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SCHOLAR Study Guide

National 5 Mathematics

Course MaterialsTopic 13: The straight line

Authored by:Margaret Ferguson

Reviewed by:Jillian Hornby

Previously authored by:Eddie Mullan

Heriot-Watt University

Edinburgh EH14 4AS, United Kingdom.

First published 2014 by Heriot-Watt University.

This edition published in 2016 by Heriot-Watt University SCHOLAR.

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SCHOLAR Study Guide Course Materials Topic 13: National 5 Mathematics

1. National 5 Mathematics Course Code: C747 75

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1

Topic 1

The straight line

Contents

13.1 Determining the gradient of a straight line . . . . . . . . . . . . . . . . . . . . 3

13.1.1 The gradients of horizontal and vertical straight lines. . . . . . . . . . . 7

13.1.2 Parallel lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

13.2 Determining the equation of a straight line . . . . . . . . . . . . . . . . . . . . 12

13.3 Identifying the gradient and the y-intercept of a straight line . . . . . . . . . . . 17

13.4 Finding equations of parallel lines . . . . . . . . . . . . . . . . . . . . . . . . . 21

13.5 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

13.6 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2 TOPIC 1. THE STRAIGHT LINE

Learning objectives

By the end of this topic, you should be able to:

• determine the gradient of a straight line;

• determine the equation of a straight line;

• determine equations of parallel lines;

• identify the gradient and y-intercept from the equation of a straight line.

© HERIOT-WATT UNIVERSITY

TOPIC 1. THE STRAIGHT LINE 3

1.1 Determining the gradient of a straight line

We can describe how steep a slope is by using the idea of gradient. The bigger thegradient the steeper the slope.

If the line slopes down from left toright the gradient will be negative.

If the line slopes up from left to rightthe gradient will be positive.

The gradient is defined by the letter m.

The change in x coordinates is called the horizontal distance and the change in ycoordinates is called the vertical distance.

We can choose any two points on the line by identifying their coordinates as (x1 , y1)and (x2 , y2).

Key point

This gives the formula for the gradient of a straight line as:

m = V ertical distanceHorizontal distance = y2 − y1

x2 − x1

Finding the gradient of a straight line

Go online

This activity shows how to find the gradient of a line using the gradient formula, m =y2 − y1x2 − x1

.



Using the formula for the gradient of a straight line we would get,

m = y2 − y1x2 − x1

= 8 − 09 − 1 = 8

8 = 1

Now try this again by drawing graphs of straight lines using the following sets ofcoordinates. Use the equation m = y2 − y1

x2 − x1to find the gradient of each line.

Q1:

a) (1,0) and (4,9)

b) (1,2) and (7,8)

c) (2,2) and (5,8)

d) (2,3) and (3,6)

e) (1,3) and (3,7)

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Notice we can choose the coordinates of any two points which lie on the line to calculatethe gradient of that line.

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Example

Problem:

Calculate the gradient of the line from A(1,5) to B(7, -2) using the gradient formula.

Solution:

Look at the diagram. A has coordinates (1,5) so x1 = 1 and y1 = 5. B hascoordinates (7, -2) so x2 = 7 and y2 = − 2.

mAB = y2 − y1x2 − x1

= −2 − 57 − 1 = −7

6 or − 76

mAB is used to represent the gradient of the line AB.

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Example

Problem:

Calculate the gradient of the line which passes through C(-1, 5) and D(3, 7)?

Solution:

C(-1,5) so x1 = − 1 and y1 = 5 and D(3,7) so x2 = 3 and y2 = 7.

mCD = y2 − y1x2 − x1

= 7 − 53 − (−1) = 2

4 = 12

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Finding the gradient of a straight line practice

Go online

Q2: Calculate the gradient of the line from A(3, 4) to B(9, 1) using the gradient formula.

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Q3: Calculate the gradient of the line which passes through E(2, 4) and F (4, 2)?

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Determining the gradient of a straight line exercise

Go online

Q4: Calculate the gradient of the line AB.

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Q5: What is the gradient of the line which passes through the points C(-6,-2) and D(-1,3) ?

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Q6: Calculate the gradient of the line which passes through E(-4,-5) and F (3,4).

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1.1.1 The gradients of horizontal and vertical straight lines.

The gradients of horizontal and vertical straight lines can be calculated using thegradient formula but extra care must be taken.

Finding the gradients of horizontal straight lines

Go online

A horizontal line has y2 − y1 = 0 giving m = y2 − y1x2 − x1

= 0x2 − x1

= 0 so thegradient of a horizontal straight line is 0.

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Finding the gradients of vertical straight lines

Go online

A vertical line has x2 − x1 = 0 giving m = y2 − y1x2 − x1

= y2 − y10 = undefined so

the gradient of a vertical straight line is undefined.

The gradient is undefined as it is not possible to divide by 0. Try to divide a number by0 on your calculator.

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Key point

The gradient of a horizontal line is 0.The gradient of a vertical line is undefined.



Examples

1.

Problem:

Calculate the gradient of the line from A(3,5) to B(-2,5) using the gradient formula.

Solution:

Look at the diagram.

A has coordinates (3,5) so x1 = 3 and y1 = 5 and B has coordinates (-2,5) sox2 = − 2 and y2 = 5. The line AB is horizontal therefore mAB = 0. We can checkthis using the gradient formula:

mAB = y2 − y1x2 − x1

= 5 − 5−2 − 3 = 0

−5 = 0

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2.

Problem:

Calculate the gradient of the line which passes through C(-1,3) and D(-1,-2)?

Solution:

C(-1,3) so x1 = − 1 and y1 = 3 and D(-1,-2) so x2 = − 1 and y2 = − 2.

A quick plot of these points would show that the line is vertical since the x coordinate inboth points is the same. Therefore the gradient of CD is undefined.

We can check this using the gradient formula:

mCD = y2 − y1x2 − x1

= (−2) − 3(−1) − (−1) = −5

0 = undefined

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Determining the gradient of horizontal and vertical straight lines practice

Go online

Q7: Calculate the gradient of the line from A(3,1) to B(3,8) using the gradient formula.

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Q8: Calculate the gradient of the line which passes through E(2,4) and F (-2,4)?

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Determining the gradient of horizontal and vertical straight lines exercise

Go online

Q9:

a) Calculate the gradient of the line GH.

b) Calculate the gradient of the line KL.

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Q10:

Which line has a gradient which is undefined PQ or RS?

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1.1.2 Parallel lines

Parallel lines have the same gradient so it is easy to prove if lines are parallel.

Examples

1.

Problem:

Are the lines on this graph parallel?

Solution:

mAB = y2 − y1x2 − x1

= (−3) − 55 − 1 = −8

4 = −2

mCD = y2 − y1x2 − x1

= (−3) − 53 − (−1) = −8

4 = −2

We can prove the lines AB and CD are parallel because the gradients are equal.

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2.

Problem:

Are the lines on this graph parallel?

Solution:

mEF = y2 − y1x2 − x1

= 3 − (−3)2 − (−1) = 6

3 = 2

mGH = y2 − y1x2 − x1

= 3 − (−3)5 − 3 = 6

2 = 3

We can prove the lines AB and CD are not parallel because the gradients are notequal.

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Identifying parallel lines practice

Go online

Q11: The gradient of the line AB is -1 and the gradient of the line CD is -1. Are theselines parallel?

a) Parallelb) Not parallel

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Q12: G(-1,-2), H(7,5), K(0,1) and L(4,7). Calculate the gradient of the line GH and theline KL.Determine if the lines are parallel.

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Key point

Parallel lines have the same gradient.

Identifying parallel lines exercise

Go online

Q13:

a) Calculate the gradient of the line VW .

b) Calculate the gradient of the line XY .

c) Are these lines parallel? Yes or No.

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Q14:

a) Calculate the gradient of the line which passes through I(-2,6) and J(8,-2).

b) Calculate the gradient of the line which passes through M (0,5) and N (5,1).

c) Are these lines parallel? Yes or No.

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1.2 Determining the equation of a straight line

Key point

The equation of a straight line takes the form y = mx + c where m is the gradientand c is the y-intercept.

The y-intercept is the y coordinate of the point where the line cuts the y-axis.

Remember, to calculate the gradient we use: m = y2 − y1x2 − x1

Example

Problem:

Find the equation of the line in the diagram in the form y = mx + c.

Solution:

The line cuts the y-axis at (0,1) ⇒ c = 1.The line passes through (0,1) and (2,5). So we have: m = 5 − 1

2 − 0 = 2

Remember we could use the coordinates of any two points that lie on the line to calculatethe gradient.

So the equation is y = 2x + 1.

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Key point

When we know the gradient, m and the y-intercept, c of a straight line we canfind the equation of the straight line by substituting the value for m and c intoy = mx + c.



Example

Problem:

Find the equation of the line in the diagram.

Solution:

This is a horizontal straight line and its equation is special.

The line cuts the y-axis at (0,3) ⇒ c = 3.

The line passes through (0,3) and (3,3). So we have: m = 3 − 33 − 0 = 0

So the equation is y = 0x + 3.

However this can be simplified to y = 3.

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Key point

The equation of all horizontal straight lines take the formy = c.

Example

Problem:

Find the equation of the line in the diagram.



Solution:

This is a vertical straight line and its equation is special too.

The line does not cut the y-axis so there is no value of c.

The line passes through (4,0) and (4,4). So we have: m = 4 − 04 − 4 = undefined

The gradient is undefined and there is no value of c so we have nothing to substituteinto y = mx + c.

However , if you look at other points on the line they all have an x coordinate of 4 i.e.(4,. . .).

So the equation is x = 4.

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Key point

The equation of all vertical straight lines take the formx = . . ..

Key point

The equation of a straight line can also be written as y − b = m(x − a) where(a, b) is a point on the line.

Example

Problem:

Find the equation of the line which passes through A(1,-1) and B(5,7).

Solution:

We don’t know the y-intercept but we can find the gradient.

mAB = 7 − ( - 1)5 − 1 = 8

4 = 2

We can find the equation of a straight line if we know the gradient and a point (a, b) usingthe formula: y − b = m(x − a).

m = 2 and the point B(5,7) gives a = 5 and b = 7 all we have to do is substitute intothe equation giving:

y − b = m (x − a)

y − 7 = 2 (x − 5) expand the brackets

y − 7 = 2x − 10 get rid of - 7 from the left by + 7

y = 2x − 3 ( - 10) + 7 = - 3

The equation of equation of the straight line is y = 2x − 3.

We could have chosen the point A(1,-1) to substitute for a and b as A is also on the line.The resulting equation would be the same.

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Top tip

If we know the gradient and the y-intercept use:

y = mx + cto find the equation of the straight line.If we know the gradient and the coordinates of any point on the line use:

y − b = m(x − a)to find the equation of the straight line.

Key point

These techniques will work for the equations of all straight lines except vertical.

Equation of a straight line practice

Go online

Q15: Find the equation of the line in the diagram in the form y = mx + c.

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Q16: Find the equation of the line in the diagram.

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Q17: Find the equation of the line in the diagram.

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Q18: Find the equation of the line which passes through C(-3,8) and D(2,-2).

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Equation of a straight line exercise

Go online


The equation of the line is y = ?

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Q21: Find the equation of the lines in the diagram.

a) What is the equation of the horizontal line?

b) What is the equation of the vertical line?

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Q22:

a) Find the equation of the line passing through A(-3,1) and B(2,6) in the formy = mx + c.

b) Find the equation of the line passing through C(-2,1) and D(4,-2) in the formy = mx + c.

c) Find the equation of the line passing through E(6,-4) and F (6,5).

d) Find the equation of the line passing through G(-4,1) and H(1,11) in the formy = mx + c.

e) Find the equation of the line passing through K(1,-5) and L(8,-5).

f) Find the equation of the line with gradient 4 that passes through the point (-5,-3) inthe formy = mx + c.

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1.3 Identifying the gradient and the y-intercept of a straightline

Equation of a straight line

Look at the equation y = mx + c.We can identify the gradient and the y-intercept:

• m is the gradient;



• c gives us the coordinates of the y-intercept (0, c).

For the equation y = 3x + 4,

• m = 3 so the gradient is 3;

• c = 4 and the y-intercept is (0,4).

For the equation y = x − 2,

• m = 1 so the gradient is 1;

• c = − 2 and the y-intercept is (0,-2).

Examples

1.

Problem:

Identify the gradient of the line with equation y = 2x + 3.

Solution:

We can see that m = 2 so the gradient is 2.

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2.

Problem:

Identify the y-intercept of the line with equation y = − 2x + 3.

Solution:

We can see that c = 3 so the y-intercept is (0,3).

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3.

Problem:

Identify the gradient of the line with equation y = − x − 4.

Solution:

We can see that m = − 1 so the gradient is -1.

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4.

Problem:

Identify the y-intercept of the line with equation y = x.

Solution:

We can see that c = 0 so the y-intercept is (0,0).

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5.

Problem:

Identify the gradient of the line with equation y − 3x = − 5.

Solution:

First we must re-arrange the equation so that it is in the form y = mx+ c.

y − 3x = −5 get rid of - 3x on the left by + 3x

y = 3x − 5

We can see that m = 3 so the gradient is 3.

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6.

Problem:

Identify the gradient of the line with equation 2y + 4x − 7 = 0.

Solution:

Again we must re-arrange the equation so that it is in the form y = mx + c.

2y + 4x − 7 = 0 get rid of - 7 on the left by + 7

2y + 4x = 7 get rid of + 4x on the left by - 4x

2y = −4x + 7 leave y on its own by ÷ 2

y = −2x + 3 · 5We can see that m = − 2 so the gradient is -2.

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Key point

The equation of a straight line must be written in the formy = mx + c before we can identify the gradient and they-intercept.

Identifying the y-intercept and gradient of a straight line practice

Go online

Q23: Identify the gradient of the line with equation y = 5x − 1.

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Q24: Identify the coordinates of the y-intercept of the line with equation y = 5x − 8.

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Q25: Identify the gradient of the line with equation y = − 3x + 2.

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Q26: Identify the coordinates of the y-intercept of the line with equation y = − 3x.

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Q27: Identify the gradient of the line with equation y + x = 7.

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Q28: Identify the coordinates of the y-intercept of the line with equationy − 4x + 6 = 0.

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Q29: Identify the gradient of the line with equation 3y − 12x = 2.

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Q30: Identify the coordinates of the y-intercept of the line with equation y = 2.

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Identifying the y-intercept and gradient of a straight line exercise

Go online

Identifying the gradient

Q31:

a) Identify the gradient of the line with equation y = 5x + 4

b) Identify the gradient of the line with equation y − 7x + 3 = 0

c) Identify the gradient of the line with equation 2y + 6x = 9

d) Identify the gradient of the line with equation y = 12

e) Identify the gradient of the line with equation x = − 10

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Q32:

a) Identify the coordinates of the y-intercept of the line with equation y = 5x + 4

b) Identify the coordinates of the y-intercept of the line with equation y + 1 · 5x = 5

c) Identify the coordinates of the y-intercept of the line with equation−2y − 8x + 12 = 0

d) Identify the coordinates of the y-intercept of the line with equation y = 13

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1.4 Finding equations of parallel lines

We already know that parallel lines have the same gradient so it is only the y-interceptwhich changes.

Equation of parallel lines

Go online

Q33:

Use this graph and the equation y = mx + c as a starting point.

Determine the gradient of the line shown and then draw a new line using the same valuefor m but increase or decrease the value of c.

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Now try this again by drawing graphs of parallel lines using the following equations ofstraight lines and then changing the value of c.

1. y = 2x − 4; increase c

2. y = − 4x − 3; increase c

3. y = − x + 6; decrease c

4. y = 4x + 2; decrease c

Show the graphs you have drawn to your teacher/tutor for feedback.

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Examples

1.

Problem:

Identify the equation of the line which is parallel to y = 2x + 3 and passes through thepoint (0,7).

Solution:

The parallel line has gradient 2 and y-intercept (0,-7).

mparallel = 2 and c = − 7 so the equation of the parallel line is y = 2x − 7.

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2.

Problem:

Identify the equation of the line which is parallel to y = − x + 3 and passes throughthe point (2,5).

Solution:

This time the parallel line has gradient -1 so mparallel = − 1.

We don’t know the y-intercept but we do have a point on the parallel line (2,5) so byusing y − b = m(x − a) with m = − 1, a = 2 and b = 5 we can find its equation.

y − 5 = −1 (x − 2)

y − 5 = −x + 2

y = −x + 7

So the equation of the parallel line is y = − x + 7.

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3.

Problem:

Identify the equation of the line which is parallel to y − 3x + 4 = 0 and passes throughthe point (-6,4).

Solution:

This time we need to re-arrange the equation to identify the gradient of the parallel line.

y − 3x + 4 = 0 ⇒ y = 3x − 4

The gradient is 3 so mparallel = 3.

We don’t know the y-intercept but we do have a point on the parallel line (-6,4) so byusing m = 3, a = − 6 and b = 4 we can find its equation by substituting intoy − b = m(x − a).

y − 4 = 3 (x − (−6))

y − 4 = 3 (x + 6)

y − 4 = 3x + 18

y = 3x + 22

So the equation of the parallel line is y = 3x + 22.

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4.

Problem:

Identify the equation of the line which is parallel to y = 1 and passes through the point(0,-2).

Solution:

The line y = 1 is a horizontal straight line with gradient 0.

The parallel line is also horizontal with y-intercept (0,-2).

mparallel = 0 and c = -2 giving y = 0x − 2

So the equation of the parallel line is y = − 2.

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5.

Problem:

Identify the equation of the line which is parallel to x = 8 and passes through the point(4,3).

Solution:

The line x = 8 is a vertical straight line with gradient undefined.

The parallel line is also vertical but passes through (4,3).

We know that all the points on a vertical straight line have the same x-coordinate so inthis case (4, . . .).

So the equation of the parallel line is x = 4.

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Finding equations of parallel lines practice

Go online

Q34: Identify the equation of the line which is parallel to y = 4x + 1 which passesthrough (0,-3).

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Q35: Identify the equation of the line which is parallel to y = −0 ·5x − 2 which passesthrough (2,6).

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Q36: Identify the equation of the line which is parallel to y + 3x −1 = 0 which passesthrough (1,-4).

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Q37: Identify the equation of the line which is parallel to 4y − 4x + 15 = 0 whichpasses through (-2,-1).

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Q38: Identify the equation of the line which is parallel to y = − 4 and passes throughthe point (0,3).

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Q39: Identify the equation of the line which is parallel to x = 1 and passes through thepoint (2,5).

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Finding equations of parallel lines exercise

Go online

Q40: Identify the line which passes through (0,4) and is parallel to the line with equationy = 3x + 1 in the form y = mx + c.

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Q41: Identify the line which passes through (1,9) and is parallel to the line with equationy = − 3x + 4 in the form y = mx + c.

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Q42: Identify the line which passes through (-5,2) and is parallel to the line with equationy − 2x + 1 = 0 in the form y = mx + c.

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Q43: Identify the line which passes through (3,3) and is parallel to the line with equation5y − 10x = 12 in the form y = mx + c.

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Q44: Identify the line which passes through (2,1) and is parallel to the line with equationy = − 4.

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Q45: Identify the line which passes through (4,3) and is parallel to the line with equationx = 1.

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1.5 Learning points• The formula for the gradient, m, of a straight line requires two points on the line

(x1, y1) and (x2, y2). The formula is m = y2 − y1x2 − x1

• A horizontal straight line has gradient 0.

• A vertical straight line has gradient undefined.

• Parallel lines have the same gradient.

• The equation of a straight line with gradient m and y-intercept (0,c) is y = mx + c.

• The equation of a vertical straight line with x-intercept (k,0) is x = k.

• The equation of a horizontal straight line with y-intercept (0,c) is y = c.

• The equation of a straight line with gradient m and passes through the point (a, b)is y − b = m(x − a).

• To be able to identify the gradient or y-intercept, the equation of a straight linemust be in the form y = mx + c. If the equation is not in that form it must bere-arranged.



1.6 End of topic test

End of topic 13 test

Go online

Gradient of a straight line

Q46:

What is the gradient of the line which passes through the points A( -5,4) and B(2, -2) ?

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Q47:

a) Find the gradient of the line which passes through C(5,7) and D(5,-2).

b) Find the gradient of the line which passes through E(-3,-3) and F (5,9).

c) Find the gradient of the line which passes through G(-2,1) and H(7,1).

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Q48:

a) Find the gradient of the line which passes through K(-4,-1) and L(6,9).

b) Find the gradient of the line which passes through M (5,-3) and N (15,7).

c) Is the line KL parallel with the line MN? (answer yes or no)

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Equations of vertical and horizontal straight lines

Q50:

1. Find the equation of the blue line in the diagram.

2. Find the equation of the red line in the diagram.

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Q51: Find the equation of the straight line passing through A(-3,-6) and B(3,0) in theform y = mx + c.

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Gradient and y-intercept

Q52:

1. Identify the gradient of the line with equation y = 4x − 5.

2. Identify the coordinates of the y-intercept of the line with equation y = 4x − 5.

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Q53:

1. Identify the gradient of the line with equation 2x + y − 3 = 0.

2. Identify the coordinates of the y-intercept of the line with equation 2y − 6x = 10.

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Equation of parallel lines

Q54: Determine the equation of the line passing through (2,1) which is parallel to theline with equation y = 7x − 2 in the form y = mx + c.

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GLOSSARY 29

Glossary

parallel lines

two lines are parallel when they are always the same distance apart and have thesame gradient

x-intercept

the x-intercept is the x coordinate of the point where the line cuts the x-axis


30 ANSWERS: TOPIC 13

Answers to questions and activities

13 The straight line

Finding the gradient of a straight line (page 3)

Q1:

a) m = 3

b) m = 1

c) m = 2

d) m = 3

e) m = 2

Finding the gradient of a straight line practice (page 5)

Q2:

A(3,4) so x1 = 3 and y1 = 4 and B(9,1) so x2 = 9 and y2 = 1 so,

mAB = y2 − y1x2 − x1

= 1 − 49 − 3 = - 3

6 = - 12 or - 1

2

Q3:

E(2,4) so x1 = 2 and y1 = 4 and F (4,2) so x2 = 4 and y2 = 2 so,

mEF = y2 − y1x2 − x1

= 2 − 44 − 2 = −2

2 = −22 = −1

Determining the gradient of a straight line exercise (page 5)

Q4: mAB = -11/3

Q5: 1

Q6: mEF = 9/7

Determining the gradient of horizontal and vertical straight lines practice (page 8)

Q7:

A(3,1) so x1 = 3 and y1 = 1 and B(3,8) so x2 = 3 and y2 = 8 so,

mAB = y2 − y1x2 − x1

= 8 − 13 − 3 = 7

0 = undefined AB is a vertical line.

Q8:

E(2,4) so x1 = 2 and y1 = 4 and F (-2,4) so x2 = − 2 and y2 = 4 so,

mEF = y2 − y1x2 − x1

= 4 − 4(−2) − 2 = 0

−4 = 0 EF is a horizontal line.


ANSWERS: TOPIC 13 31

Determining the gradient of horizontal and vertical straight lines exercise (page9)

Q9:

a) undefined

b) 0

Q10: PQ

Identifying parallel lines practice (page 11)

Q11: a) Parallel

Q12: mGH = y2 − y1x2 − x1

= 5 − ( - 2)7 − ( - 1) = 7

8

mKL = y2 − y1x2 − x1

= 7 − 14 − 0 = 6

4 = 32

since the gradients of GH and KL are not equal, the lines are not parallel.

Identifying parallel lines exercise (page 11)

Q13:

a) 3/2b) 2

c) No

Q14:

a) −4/5

b) −4/5

c) Yes

Equation of a straight line practice (page 15)

Q15:

The line cuts the y-axis at (0,3) ⇒ c = 3

The line passes through (0,3) and (3,0). So we have: m = 3 − 00 − 3 = −1

Remember we could use the coordinates of any two points that lie on the line to calculatethe gradient.

So the equation is y = − x + 3.

Q16:

The line cuts the y-axis at (0,-1) ⇒ c = − 1

The line passes through (0,-1) and (4,-1). So we have: m = (−1) − (−1)4 − 0 = 0

So the equation is y = 0x + (−1). This can be simplified to y = − 1.



Q17:

The line does not cut the y-axis so has no value of c and it is vertical.

The line passes through (-2,0) and (-2,2). So we have: m = 2 − 0(−2) − (−2) =

undefined

All points on the line have an x coordinate of -2 i.e. (-2,. . .). So the equation is x = −2.

Q18:

mCD = ( - 2) − 82 − ( - 3) = - 10

5 = - 2

Using the formula y − b = m(x − a)

Since m = − 2 and the point C(-3,8) would give us a = − 3 and b = 8 we cansubstitute into the formula:y − b = m (x − a)

y − 8 = - 2 (x − ( - 3)) deal with double negative

y − 8 = - 2 (x + 3) expand the brackets

y − 8 = - 2x − 6 get rid of - 8 on the right by + 8

y = - 2x + 2 ( - 6) + 8 = 2

The equation of equation of the straight line is y = − 2x + 2.

Equation of a straight line exercise (page 16)

Q19:

Hint:

• To find the equation of a line, you must first identify the gradient of the line andwhere it crosses the y-axis.

Steps:

• What is the gradient of the line? 1/2

• What are the coordinates of the y-intercept? (0,-5)

Answer: y = 12x − 5

Q20:

Hint:


Steps:

• What is the gradient of the line?−3

• What are the coordinates of the y-intercept? (0,-2)

Answer: y = −3x − 2

Q21:



a) The equation of the horizontal line is y = 1

b) The equation of the vertical line is x = − 3

Q22:

a) y = x + 4

b) y = 0 · 5xc) x = 6

d) y = 2x + 9

e) y = − 5

f) y = 4x + 17

Identifying the y-intercept and gradient of a straight line practice (page 19)

Q23: 5

Q24: (0,-8)

Q25: -3

Q26: (0,0)

Q27: -1

Q28: (0,-6)

Q29: 4

Q30: (0,2)

Identifying the y-intercept and gradient of a straight line exercise (page 20)

Q31:

a) 5

b) 7

c) -3

d) 0

e) undefined

Q32:

a) (0,4)

b) (0,5)

c) (0,6)

d) (0,13)



Equation of parallel lines (page 21)

Q33:

You may have drawn a line which has a gradient of 1 but with a different value of c thanthe example lines shown here.

m = 1The equation of the original line is y = x + 1.

Finding equations of parallel lines practice (page 23)

Q34: y=4x-3

Q35: y=-0.5x+7

Q36: y=-3x-1

Q37: y=x+1

Q38: y=3

Q39: x=2

Finding equations of parallel lines exercise (page 23)

Q40: y = 3x + 4

Q41: y = -3x + 12

Q42: y = 2x + 12

Q43: y = 2x - 3

Q44: y = 1

Q45: x = 4



End of topic 13 test (page 26)

Q46: -6/7

Q47:

a) undefined

b) 2/3

c) 0

Q48:

a) 1

b) 1

c) yes

Q49:

Hint:


Steps:

• What is the gradient of the line? 2/3

• What is the y-intercept? -1

Answer: y = 2/3x − 1

Q50:

1. y = − 4

2. x = 2

Q51: y = x − 3

Q52:

1. 4

2. (0,-5)

Q53:

1. -2

2. (0,5)

Q54: y = 7x − 13


SCHOLAR Study Guide National 5 Mathematics Course ... · The Scottish Qualiﬁcations Authority for...

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