A-level Maths Chapter 4 Sample Chapter

Edexcel A-level Mathematics Year 1 and AS Student Book Helen Ball Kath Hipkiss Michael Kent Chris Pearce

Contents 1–A lgebra and functions 1: Manipulating algebraic expressions 00 1.1 Manipulating polynomials algebraically 00 00

1.3 The binomial expansion

00

1.4 Factorisation

00

1.5 Algebraic division

00

1.6 Laws of indices

00

1.7 Manipulating surds

00

1.8 Rationalising the denominator

00

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m

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1.2 Expanding multiple binomials

00

4.1 Writing the equation of a straight line in the form ax + by + c = 0

00

4.2 Finding the equation of a straight line using the formula y – y1 = m(x – x1)

00

4.3 Finding the gradient of the straight line between two points

00

4.4 Finding the equation of a straight line using the formula y – y1 / y2 – y1 = x – x1 / x2 – x1 00 4.5 Parallel and perpendicular lines

00

4.6 Straight line models

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2 – Algebra and functions 2: Equations and inequalities

4 – Coordinate geometry 1: Equations of straight lines

00

5 – Coordinate geometry 2: Circles 00 5.1 Equations of circles

00

2.2 The discriminant of a quadratic function 00

5.2 Angle in a semicircle

00

2.3 Completing the square

00

5.3 Perpendicular from the centre to a chord 00

2.4 Solving quadratic equations

00

5.4 Radius perpendicular to the tangent

2.5 Solving simultaneous equations

00

2.6 Solving linear and quadratic simultaneous equations

00

6.1 Sine and cosine

00

2.7 Solving linear inequalities

00

6.2 The sine rule and the cosine rule

00

2.8 Solving quadratic inequalities

00

6.3 Trigonometric graphs

00

6.4 The tangent function

00

2.1 Quadratic functions

3.1 Sketching curves of quadratic functions 00 00

00

6.5 Solving trigonometric equations

00

6.6 A useful formula

00

7 – Exponentials and logarithms

3.3 Sketching curves of quartic functions 00

7.1 The function

3.4 Sketching curves of reciprocal functions 00

7.2 Logarithms

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3.2 Sketching curves of cubic functions

6 – Trigonometry

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00

00

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3 – Algebra and functions 3: Sketching curves

00

ax

00 00 00

3.5 Intersection points

00

7.3 The equation

=b

00

3.6 Proportional relationships

00

7.4 Logarithmic graphs

00

3.7 Translations

00

7.5 The number e

00

3.8 Stretches

00

7.6 Natural logarithms

00

7.7 Exponential growth and decay

00

ax

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Contents

8 – Differentiation 00

12.2 Variance and standard deviation 00

8.1 The gradient of a curve 00

12.3 Displaying and interpreting data 00

8.2 The gradient of a quadratic curve 00

8.3 Differentiation of x² and x³ 00

8.4 Differentiation of a polynomial 00

8.5 Differentiation of xn 00

8.6 Stationary points and the second derivative 00

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13 – Probability and statistical distributions 00

13.1 Calculating and representing probability 00

13.2 Discrete and continuous distributions 00

13.3 The binomial distribution 00

8.7 Tangents and normals 00

9.1 Indefinite integrals 00

14.1 Populations and samples 00

9.2 The area under a curve 00

14.2 Hypothesis testing 00

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9 – Integration 00

14 – Statistical sampling and hypothesis testing 00

10 – Vectors 00

15 – Kinematics 00

10.1 Definition of a vector 00

10.2 Adding vectors 00

15.2 Equations of constant acceleration 00

10.3 Vector geometry 00

15.3 Vertical motion 00

10.4 Position vectors 00

15.4 Displacement-time and velocity-time graphs 00

11 – Proof 00

15.5 Variable acceleration 00

11.1 Proof by deduction 00

11.2 Proof by exhaustion 00

16 – Forces 00

11.3 Disproof by counter example 00

15.1 The language of kinematics 00

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16.1 Forces 00

16.2 Newton’s laws of motion 00

12 – Data presentation and interpretation 00

16.3 Vertical motion 00

16.4 Connected particles 00

16.5 Pulleys 00

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12.1 Measures of central tendency and spread 00

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4

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.

the tariff with the highest upfront cost

300

the tariff with the lowest monthly charge.

You will learn how to:

200

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LEArninG oBJECtiVES

Option A

250 Cost (£)

› ›

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From this graph you could identify important features like:

Option C

150

Option B

100

›

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

50

0

0

1

2

3

4

5 6 7 Time (months)

8

9

10

11

12

be able to use straight line models in a variety of contexts.

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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.

Prior KnoWLEdGE 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

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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES

›

find the equation of the line through two given points or through one point with a given gradient

›

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.

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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?

2 Here are the equations of six lines:

m

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. b Identify the lines that are parallel.

c Identify the lines that are perpendicular.

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3 Find the equation of the line through the points (1, 8) and (2, 0) by sketching the graph.

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

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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.

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The equation of a straight line may be written in the form ax + by + c = 0, where a, b and c are integers.

Example 1

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Write y = 2x + 3 in the form ax + by + c = 0, where a, b and c are integers.

Solution Subtract y from both sides of the equation.

0 = 2x + 3 − y Rearrange.

2x − y + 3 = 0

2

M

Modelling

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a, b and c are integers, so this is the final answer.

PS

Problem solving

PF

Proof

CM

Communicating mathematically

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4.1

Writing the equation of a straight line in the form ax + by + c = 0

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

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1 Write y = − x + 3 in the form ax + by + c = 0, where a, b and c are 2 integers.

Solution

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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.

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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2y = − x + 6

Stop and think As a mathematician, which format of the equation do you find most useful and why?

3

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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES

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

4

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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.

4

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4.2

Finding the equation of a straight line using the formula y – y1 = m(x – x1)

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.

KEY INFORMATION 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.

Example 3

You need to learn the formula y − y1 = m ( x − x1 ) as it is NOT given in the formula booklet.

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Find the equation of the line with gradient 2 that goes through the point (3, 7).

m

Solution

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

Add 7 to both sides of the equation. y = 2x + 1

Example 4

A line with a gradient of −1 and the line y = 2 meet the y axis at the same point.

Technology 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)?

ed

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.

5

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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES

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

m

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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PS

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

6

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4.3

Finding the gradient of the straight line between two points

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

x

0

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 7

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4 Coordinate Geometry 1: Equations of Straight Lines

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. m = −6 4

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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

Example 6

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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?

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)

ed

Write down what you know.

You might find if you make a quick sketch of this scenario it aids your understanding.

Technology

State the formula you are going to use. y −y m= 2 1

x 2 − x1

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.

2 = 6 −0 5−x

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Substitute the values into the formula.

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x 2 = 5, y 2 = 6

Multiply both sides of the equation by (5 − x).

2( 5 − x ) = 6

Expand the bracket.

10 − 2x = 6 8

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4.3

Finding the gradient of the straight line between two points

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.

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x=2

The coordinates of the x intercept are (2, 0).

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m exercise 4.3A 1

Find the gradient of the line passing through the given points.

b (5, 9) and (3, 3) c (1, −3) and (3, −9)

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a (2, 3) and (7, 8)

2

p xx

Find the gradient of the line passing through the given points.

a (8a, 5a) and (3a, 3a) b (a, a) and (3a, −5a) 3

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)

9

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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES

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.

M

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 ).

7

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Which is steepest part of the ride − the ascent or the descent?

Find the gradient of the line passing through ( 12 , 13 ) and ( 34 , − 32 ).

m

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

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

x

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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 10

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4.4

y – y1 x – x1 Finding the equation of a straight line using the formula y – y = x – x 2 1 2 1

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

You might find it aids your understanding to make a quick sketch of this scenario.

Write the equation of the line in the form ax + by + c = 0, where a, b and c are integers.

e, pl

Solution

State the formula you are going to use.

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y − y1 x − x1 = y 2 − y1 x 2 − x1

Write down what you know.

Let (x1, y1) = (0, − 5) Let (x 2, y 2) = (9, − 8)

Substitute the values into the formula. y − −5 x − 0 =

−8 − −5 9 − 0

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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.

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

0 = x + 3y + 15

Reorder.

x + 3y + 15 = 0

11

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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES

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) 3

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b (−3, −4) and (−7, 6)

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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 )

12

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4.5

Parallel and perpendicular 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

The lines y = 7 − 3x and y = 2x + 8 intersect at point B.

sa

CM

10 The lines y = x − 9 and y = 5 − x intersect at point A. 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.

13

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4 Coordinate Geometry 1: Equations of Straight 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

sa

Find the equation of the line that is parallel to 3x − y + 7 = 0 and passes through the point (1, 8).

e, pl

Solution

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

Technology Using a graphing software package, try finding the equations of lines passing through other points but that are parallel to the given equation.

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

The equation is not yet in the required format.

o pr

Rearrange.

ed

y = 3x + 5

3x − y + 5 = 0

of 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?

14

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4.5

Parallel and perpendicular 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.

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.

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) .

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.

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

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

15

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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht 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).

PS

3

sa

Show that lines A and B are parallel.

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

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.

PF

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.

16

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4.5

Parallel and perpendicular 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

m2 = − 1 . m1

(1, m1)

Two straight lines with gradients m1 and m2 are perpendicular when m1m2 = −1 .

2 2 1 + m1

m1 – m2

e, pl

m m1

A line with a gradient of m1 is perpendicular to a line with a gradient of m2 when

x

ct rre co un

2 2 1 + m2

You need to learn the conditions for perpendicular lines as they are NOT given in the formula booklet.

(1, m2)

Line 1 has a gradient of m1 .

PROOF Using Pythagoras’ theorem:

Line 2 has a gradient of m2.

12 + m12 + 12 + m22 = (m1 − m2 )

If m1m2 = −1 , then the lines are perpendicular.

2 + m12 + m22 = m12 − 2m1m2 + m22 2 = −2m1m2 − 1 = −m1 m2

ed

Example 10

2

Find the equation of the line that is perpendicular to 4x − y − 9 = 0 and passes through the point (3, 5).

Solution 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.

Find the gradient of the line perpendicular to it.

m1m2 = −1

4m2 = −1

m2 = − 14

( x1, y1) = (3, 5) 17

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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht 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

( 12 ) = −1

m1m2 = ( −2)

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?

of

So the lines 2x + y − 5 = 0 and x − 2y + 4 = 0 are perpendicular.

Technology

o pr

Show that m1m2 = −1

ed

y = 12 x + 2

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 18

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4.6

Straight line models

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

sa

So y = 12 x − 1

CM

3

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 CM

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

o pr

By considering gradients, work out which of the following the quadrilateral could possibly be and state your reasons. 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

Key InFoRMATIon

You need to be able to use straight line models in a variety of contexts.

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4 Coordinate Geometry 1: Equations of Straight Lines

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

e, pl

Solution

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

20

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4.6

Straight line models

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

m

Option A

250

Cost (£)

150 100

0

1

2

3

Option B

ct rre co un

50 0

Option C

e, pl

200

4

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.

21

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4 Coordinate Geometry 1: Equations of Straight Lines

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

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

ed

50 x 7

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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4.6

Straight line models

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 CM

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.

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?

23

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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES

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

Example charge for 10 miles would be £40.

taxi firm C

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

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.

M

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

24

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4

Key points and Practice questions

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

p xx

ct rre co un

Practice questions

Line A is parallel to y = 3x − 5 and passes through the point (1, 6).

[3 marks]

Find the equation of line A.

2

a Line A passes through (0, 7) and (2, 3) and line B passes through (6, 4) and (8, 5). 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.

Using m =

[2 marks]

Give a reason for your answer.

Line A is perpendicular to x − y − 8 = 0 and passes through the y intercept of y = 9x + 5.

ed

3

Find the equation of line A. CM

[4 marks]

[4 marks]

4 The graph shows the journeys of two different runners.

o pr

Runner A lives 3 miles away from runner B.

M

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] 25

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4 CoordinAtE GEomEtry 1: EquAtionS of StrAiGht LinES

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

7

M

sa

PS

[5 marks]

Show that ABC is a right-angled triangle.

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.

[10 marks]

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

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.

What do you notice?

b Find the point of intersection of lines A and B.

of

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]

26

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Colour Palette Pure Maths Chapter 1

C = 0; M = 100; Y = 100; K = 0

Background color

C = 0; M = 40; Y = 25; K = 5

m

sa

Tab color

Chapter 2

e, pl

C = 0; M = 71; Y = 92; K = 0

Background color

C = 0; M = 50; Y = 60; K = 0

Statistics Chapter 1 Tab color

C = 85; M = 0; Y = 100; K = 0

ed

Background color

ct rre co un

Tab color

C = 40; M = 0; Y = 60; K = 0

C = 80; M = 0; Y = 50; K = 0

Background color

C = 40; M = 0; Y = 30; K = 0

227878 A-Level Maths_Chapter 4.indd 27

of

Tab color

o pr

Chapter 2

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Mechanics Chapter 1 Tab color

C = 63; M = 3; Y = 0; K = 0

Background color

C = 20; M = 0; Y = 0; K = 0

sa

Chapter 2

m

Tab color

e, pl

Background color

C = 100; M = 40; Y = 0; K = 30

Tab color Background color

ct rre co un

FM & BM

C = 40; M = 10; Y = 0; K = 10

C = 60; M = 0; Y = 30; K = 20

C = 60; M = 0; Y = 30; K = 20

ed of

o pr

227878 A-Level Maths_Chapter 4.indd 28

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Investigating the large data set

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?

e, pl

m

sa ed

ct rre co un of

o pr

227878 A-Level Maths_Chapter 4.indd 29

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