 9). Line B passes through (1, −3) and (5, −31). Find the equations of lines A and B and find the point of intersection of lines A and B.
PS
10 The lines y = x − 9 and y = 5 − x intersect at point A. The lines y = 7 − 3x and y = 2x + 8 intersect at point B.
sa
CM
Find the equation of the line between A and B.
m
Does this line pass through the point (1, 2)?
Justify your answer.
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4.5 Parallel and perpendicular lines
y
m1
m2 m3
m4
x
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0
ed
0
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Gradient conditions for parallel lines In your GCSE mathematics course you should have learned that if two straight lines are parallel then the gradient of each line will be the same. More formally, if you know the equations or gradients of two or more straight lines, and if the value of m in each case is the same, then the lines are parallel.
Key InFoRMATIon Line 1 has a gradient of m1. Line 2 has a gradient of m2 . Line 3 has a gradient of m3 . Line 4 has a gradient of m4 . If m1 = m2 = m3 = m4 then all the lines are parallel.
Two or more lines are parallel if their gradients, m, are the same. You need to learn the condition for parallel lines as it is NOT given in the formula booklet.
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4.5
Parallel and perpendicular lines
If you know that two straight lines are parallel then you can use this information to identify different shapes. For example, a quadrilateral with two pairs of parallel sides must be a square, rectangle, rhombus or parallelogram. If you also know that a pair of lines are perpendicular to one another, then the shape must be either a square or rectangle.
Example 8 Find the equation of the line that is parallel to 3x − y + 7 = 0 and passes through the point (1, 8).
sa
Technology Using a graphing software package, try finding the equations of lines passing through other points but that are parallel to the given equation.
Solution
e, pl
m
Write the equation of the line in the form ax + by + c = 0, where a, b and c are integers.
Rearrange the equation into the form y = mx + c . y = 3x + 7
m = 3 and (x1, y1) = (1, 8)
ct rre co un
State the formula you are going to use.
y − y1 = m ( x − x1 )
Substitute the values into the formula.
y − 8 = 3 ( x − 1)
Expand the bracket.
y − 8 = 3x − 3
Add 8 to both sides of the equation.
Subtract y from both sides of the equation. 0 = 3x − y + 5
3x − y + 5 = 0
The equation is not yet in the required format.
of
o pr
Rearrange.
ed
y = 3x + 5
Stop and think If you are asked to show that two lines, between two different pairs of coordinates, are parallel do you need to work out the equation of each line or is there a slightly simpler approach?
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4 Coordinate Geometry 1: Equations of Straight Lines
Example 9 Line A passes through (1, 5) and (3, 11). Line B passes through (4, 5) and (5, 8). Show that lines A and B are parallel.
Solution State the formula you are going to use. y −y m= 2 1 x 2 − x1
sa
m
Line A:
Let ( x1, y1) = (1, 5) .
e, pl
Let ( x 2, y 2) = (3, 11) .
It is easier (because there are fewer manipulations so it is less prone to error) to determine the gradient of the line joining two points than to find the equation of the line passing through two points.
Substitute the values into the formula.
m = 11 − 5 3 −1
m = 62
Simplify. m = 3 for line A. Line B: Let ( x1, y1) = (4, 5). Let ( x 2, y 2) = (5, 8).
ct rre co un
Do the calculations.
m = 3 for line B.
The gradients of lines A and B are the same therefore lines A and B are parallel.
of
o pr
You can also use a graphing software package to check the answer. Plot the points (1, 5) and (3, 11). Draw a line between the two points and determine the equation of the line. On the same graph, plot the points (4, 5) and (5, 8). Use the software to draw a line between the two points and determine the equation of the line. Do the lines look parallel? Is the value of m the same in both equations?
ed
Substitute the values into the formula. m = 8 −5 5−4 Simplify the equation by manipulating the numeric values.
Technology
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4.5
Parallel and perpendicular lines
exercise 4.5A
p xx
1 Work out whether the pairs of lines are parallel. a y = 2x − 6 and y = 2x + 9 b y = −3x − 5 and y = 5 − 3x 2
Line A passes through (1, 7) and (3, 11). Line B passes through (2, −3) and (5, 3).
sa
Show that lines A and B are parallel.
PS
3
m
Find the equation of the line that is parallel to line y = 7x − 2 and passes through the x intercept of line y = x − 3.
e, pl
4 Work out whether the pairs of lines are parallel. a 4x − y + 2 = 0 and −y = 4x + 3 b 6x − 2y + 2 = 0 and y = 3x − 3 5
Line A passes through (−1, 3) and (3, −1).
ct rre co un
Line B passes through (−2, 2) and (−3, 3). Show that lines A and B are parallel. PS
6
Find the equation of the line that is parallel to line −4x + y + 7 = 0 and passes through the y intercept of line 2x − y + 3 = 0.
7 Work out whether the pairs of lines are parallel. a 5x − 3y − 7 = 0 and 6y = 10x + 3
b 8x − 3y − 7 = 0 and 6y − 8x + 2 = 0 PF
8 The four sides of a rectangle have the equations 2x − y + 6 = 0 , 2x + 4y − 44 = 0, 2x − y − 4 = 0 and 2x + 4y − 24 = 0.
Show which pairs of lines form parallel sides.
ed
PS
of
o pr
Gradient conditions for perpendicular lines In your GCSE mathematics course you should have learned that if two straight lines are perpendicular (intersect at right angles), then the gradient of one line is the negative reciprocal of the other. For example, if the gradient of a line is 2 then the gradient of a line perpendicular to it will be − 12 . The product of a number and its negative reciprocal is −1.
− 12 is the negative reciprocal of 2 and vice versa.
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4 Coordinate Geometry 1: Equations of Straight Lines
More formally you can say:
›› a line with a gradient of m1 is perpendicular to a line with a
KEY INFORMATION
gradient of m2 when m2 = − 1 m1 ›› two straight lines with gradients m1 and m2 are perpendicular when m1m2 = −1.
y
sa m2
A line with a gradient of m1 is perpendicular to a line with a gradient of m2 when m2 = − 1 . m1
(1, m1)
Two straight lines with gradients m1 and m2 are perpendicular when m1m2 = −1 .
2 2 1 + m1
You need to learn the conditions for perpendicular lines as they are NOT given in the formula booklet.
x
e, pl
m
m1 – m2
m1
Line 1 has a gradient of m1 . Line 2 has a gradient of m2.
ct rre co un
2 2 1 + m2
(1, m2)
PROOF Using Pythagoras’ theorem: 12 + m12 + 12 + m22 = (m1 − m2 )
2 + m12 + m22 = m12 − 2m1m2 + m22
If m1m2 = −1 , then the lines are perpendicular.
Find the equation of the line that is perpendicular to 4x − y − 9 = 0 and passes through the point (3, 5).
2 = −2m1m2
y = 4x − 9
m1 = 4
of
Rearrange the equation into the form y = mx + c .
o pr
Write the equation of the line in the form ax + by + c = 0, where a, b and c are integers.
Solution
− 1 = −m1 m2
ed
Example 10
2
Find the gradient of the line perpendicular to it.
m1m2 = −1
4m2 = −1
m2 = − 14
( x1, y1) = (3, 5) 118
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4.5
Parallel and perpendicular lines
State the formula you are going to use. y − y1 = m ( x − x1 ) Substitute the values into the formula. y − 5 = − 14 ( x − 3) Multiply both sides of the equation by −4. −4( y − 5) = x − 3
sa
Expand the bracket. −4y + 20 = x − 3
m
Add 4y and subtract 20 from both sides of the equation. x + 4y − 23 = 0
e, pl
Example 11
Solution
ct rre co un
Show that the lines 2x + y − 5 = 0 and x − 2y + 4 = 0 are perpendicular. Rearrange the first equation into the form y = mx + c . y − 5 = −2x y = −2x + 5 m1 = −2
Rearrange the second equation into the form y = mx + c . 2y = x + 4
m2 = 12
ed
y = 12 x + 2
( 12 ) = −1
m1m2 = ( −2)
of
So the lines 2x + y − 5 = 0 and x − 2y + 4 = 0 are perpendicular.
Using a graphing software package, try plotting lines that intersect but are not perpendicular to each other. What do you notice about the products of their gradients?
o pr
Show that m1m2 = −1
Technology
exercise 4.5B 1 Work out whether the pairs of lines are perpendicular. a y = 2x + 6 and y = 12 x + 9 b y = −3x − 5 and y = 13 x + 5 119
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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES
2
Review and correct this method to find the equation of the line that is perpendicular to y = − 12 x + 3 and passes through the point (1, 0). m = − 12 Let (x1, y1) = (1, 0) y − y1 = m ( x − x1 ) y − 1 = 12 x − 0
CM
3
sa
So y = 12 x − 1
Are the following pairs of lines perpendicular? You must provide justification for your reasons.
m
a 3x − y + 2 = 0 and −y = 13 x + 3 PS
4
e, pl
b 12x − 4y + 4 = 0 and y = 12x − 12
Line A passes through (1, 1) and (3, 9). Line B passes through (4, 0) and (8, −1). Show that lines A and B are perpendicular.
ct rre co un
PS
5 The four sides of a rectangle have the equations 2x − y + 6 = 0, 2x + 4y − 44 = 0, 2x − y − 4 = 0 and 2x + 4y − 24 = 0.
M
Show which lines are perpendicular. PS
6
Find the equation of the line that is perpendicular to line 8x − 2y − 7 = 0 and passes through the y intercept of the line 4x − 2y + 5 = 0.
CM
7
In each of the following cases, explain the mathematical manipulation steps required to show whether or not each pair of lines are perpendicular.
a 5x − 3y − 7 = 0 and −10y = 6x + 3 PS
8
ed
b 9x − 10y − 7 = 0 and 9y − 10x + 2 = 0
A quadrilateral is drawn with vertices at (0, 6), (2, 10), (4, 4) and (6, 8).
rhombus
kite
parallelogram
square
trapezium
rectangle
You need to be able to apply everything that you have learned about straight line models in this chapter in a variety of different contexts including real-world scenarios.
of
4.6 Straight line models
o pr
By considering gradients, work out which of the following the quadrilateral could possibly be and state your reasons.
CM
Key InFoRMATIon
You need to be able to use straight line models in a variety of contexts.
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4.6
Straight line models
Example 12 On an old mercury thermometer, a temperature of 0 °C reads as 32 °F. Similarly, a temperature of 20 °C reads as 68 °F. The relationship between degrees Celsius (°C) and degrees Fahrenheit (°F) can be modelled as a straight line. If the temperature in degrees Celsius is taken as x and the temperature in degrees Fahrenheit as y, using y − y1 x − x1 = , y 2 − y1 x 2 − x1
sa
Solution
e, pl
m
find the equation of the line in the form ax + by + c = 0, where a, b and c are integers.
State the formula you are going to use. y − y1 x − x1 = y 2 − y1 x 2 − x1
Let ( x1, y1) = (0, 32) Let ( x 2, y 2) = (20, 68)
ct rre co un
Write down what you know.
Substitute the values into the formula. y − 32 = x −0 68 − 32 20 − 0
Simplify the equation by manipulating the numeric values. y − 32 x = 36 20
y − 32 = 36x 20
y − 32 = 9x 5 Multiply both sides of the equation by 5.
Expand the bracket.
5y − 160 = 9x
Subtract 5y and add 160 to both sides of the equation. 0 = 9x − 5y + 160
Technology
You can also use a graphing software package to check the answer. Plot the points (0, 32) and (20, 68). Draw a line between the two points and determine the equation of the line.
of
5 ( y − 32) = 9x
o pr
Simplify.
ed
Multiply both sides of the equation by 36.
Rearrange.
9x − 5y + 160 = 0
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4 Coordinate Geometry 1: Equations of Straight Lines
Example 13 Anna has a very old mobile phone and is not very pleased with the service that she is receiving from her current mobile phone provider. The number of different phones and tariffs available are bewildering but she manages to narrow her choice down to three options.
sa
She represents the three different tariffs on a graph. 300
150 100
0
1
2
3
4
Option B
ct rre co un
50 0
Option C
e, pl
200 Cost (£)
Option A
m
250
5 6 7 Time (months)
8
9
10
11
12
The relationship between time (months) and cost (£) can be modelled as a straight line.
By finding the equations for options A and C, work out at what point in time option C becomes cheaper than option A.
Solution
ed
You are being asked to find the point of intersection, specifically the x coordinate of the point of intersection, of two straight lines.
You need to break down the problem into mathematical steps. It might be easier to work backwards.
of
›› State the x coordinate of the point of intersection. ›› Find the point of intersection of line A and line C. ›› Find the equation of line C. ›› Find the equation of line A. ›› Find two pairs of coordinates on line C. ›› Find two pairs of coordinates on line A.
o pr
How are you going to do this?
So you need to work through this list of steps from the bottom to the top.
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4.6
Straight line models
State the formula you are going to use. y − y1 x − x1 = y 2 − y1 x 2 − x1
For line A write down what you know. Let ( x1, y1) = (0, 0) Let ( x 2, y 2) = (8, 200)
sa
Substitute the values into the formula. y −0 = x −0 200 − 0 8 − 0
m
Simplify the equation by manipulating the numeric values. y =x 200 8
e, pl
Multiply both sides of the equation by 200. y = 25x
For line C write down what you know. Let ( x 2, y 2) = (7, 150)
ct rre co un
Let ( x1, y1) = (0, 100)
Substitute the values into the formula. y − 100 = x −0 150 − 100 7 − 0 y − 100 x = 50 7
y − 100 = 50x 7
y=
+ 100
ed
50 x 7
Make the equations of the two straight lines equal to each other and solve this equation to find the x coordinate of the point of intersection. 25x = 50 x + 100 7
Multiply both sides of the equation by 7.
175x = 50x + 700
125x = 700
of
Subtract 50x from both sides of the equation.
o pr
Divide both sides of the equation by 125. x = 5.6 Option C becomes cheaper than option A 5.6 months after the start of both contracts.
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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES
exercise 4.6A M
1
p xx
At a post office a holiday maker who is travelling to the USA exchanges £500 for $800. Another holiday maker exchanges £100 for $160. The relationship between pounds (£) and US dollars ($) can be modelled as a straight line. If the number of pounds is taken as x and the number of dollars as y, using y − y1 x − x1 = y 2 − y1 x 2 − x1
sa
find the equation of the line in the form y = mx + c.
PS
2
m
On a distance−time graph the equations for two different sections of the journey are y = 2x + 6, 0 ⩽ x ⩽ 1 and y = −2x + 12, 4 ⩽ x ⩽ 5 where x is time in hours and y is distance travelled.
M
e, pl
Decide which of the following statements are true and which are false. Give a reason for each of your decisions.
a The lines for the two equations are perpendicular to each other.
ct rre co un
b The two lines cross the y axis at the same point.
c The line for y = −2x + 12 is travelling away from the start. d The speed is the same in the two sections of the journey.
e The lines for the two equations are parallel to each other. f The line for y = 2x + 6 is travelling back to the start. M
3
At the start of Year 12, two students, Anna and Bhavini, buy new printers.
Anna pays £50 for her new printer and estimates that within five years her printer will have cost £950 to buy and run.
CM
ed
Bhavini pays £67.50 for her new printer and estimates that within five years her printer will have cost £667.50 to buy and run.
For each printer, interpret the values of m and c.
o pr
Find the equation of the costs of each printer in the form y = mx + c where x is years and y is the overall cost.
If they both use four ink cartridges per year, what is the cost of one ink cartridge in each case?
of
What is significant about the x intercept in each case?
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4.6
Straight line models
CM
4
M
A group of friends are going to a party and need to hire a taxi to take them there. They call three different taxi firms find the taxi firm with the cheapest fare. taxi firm A
No callout charge. Charge per mile. Example charge for 10 miles would be £50.
taxi firm B
Callout charge of £10. Additionally, charge per mile.
sa
taxi firm C
Example charge for 10 miles would be £40. Callout charge of £15. Additionally, charge per mile.
m
Example charge for 10 miles would be £35.
e, pl
The relationship between miles travelled and overall cost can be modelled as a straight line. If x is the number of miles travelled and y is the cost, state, for each taxi firm, two pairs of x and y values.
PS M
5
ct rre co un
y − y1 x − x1 = , find the equation of the line for each taxi firm in the form y = mx + c. y 2 − y1 x 2 − x1 Hence work out which taxi firm is cheapest over 4 miles and 7 miles. Using
A litre is equivalent to 1.75 pints and 1 pint is equivalent to 570 millilitres. The relationship between pints and litres can be modelled as a straight line. If the number of litres is taken as x and the number of pints as y, find the equation of the line in the form y = mx + c.
ed of
o pr 125
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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES
Summary of key points The equation of a straight line may be written in the form ax + by + c = 0 where a, b and c are integers.
›
You can find the equation of a line using the formula y − y1 = m ( x − x1 ) where m is the gradient of the line and ( x1, y1) is a point on the line. (You need to learn this.)
›
You can find the gradient m of the line joining two points with coordinates (x1, y1) and ( x 2, y 2) using y −y the formula m = 2 1 . x 2 − x1 You can find the equation of the line joining two points with coordinates (x1, y1) and ( x 2, y 2) using y − y1 x − x1 the formula . = y 2 − y1 x 2 − x1 Two or more lines are parallel if their gradients, m, are the same. (You need to learn this.) A line with a gradient of m1 is perpendicular to a line with a gradient of m2 when m2 = − 1 . m1 Two straight lines with gradients m1 and m2 are perpendicular when m1m2 = −1. (You need to learn this.)
›
e, pl
m
› › ›
sa
›
1
y 2 − y1 , find the gradients of lines A and B. x 2 − x1 b Compare the gradients of lines A and B and decide which line is steeper. Give a reason for your answer.
ed
4 The graph shows the journeys of two different runners.
[2 marks] [4 marks]
o pr
Runner A lives 3 miles away from runner B.
M
[4 marks]
Line A is perpendicular to x − y − 8 = 0 and passes through the y intercept of y = 9x + 5. Find the equation of line A.
CM
[3 marks]
a Line A passes through (0, 7) and (2, 3) and line B passes through (6, 4) and (8, 5). Using m =
3
p xx
Line A is parallel to y = 3x − 5 and passes through the point (1, 6). Find the equation of line A.
2
ct rre co un
Practice questions
They both go out for a run one morning and meet 3 hours later 9 miles away from runner B’s home.
Distance (miles)
Runner B
16
Runner A
12
[1 mark]
b If x is time in hours and y is distance in miles, for each runner, find two pairs of x and y values.
y − y1 x − x1 = find the equation of the line y 2 − y1 x 2 − x1 for each runner in the form y = mx + c. [6 marks] Using
8 4 0
runner A have run?
of
a When the runners meet how far will
20
c Using the values of m from part b, decide which 0
1
2 3 4 Time (hours)
5
6
runner runs faster. Give a reason for your answer.
[2 marks]
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4
Key points and Practice questions
PS
5
M
Line A is perpendicular to y = 2x and passes through the point (12, 4).
a Find the equation of line A.
[3 marks]
b Find the coordinates of the axes intercepts.
[2 marks]
c Find the area of the triangle with vertices at the origin, the x intercept of line A and [2 marks]
the y intercept of line A. PS
6 The points A(4, 5), B(−2, 8) and C(−10, −8) are the vertices of a triangle.
PF
[5 marks]
Show that ABC is a right-angled triangle.
sa
PS
7
M
At a fun fair you pay an entrance fee of £5 and then can either pay £2 a time to go on a ride or buy a book of tickets for six rides costing £10.
e, pl
m
The relationship between number of rides and overall cost (including the entrance fee) can be modelled as a straight line.
ct rre co un
If x is number of rides and y is the cost, state, for the first payment option, two pairs of x and y values. y − y1 x − x1 = Using , find the equation of the line for the first option in the form y 2 − y1 x 2 − x1 y = mx + c. Hence work out which payment option is cheaper for 4 rides and for 17 rides. PS
8
Line A has a gradient of
1 2
[10 marks]
and passes through the points (8, 5) and (−12, a).
Find the equation of line B which is perpendicular to line A and passes through (0, a). PS
9
Find the equation of the line, in the form y = mx + c, that is perpendicular to the
(
line which passes through the points 2 2, at (0, 3).
)
2 and
(
)
2, 2 3 and meets the y axis [5 marks]
10 The lines x + 4y − 8 = 0 and 3x + 5y + 15 = 0 intersect at point A.
ed
PS
[5 marks]
Find the equation of the line that passes through the point of intersection and is perpendicular to 3x + 5y + 15 = 0 .
11 A straight line A passes through the points (0, 32) and (5, 41).
CM
A straight line B passes through the points (32, 0) and (41, 5).
a Find the equations of lines A and B in the form y = mx + c. b Find the point of intersection of lines A and B.
of
What do you notice?
o pr
PS
[7 marks]
[4 marks] [3 marks]
c On the same pair of axes sketch the graphs of lines A and B including the point of intersection. What transformation links lines A and B?
[3 marks]
d If you are told x in the equation for line A is degrees Celsius and y is degrees Fahrenheit and vice versa for Line B what can you deduce about the point of intersection?
[1 mark]
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e, pl m sa
ed ct rre co un
of o pr
4
Worked solutions
4 Coordinate Geometry 1: Equations of Straight Lines Exercise 4.1A
p xx
1 The lines not in the form ax + by + c = 0, where a, b and c are integers are: c because it does not equal zero
sa
d because it does not equal zero e because c is not an integer f because a, b and c are not integers.
m
2 The lines in the form ax + by + c = 0, where a, b and c are integers are:
e, pl
a Rearrange y = 4 + 5x.
5x − y + 4 = 0 where a = 5, b = −1 and c = 4
b Rearrange y = 3 − 2x.
2x + y − 3 = 0 where a = 2, b = 1 and c = −3
5y = 3x − 12 3x − 5y − 12 = 0 where a = 3, b = −5 and c = −12 7 When the line intercepts the x axis, y = 0. Substituting y = 0 −3x + 2 = 0 2 = 3x x = 32 So the coordinates of the x intercept are
a Multiply by 3: 3y = x − 21 , and rearrange.
x − 3y − 21 = 0 where a = 1, b = −3 and c = −21
Substituting x = 0 −5y + 2 = 0 2 = 5y y = 52
( 52 ).
So the coordinates of the x intercept are 0,
Exercise 4.2A
y − y1 = m ( x − x1 ) y − 0 = 2( x − 3)
2x + 5y − 30 = 0 where a = 2, b = 5 and c = −30
c Multiply by 3: 3y = 4x + 21 2
Multiply by 2: 6y = 8x + 21, and rearrange.
8x − 6y + 21 = 0 where a = 8, b = −6 and c = 21
b m = 3 and (x1, y 1) = (0, 3) y − y1 = m ( x − x1 ) y − 3 = 3( x − 0 )
4 Rearrange: 8x + 3 = 2y Rearrange and divide by 2: y = 4x + 32
c m = 2 and (x1, y 1) = (3, 4) y − y1 = m ( x − x1 )
5 Equation written down incorrectly. It should be y = 3 − 52 x
Although the values are wrong this step is actually correct. With the correct values it should be 5x + 2y = 6 . Although the values are wrong this step is actually correct. With the correct values it should be 5x + 2y − 6 = 0.
y − 4 = 2( x − 3)
y − 4 = 2x − 6 y = 2x − 2
of
Not all expressions in the equation have been multiplied by 2. It should be 2y = 6 − 5x
y − 3 = 3x y = 3x + 3
o pr
gradient = 4 and y intercept = 32 ; coordinates (0, 32 )
y = 2x − 6
ed
p xx
1 a m = 2 and (x1, y1) = (3, 0)
b Multiply by 5: 5y = −2x + 30, and rearrange.
( 32 , 0).
When the line intercepts the y axis, x = 0.
ct rre co un
3 The lines in the form ax + by + c = 0, where a, b and c are integers are:
6 The line in the form ax + by + c = 0, where a, b and c are integers is: 5y = x−4 3 5y = 3( x − 4 )
d m = −5 and (x1, y 1) = (2, 3) y − y1 = m ( x − x1 ) y − 3 = −5( x − 2) y − 3 = −5x + 10
y = −5x + 13
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4 Coordinate Geometry 1: Equations of Straight Lines
Find the point of intersection. 3x − 2 = −x + 10 4x = 12 x=3 y = −3 + 10 = 7
2 a m = −4 and (x1, y 1) = (−2, −5) y − y1 = m ( x − x1 )
y − −5 = −4( x − −2) y + 5 = −4x − 8 y = −4x − 13
Point of intersection is (3, 7)
b m = −1 and (x1, y 1) = (2, −2) y − y1 = m ( x − x1 )
7 m = 52 and ( x1, y 1) = (2, 3)
y − −2 = −1( x − 2)
sa
y − y 1 = m ( x − x1 )
y + 2 = −x + 2 y = −x
y −3 =
m
y − y1 = m ( x − x1 )
y −3 =
e, pl
y − 3 = −2( x − 0 )
4 m = 3 and (x1, y 1) = (−1, 0) y − 0 = 3 ( x − −1)
y = 3x + 3 3x − y + 3 = 0
5 First find the point of intersection. 2x + 4 = 7 − x 3x = 3 x =1 y = 2+4 = 6 y − y1 = m ( x − x1 )
y − 3 = −5 y = −2 The line does not go through the origin.
Exercise 4.3A 1 a m =
m =1 y −y b m = 2 1 x 2 − x1
(x1, y 1) = (5, 9)
y − y1 = m ( x − x1 )
m=3
y 2 − y1 x 2 − x1
(x1, y 1) = (1, −3)
(x 2, y 2) = (3, −9)
m = −9 − −3 3 −1 −6 m= 2
m = −3
of
Find the equation of the second line.
m = 9−3 5−3 m = 62
c m =
y − 1 = 3( x − 1) y − 1 = 3x − 3 y = 3x − 2
(x 2, y 2) = (3, 3)
o pr
y − y1 = m ( x − x1 )
(x 2, y 2) = (7, 8)
y − 6 = 3x − 3 y = 3x + 3 m = 3 and (x1, y 1) = (1, 1)
y 2 − y1 x 2 − x1
m = 8 −3 7 −2 m = 55
6 Find the equation of the first line.
p xx
(x1, y 1) = (2, 3)
y − 6 = 3( x − 1)
m = −1 and (x1, y 1) = (4, 6)
( 0 − 2) ( −2 )
ed
m = 3 and (x1, y 1) = (1, 6)
y −3 =
5 2 5 2
ct rre co un
y − y1 = m ( x − x1 )
( x − 2)
When you substitute in x = 0, y will equal 0 if the line goes through the origin.
3 m = −2 and (x1, y 1) = (0, 3)
y − 3 = −2x 2x + y − 3 = 0
5 2
y − 6 = −1( x − 4 ) y − 6 = −x + 4 y = −x + 10
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4
Worked solutions
y 2 − y1 x 2 − x1 (x1, y1) = (8a, 5a)
2 a m =
c m =
(x1, y1) = (1, −7)
(x 2, y 2) = (3a, 3a)
m = 5a − 3a 8a − 3a 2a m = 5a m = 25 y −y b m = 2 1 x 2 − x1
sa
y 2 − y1 x 2 − x1 y coordinate incorrect in (x2, y2). It should be (5, −8).
m = −8 − 2 5 −1 m = −10 4 m = −5 2
(x 2, y 2) = (4, − 8) m = −8 − 1 4 −1 m = −9 3 m = −3 Lies on a straight line with a gradient of −3. y 2 − y1 x 2 − x1
(x1, y1) = (1, 3)
y 2 − y1 x 2 − x1
(x 2, y 2) = (3, 1)
m = 1− 7 3 −1
m = −6 2
m = −3
Lies on a straight line with a gradient of −3. y 2 − y1 x 2 − x1 (x1, y 1) = ( 5, 12 )
5 m =
(x 2, y 2) = (2, 2) m=
(x 2, y 2) = (4, 9) m = 9−3 4 −1 m = 63
m=2 Does not lie on a straight line with a gradient of −3.
2 − 12 2−5 3 2
−3 m = − 12 m=
y 2 − y1 x 2 − x1
(
(x1, y 1) = 11, − 52
)
(x 2, y 2) = (2, 2) m= m=
2 − − 52 2 − 11 9 2
−9 m = − 12
of
b m =
m = −3 Lies on a straight line with a gradient of −3.
o pr
ed
y 2 − y1 x 2 − x1 (x1, y 1) = (1, 1)
m = −9 3
m=
4 a m =
ct rre co un
Numerator and denominator confused. Should be:
m = −16 − −7 4 −1
(x1, y1) = (1, 7)
m = −5a − a 3a − a −6 m= a 2a m = −3
3 Should be m =
(x 2, y 2) = (3a, −5a)
e, pl
(x 2, y 2) = (4, −16)
d m =
m
(x1, y1) = (a, a)
y 2 − y1 x 2 − x1
T he gradient, m, between the two different pairs of coordinates is the same therefore we can conclude that all the points lie on a straight line. 6 Ascent: y −y m= 2 1 x 2 − x1 (x1, y 1) = (−10, 0) (x 2, y 2) = (0, 1 ) 2
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4 Coordinate Geometry 1: Equations of Straight Lines
m= m= m=
2 a (x1, y 1) = (1, −3)
1 −0 2 0 − −10
(x 2, y 2) = (3, −9)
1 2
y − y1 x − x1 = y 2 − y1 x 2 − x1
10 1 20
y − −3 x − 1 = −9 − −3 3 − 1 y + 3 x −1 = −6 2 y + 3 = −3( x − 1)
Descent: m=
y 2 − y1 x 2 − x1
sa
(x1, y 1) = (0, 12 ) (x 2, y 2) = (2, − 12 m= 2−0 − 18 m= 2 1 m = − 16
3) 8
3 8
b (x1, y 1) = (−3, −4)
m
(x 2, y 2) = (−7, 6) y − y1 x − x1 = y 2 − y 1 x 2 − x1
e, pl
The descent is the steepest because 7 m =
y 2 − y1 x 2 − x1
( 12 , 13 ) (x 2, y 2) = ( 34 ,− 23 )
1 16
>
y − −4 x − −3 = 6 − −4 −7 − − 3 y +4 x +3 = −4 10 y + 4 = − 52 ( x + 3)
1 . 20
ct rre co un
(x1, y 1) =
y + 4 = − 52 x − 15 2
−2 − 1 m = 33 13 −2 4
3
m = −1 1
Exercise 4.4A
y1 and y2 and x1 and x2 confused. The next line should be:
p xx
y − 2 x −1 = −8 − 2 5 − 1 With above correction the remainder of the calculation should be:
ed
1 a (x1, y 1) = (2, 3) (x 2, y 2) = (7, 8)
y − 2 x −1 = −10 4 4( y − 2) = −10( x − 1) 4y − 8 = −10x + 10 4y = −10 0x + 18 − 5 So y = x + 9 2
b (x1, y 1) = (3, 3)
4
(x 2, y 2) = (5, 9)
y = 6 − 2x
of
y −3 x −2 = 8−3 7−2 y −3 x −2 = 5 5 y −3= x −2 y = x +1
o pr
y − y1 x − x1 = y 2 − y1 x 2 − x1
y intercept when x = 0 so when y = 6
y − y1 x − x1 = y 2 − y 1 x 2 − x1 y −3 x −3 = 9 −3 5 −3 y −3 x −3 = 6 2 y − 3 = 3( x − 3)
y − y1 x − x1 = y 2 − y1 x 2 − x1
Let (x 2, y 2) = (5, −8)
m = −4
502
y = − 25 x − 23 2
Let (x1, y 1) = (1, 2)
4
y + 3 = −3x + 3 y = −3x
(x1, y 1) = (0, 6) (x 2, y 2) = (−1, −2)
y − 3 = 3x − 9 y = 3x − 6
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4
Worked solutions
Line B is steeper because 1 > 13 . (The negatives can be ignored because they just indicate the direction of the line.)
y − y1 x − x1 = y 2 − y 1 x 2 − x1 y −6 = x −0 −2 − 6 −1 − 0 y −6 = −x −8 y − 6 = 8x 8x − y + 6 = 0
y − y1 x − x1 = y 2 − y 1 x 2 − x1 y − 13 x− 1 = 3 12 2 1 −3−3 −2 4
5 2x + y − 4 = 0
sa
x intercept when y = 0
y − 13 x − 12 = 1 −1 4
2x − 4 = 0 so x = 2
(x1, y 1) = (2, 0)
m
(
y − 13 = −4 x − 12
(x 2, y 2) = (3, −7)
y − = −4x + 2
e, pl
(x1, y 1) = (2, 7) (x 2, y 2) = (5, 6) y − y1 x − x1 = y 2 − y1 x 2 − x1
y − y1 x − x1 = y 2 − y 1 x 2 − x1
y − − 15 x − 72 = − 12 − − 15 − 13 − 72
y + 15 x−2 = 7 76 5 2 − 10 + 10 − 21 − 21 y + 15 x − 72 = 13 3 − 10 − 21 y + 15 21 x − 2 = − 13 3 7 − 10
( ) 63 x − 2 y + 15 = 130 ( 7) 63 x − 9 y + 15 = 130 65
8
63 x − 22 y = 130 65
2x − 5y + 7 = 0
o pr
m = − 13
( 72 , − 15 ) (x 2, y 2) = ( − 13 , − 12 )
b (x1, y 1) =
ed
y −7 x −2 = 6 −7 5−2 y −7 x −2 = −1 3 2 x − y −7 = 3 − 13 x y = 23 3
y = −4x + 73
ct rre co un
6 Line A
)
1 3
y − y1 x − x1 = y 2 − y1 x 2 − x1
y −0 x −2 = −7 − 0 3 − 2 y −0 x −2 = −7 1 y = −7(x − 2) y = −7x + 14 7x + y − 14 = 0
( 12 , 13 ) (x 2, y 2) = ( 34 , − 23 )
7 a (x1, y 1) =
y intercept when x = 0
Line B
−5y + 7 = 0 so y =
(x1, y 1) = (5, − 4)
y = −3x + 5
(x 2, y 2) = (7, −6)
x intercept when y = 0
y − y1 x − x1 = y 2 − y 1 x 2 − x1
−3x + 5 = 0 so x = (x 2, y 2) =
( ) ( 53 , 0)
5 3
of
y − −4 x − 5 = −6 − −4 7 − 5 y + 4 x −5 = −2 2 y + 4 = −x + 5 y =1−x m = −1
(x1, y 1) = 0,
7 5
7 5
y − y1 x − x1 = y 2 − y1 x 2 − x1 y − 75 x − 0 = 0 − 75 53 − 0
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4 Coordinate Geometry 1: Equations of Straight Lines
y − 75 3 = 5x − 75 y −
7 5
(x1, y1) = (7, −2)
(
(x 2, y 2) = − 15 , 38 5
( )
= − 75 35 x
y − y1 x − x1 = y 2 − y 1 x 2 − x1
21 x y − 75 = − 25
y − −2 = x −7 − −2 − 15 − 7
21 x + 7 y = − 25 5
38 5
9 Line A
y +2
(x1, y 1) = (3, 5)
48 5
sa
y − y1 x − x1 = y 2 − y1 x 2 − x1
y +2= y +2= y=
y − 5 = 4x − 12 y = 4x − 7
e, pl
m
y −5 x −3 9−5 = 4−3 y −5 x −3 4 = 1 y − 5 = 4( x − 3 )
Line B
= x −367 −5
− 36 y + 2) = 5(
(x 2, y 2) = (4, 9)
( (
) )
48 x − 7 5 − 43 x − 7 − 43 x + 28 3 − 43 x + 22 3
When x = 1, y = 6 so this line does not pass through the point (1, 2).
Exercise 4.5A
p xx
1 a m = 2 in both equations so the lines are parallel.
(x 2, y 2) = (5, −31) y − y1 x − x1 = y 2 − y1 x 2 − x1
At the point of intersection: 4x − 7 = 4 − 7x
2 Line A (x1, y1) = (1, 7)
(x 2, y 2) = (3, 11)
(x 2, y 2) = (5, 3) m=
y 2 − y1 x 2 − x1
3 − −3 m = 5−2 m=2 m = 2 in both equations so the lines are parallel. 3 From y = 7x − 2, m = 7
10 Point A x −9 = 5− x 2x = 14 x = 7, y = −2
of
Substitute into either equation and y = −3. So point of intersection is (1, −3).
m = 11 − 7 3 −1 m=2
o pr
x =1
y 2 − y1 x 2 − x1
Line B (x1, y1) = (2, −3)
11x = 11
m=
ed
y − −3 = x −1 −31 − −3 5 − 1 y + 3 x −1 = 4 −28 y + 3 = −7( x − 1) y + 3 = −7x + 7 y = 4 − 7x
ct rre co un
b m = −3 in both equations so the lines are parallel.
(x1, y1) = (1, −3)
The x intercept of y = x − 3 is when y = 0 so is at (3, 0). (x1, y 1) = (3, 0)
y − y1 = m ( x − x1 )
Point B 7 − 3x = 2x + 8 5x = −1 x = − 15 , y =
)
y − 0 = 7( x − 3 )
38 5
y = 7x − 21
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4
Worked solutions
4 a Arrange the equations in the form y = mx + c: y = 4x + 2, y = −4x − 3
m is not equal so the lines are not parallel.
b Arrange the equations in the form y = mx + c: y = 3x + 1, y = 3x − 3
m = 3 in both equations so the lines are parallel.
5 Line A
sa (x 2, y 2) = (3, −1) y 2 − y1 x 2 − x1
e, pl
−1 − 3 m = 3 − −1 m = −1 Line B
(x1, y 1) = (−2, 2)
m=
y 2 − y1 x 2 − x1
3−2 m = −3 − −2 m = −1 m = −1 in both equations so the lines are parallel. 6 From −4x + y + 7 = 0, y = 4x − 7, so m = 4.
The y intercept of 2x − y + 3 = 0 is when x = 0 so is at (0, 3). (x1, y1) = (0, 3)
y − 3 = 4( x − 0 )
Let (x1, y 1) = (1, 0)
y − y1 = m ( x − x1 )
(x1, y 1) incorrectly substituted. Correcting this and using correct value of m2 gives: y − 0 = 2x − 2 So y = 2x − 2
m is not equal so the lines are not parallel.
8 Arrange the equations in the form y = mx + c: 2x − y + 6 = 0, y = 2x + 6, m = 2 2x + 4y − 44 = 0, 4y = 44 − 2x, y = 11 − 12 x, m = − 12 2x − y − 4 = 0, y = 2x − 4, m = 2 2x + 4y − 24 = 0, 4y = 24 − 2x, y = 6 − 12 x, m = − 12
b m1 = 3 and m2 = 12 so m1m2 ≠ −1 so the lines are not perpendicular.
4 Line A
(x1, y 1) = (1,1)
(x 2, y 2) = (3, 9)
m=
y 2 − y1 x 2 − x1
9 −1 m = 3 −1 m=4 Line B
(x1, y1) = (4, 0)
(x 2, y 2) = (8, −1)
m=
y 2 − y1 x 2 − x1
−1 − 0 m = 8−4 m = − 14
of
b Arrange the equations in the form y = mx + c: y = 83 x − 7, y = 43 x − 13
3 a m1 = 3 and m2 = − 13 so m1m2 ≠ −1 so the lines are perpendicular.
o pr
y = 4x + 3 7 a Arrange the equations in the form y = mx + c: y = 53 x − 73 , y = 53 x + 12 m = 53 in both equations so the lines are parallel.
1 a m1 = 2 and m2 = 12 so m1m2 ≠ −1 so the lines are not perpendicular.
ed
y − y1 = m ( x − x1 )
p xx
ct rre co un
(x 2, y 2) = (−3, 3)
Exercise 4.5B
2 m1 = − 12 for the given line but the gradient of a line perpendicular to this will be m2 = 2.
m
m=
And 2x + 4y − 44 = 0 and 2x + 4y − 24 = 0 are parallel lines as m = − 12 for both lines.
b m1 = −3 and m2 = 13 so m1m2 = −1 so the lines are perpendicular.
(x1, y 1) = (−1, 3)
So 2x − y + 6 = 0 and 2x − y − 4 = 0 are parallel lines as m = 2 for both lines.
m1 = 4 and m2 = − 1 so m1m2 = −1 so the lines are 4 perpendicular. 5 Arrange the equations in the form y = mx + c: 2x − y + 6 = 0, y = 2x + 6, m = 2 2x + 4y − 44 = 0, 4y = 44 − 2x, y = 11 − 12 x, m = − 12 2x − y − 4 = 0, y = 2x − 4, m = 2
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4 Coordinate Geometry 1: Equations of Straight Lines
2x + 4y − 24 = 0, 4y = 24 − 2x, y = 6 − 12 x, m = − 12 So 2x − y + 6 = 0 and 2x + 4y − 44 = 0 are perpendicular lines as m1m2 = −1. And 2x − y + 6 = 0 and 2x + 4y − 24 = 0 are perpendicular lines as m1m2 = −1. And 2x − y − 4 = 0 and 2x + 4y − 44 = 0 are perpendicular lines as m1m2 = −1. And 2x − y − 4 = 0 and 2x + 4y − 24 = 0 are perpendicular lines as m1m2 = −1.
sa
6 From 8x − 2y − 7 = 0, 8x − 7 = 2y , 4x − 72 = y , m1 = 4
m
So m2 = − 14
e, pl
The y intercept of 4x − 2y + 5 = 0 is when x = 0 so is at (0, 52 ) y − y 1 = m ( x − x1 ) y− = 5 2
y=
(
)
m1 = 53 and m2 = − 53 so m1m2 = −1 so the lines are 7 a perpendicular. 9 so m1m2 ≠ −1 so the lines are and m2 = 10 b m1 = 10 9 not perpendicular.
8 Describe as line A (x1, y1) = (0, 6) (x 2, y 2) = (4, 4) y −y m= 2 1 x 2 − x1
Describe as line B (x1, y1) = (0, 6)
10 − 6 m = 2−0 m = 42 m=2 Describe as line C (x1, y1) = (6, 8)
(x1, y1) = (4, 4) (x 2, y 2) = (6, 8) m=
y 2 − y1 x 2 − x1
8−4 m = 6−4 m = 42 m=2 o A is parallel to C, B is parallel to D, A is S perpendicular to B and D and C is perpendicular to B and D. o the quadrilateral can only be a square or a S rectangle.
Exercise 4.6A
p xx
1 (x1, y1) = (100, 160)
(x 2, y 2) = (500, 800)
y − y1 x − x1 = y 2 − y 1 x 2 − x1
y − 160 = x − 100 800 − 160 500 − 100 y − 160 x − 100 = 640 400 y − 160 = 640 x − 100 ) 400 ( y − 160 = 85 x − 160
y = 85 x
2 a False: m1 = 2 and m2 = −2 so m1m2 ≠ −1 so the lines are not perpendicular. b False: c1 = 6 and c 2 = 12 so the lines do not share the same y intercept.
of
(x 2, y 2) = (2, 10) y −y m= 2 1 x 2 − x1
Describe as line D
o pr
m = − 12
m = − 12
ed
4−6 m = 4−0 m = − 24
y 2 − y1 x 2 − x1
ct rre co un
− 14 x − 0 5 − 1x 2 4
m=
10 − 8 m = 2−6 m = −24
m1m2 = −1
(x1, y 1) = (0, 52 )
c False: Negative gradient means the object is travelling back to the start.
d True: Speed is the gradient and the gradients are the same (the difference in the signs just indicates direction) so they are travelling at the same speed.
(x 2, y 2) = (2, 10)
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4
Worked solutions
e False: m1 ≠ m2 so the lines are not parallel.
y − y1 x − x1 = y 2 − y1 x 2 − x1
f False: Positive gradient means the object is travelling away from the start. 3 We have been given two pairs of coordinates for each printer. The cost of the printer is actually the start of the costs when x = 0 i.e. the y intercept. We have also been given the y coordinates when x = 5. Anna’s printer:
When x = 4, y = 20. When x = 7, y = 35. Taxi firm B
sa
e.g. (x1, y1) = (0, 10)
(x1, y1) = (0, 50)
e.g. (x 2, y 2) = (10, 40)
(x 2, y 2) = (5, 950)
y − y1 x − x1 = y 2 − y1 x 2 − x1
m
y − y1 x − x1 = y 2 − y1 x 2 − x1
y − 50 = x−0 950 − 50 5 − 0 y − 50 x = 900 5 y − 50 = 180x y = 180x + 50
e, pl
(x1, y1) = (0, 67.50) (x 2, y 2) = (5, 667.50)
y − y1 x − x1 = y 2 − y1 x 2 − x1
y − 10 = x −0 40 − 10 10 − 0 10y − 10 00 = 30x y = 3x + 10 When x = 4, y = 22.
When x = 7, y = 31.
ct rre co un
Bhavini’s printer:
y − 67.50 = x −0 667.50 − 67.50 5 − 0 y − 67.50 x = 600 5 y − 67.50 = 120x y = 120x + 67.5 50
The cost of one ink cartridge for Anna’s printer is £180 ÷ 4 = £45.
e.g. (x 2, y 2) = (10, 35) y − y1 x − x1 = y 2 − y1 x 2 − x1
y − 15 = x −0 35 − 15 10 − 0 y − 15 x = 20 0 10 10y − 150 = 20x y = 2x + 15
When x = 4, y = 23. When x = 7, y = 29.
Taxi firm A is the cheapest on a 4 mile journey. Taxi firm C is the cheapest on a 7 mile journey. 5 (x1, y1) = (0.57, 1)
(x 2, y 2) = (1, 1.75)
y − y1 x − x1 = y 2 − y 1 x 2 − x1
Nothing is significant about the x intercepts. Although you can mathematically work out the x intercept in each case this would actually be a negative value which would not make sense as you cannot have a negative number of years.
of
The cost of one ink cartridge for Bhavini’s printer is £120 ÷ 4 = £30.
e.g. (x1, y1) = (0, 15)
o pr
c is the cost to buy the printer and is £50 in Anna’s case and £67.50 in Bhavini’s case.
Taxi firm C
ed
m is the cost per year to run the printer and is £180 in Anna’s case and £120 in Bhavini’s case.
y −1 = x − 0.57 1.75 − 1 1 − 0.57 y − 1 x − 0.57 = 0.75 0.43 75 y − 1 = 43 ( x − 0.57 )
4 Taxi firm A
y −1 =
e.g. (x1, y1) = (0, 0) e.g. (x 2, y 2) = (10, 50)
y −0 = x −0 50 − 0 10 − 0 10y = 50x y = 5x
y=
75 x 43 75 x 43
4275 − 4300 1 + 172
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4 Coordinate Geometry 1: Equations of Straight Lines
Practice questions
p xx
y − y1 x − x1 = y 2 − y1 x 2 − x1
1 m = 3, (x1, y 1) = (1, 6)
y − y1 = m ( x − x1 ) y − 6 = 3( x − 1) y − 6 = 3x − 3 y = 3x + 3
2 a Line A
(x1, y1) = (0, 7)
(x 2, y 2) = (2, 3)
(x 2, y 2) = (3, 9)
y −y m= 2 1 x 2 − x1
y − y1 x − x1 = y 2 − y1 x 2 − x1
Line B (x1, y1) = (6, 4) (x 2, y 2) = (8, 5)
m=
y 2 − y1 x 2 − x1
5−4 m = 8 − 6 m = 12
b Line A is steeper because 2 > 12 .
5 a m1 = 2
m1m2 = −1
so m2 = − 12
(x1, y 1) = (12, 4)
y − y1 = m ( x − x1 )
y − 4 = − 12 ( x − 12) 2
m1m2 = −1 so m2 = −1 y − y1 = m ( x − x1 ) y − 5 = −1( x − 0 )
4 a Runner A starts 3 miles away from runner B’s home.
b Runner A (x1, y1) = (0, 3) (x 2, y 2) = (3, 9)
y intercept = (0, 10)
When y = 0, x = 20
x intercept = (20, 0)
c Area of triangle = 12 bh = 12 × 20 × 10 = 100 units2 6 Pay as you go (x1, y 1) = (0, 5) (x 2, y 2) = (1, 7)
y − y1 x − x1 = y 2 − y1 x 2 − x1
of
The two runners meet 9 miles away from runner B’s home so runner A will have run 6 miles.
b When x = 0, y = −10
o pr
y = −x + 5
y = − 12 x + 10
ed
(x1, y1) = (0, 5)
c Runner B runs faster. Gradient = speed and 3 > 2.
y − 4 = − 1 x + 6
3 m1 = 1
y −0 x −0 = 9−0 3−0 y x = 9 3 y = 3x
ct rre co un
e, pl
m
3−7 m = 2 − 0 m = − 4 2 m = −2
Runner B (x1, y1) = (0, 0)
sa
y −3 x −0 = 9−3 3−0 y −3 x = 6 3 y − 3 = 2x y = 2x + 3
y −5 x −0 = 7 − 5 1− 0 y −5 =x 2 y = 2x + 5
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4
Worked solutions
For this option
m=
4 rides: x = 4 so y = £13 17 rides: x = 17 so y = £39 For other option (book of tickets) Cost of 4 rides = £5 + £10 = £15 so Pay as you go option is cheaper. Cost of 17 rides = £5 + (3 × £10) = £35 so book of tickets option is cheaper.
(x1, y1) = (−2, 8) (x 2, y 2) = (−10, −8)
sa
m=
m e, pl
1 = a −5 2 −12 − 8 −20 = 2a − 10 a = −5
m1m2 = − 12 ( 2) = −1
y − −5 = −2( x − 0 ) y = −2x − 5
8 (x1, y1) = (2 2, 2) (x 2, y 2) = ( 2, 2 3) y −y m= 2 1 x 2 − x1
m= 2 3− 2 2−2 2
y − y 1 = m ( x − x1 )
9 Line AB (x1, y1) = (−2, 8)
x = 8 − 4y
3( 8 − 4y ) + 5y + 15 = 0
Solving gives y =
39 7
Substituting this value back into the original equation gives x = − 100 7
(
, 39 Point of intersection (x1, y 1) = − 100 7 7
3x + 5y + 15 = 0
)
Rearranging gives m1 = − 35
So m2 =
5 3
y − y 1 = m ( x − x1 ) y − 39 = 7
5 3
( x − − 1007 )
y = 53 x + 617 21
11 a Line A
(x1, y1) = (0, 32)
(x 2, y 2) = (5, 41)
y − y1 x − x1 = y 2 − y 1 x 2 − x1 y − 32 = 41 − 32 y − 32 = 9 y=
x −0 5−0 x 5 9 x + 32 5
of
2 ( x − 0) 2 3− 2
y = 2 6 +2x +3 10
10 x + 4y − 8 = 0
o pr
2 2 3− 2 (x1, y1) = (0, 3) m=
so AB and BC are perpendicular to each other and ABC is a right-angled triangle.
ed
m= 2 3− 2 − 2 Gradient perpendicular to this:
y −3=
( )
ct rre co un
m = −2, (x1, y1) = (0, −5)
y 2 − y1 x 2 − x1
m = −8 − 8 −10 − −2 m2 = 2
(x1, y1) = (8, 5)
y −y m= 2 1 x 2 − x1
m = 5−8 4 − −2 m1 = − 12
Line BC
7 Line A
(x 2, y 2) = (−12, a)
y 2 − y1 x 2 − x1
(x 2, y 2) = (4, 5)
509
227878 A-Level Maths_Workedsols_CH4-6.indd 509
10/5/16 11:23 AM
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