SCHOLAR Study Guide
National 5 Mathematics
Course Materials
Topic 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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Distributed by the SCHOLAR Forum.
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. . . . . . . . . . .
13.1.2 Parallel lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
10
13.2 Determining the equation of a straight line . . . . . . . . . . . . . . . . . . . .
13.3 Identifying the gradient and the y-intercept of a straight line . . . . . . . . . . .
12
17
13.4 Finding equations of parallel lines . . . . . . . . . . . . . . . . . . . . . . . . .
13.5 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
25
13.6 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
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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.
© H ERIOT-WATT U NIVERSITY
TOPIC 1. THE STRAIGHT LINE
1.1
3
Determining the gradient of a straight line
We can describe how steep a slope is by using the idea of gradient. The bigger the
gradient the steeper the slope.
If the line slopes down from left to
right the gradient will be negative.
If the line slopes up from left to right
the 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 y
coordinates is called the vertical distance.
We can choose any two points on the line by identifying their coordinates as (x 1 , y1 )
and (x2 , y2 ).
Key point
This gives the formula for the gradient of a straight line as:
m =
V ertical distance
Horizontal distance
=
y2 − y1
x2 − x1
Finding the gradient of a straight line
This activity shows how to find the gradient of a line using the gradient formula, m
y2 − y1
x2 − x1 .
© H ERIOT-WATT U NIVERSITY
=
Go online
4
TOPIC 1. THE STRAIGHT LINE
Using the formula for the gradient of a straight line we would get,
m =
y2 − y1
x2 − x1
8−0
9−1
=
=
8
8
= 1
Now try this again by drawing graphs of straight lines using the following sets of
y1
coordinates. Use the equation m = xy22 −
− x1 to 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)
..........................................
Notice we can choose the coordinates of any two points which lie on the line to calculate
the gradient of that line.
..........................................
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 x 1 = 1 and y1 = 5. B has
coordinates (7, -2) so x2 = 7 and y2 = − 2.
mAB =
y2 − y1
x2 − x1
=
−2 − 5
7−1
=
−7
6
or −
7
6
mAB is used to represent the gradient of the line AB.
..........................................
© H ERIOT-WATT U NIVERSITY
TOPIC 1. THE STRAIGHT LINE
5
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 − y1
x2 − x1
=
7−5
3 − (−1)
=
2
4
=
1
2
..........................................
Finding the gradient of a straight line practice
Q2: Calculate the gradient of the line from A(3, 4) to B(9, 1) using the gradient formula.
..........................................
Go online
Q3: Calculate the gradient of the line which passes through E(2, 4) and F (4, 2)?
..........................................
Determining the gradient of a straight line exercise
Q4: Calculate the gradient of the line AB.
Go online
..........................................
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TOPIC 1. THE STRAIGHT LINE
Q5: What is the gradient of the line which passes through the points C(-6,-2) and D(1,3) ?
..........................................
Q6:
Calculate the gradient of the line which passes through E(-4,-5) and F (3,4).
..........................................
© H ERIOT-WATT U NIVERSITY
TOPIC 1. THE STRAIGHT LINE
1.1.1
7
The gradients of horizontal and vertical straight lines.
The gradients of horizontal and vertical straight lines can be calculated using the
gradient formula but extra care must be taken.
Finding the gradients of horizontal straight lines
A horizontal line has y2 − y1 = 0 giving m =
gradient of a horizontal straight line is 0.
y2 − y1
x2 − x1
=
0
x2 − x1
= 0 so the
Go online
..........................................
Finding the gradients of vertical straight lines
A vertical line has x2 − x1 = 0 giving m = xy22
the gradient of a vertical straight line is undefined.
− y1
− x1
=
y2 − y1
0
= undef ined so
The gradient is undefined as it is not possible to divide by 0. Try to divide a number by
0 on your calculator.
..........................................
Key point
The gradient of a horizontal line is 0.
The gradient of a vertical line is undefined.
© H ERIOT-WATT U NIVERSITY
Go online
8
TOPIC 1. THE STRAIGHT LINE
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) so
x2 = − 2 and y2 = 5. The line AB is horizontal therefore m AB = 0. We can check
this using the gradient formula:
mAB =
y2 − y1
x2 − x1
=
5−5
−2 − 3
=
0
−5
= 0
..........................................
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 in
both points is the same. Therefore the gradient of CD is undefined.
We can check this using the gradient formula:
mCD =
y2 − y1
x2 − x1
=
(−2) − 3
(−1) − (−1)
=
−5
0
= undefined
..........................................
Determining the gradient of horizontal and vertical straight lines practice
Q7:
Go online
Calculate the gradient of the line from A(3,1) to B(3,8) using the gradient formula.
..........................................
Q8:
Calculate the gradient of the line which passes through E(2,4) and F (-2,4)?
..........................................
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TOPIC 1. THE STRAIGHT LINE
9
Determining the gradient of horizontal and vertical straight lines exercise
Q9:
Go online
a) Calculate the gradient of the line GH.
b) Calculate the gradient of the line KL.
..........................................
Q10:
Which line has a gradient which is undefined P Q or RS?
..........................................
© H ERIOT-WATT U NIVERSITY
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TOPIC 1. THE STRAIGHT LINE
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 − y1
x2 − x1
=
(−3) − 5
5−1
=
−8
4
= −2
mCD =
y2 − y1
x2 − x1
=
(−3) − 5
3 − (−1)
=
−8
4
= −2
We can prove the lines AB and CD are parallel because the gradients are equal.
..........................................
2.
Problem:
Are the lines on this graph parallel?
Solution:
mEF =
y2 − y1
x2 − x1
=
3 − (−3)
2 − (−1)
=
6
3
= 2
mGH =
y2 − y1
x2 − x1
=
3 − (−3)
5−3
=
6
2
= 3
We can prove the lines AB and CD are not parallel because the gradients are not
equal.
..........................................
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TOPIC 1. THE STRAIGHT LINE
11
Identifying parallel lines practice
Q11: The gradient of the line AB is -1 and the gradient of the line CD is -1. Are these
lines parallel?
Go online
a) Parallel
b) Not parallel
..........................................
Q12: G(-1,-2), H(7,5), K(0,1) and L(4,7). Calculate the gradient of the line GH and the
line KL.
Determine if the lines are parallel.
..........................................
Key point
Parallel lines have the same gradient.
Identifying parallel lines exercise
Q13:
Go online
a) Calculate the gradient of the line V W .
b) Calculate the gradient of the line XY .
c) Are these lines parallel? Yes or No.
..........................................
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.
..........................................
© H ERIOT-WATT U NIVERSITY
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TOPIC 1. THE STRAIGHT LINE
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 gradient
and 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 − y1
x2 − 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 calculate
the gradient.
So the equation is y = 2x + 1.
..........................................
Key point
When we know the gradient, m and the y-intercept, c of a straight line we can
find the equation of the straight line by substituting the value for m and c into
y = mx + c.
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TOPIC 1. THE STRAIGHT LINE
13
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 =
3−3
3−0
= 0
So the equation is y = 0x + 3.
However this can be simplified to y = 3.
..........................................
Key point
The equation of all horizontal straight lines take the form
y = c.
Example
Problem:
Find the equation of the line in the diagram.
© H ERIOT-WATT U NIVERSITY
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TOPIC 1. THE STRAIGHT LINE
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 =
4−0
4−4
= undef ined
The gradient is undefined and there is no value of c so we have nothing to substitute
into 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.
..........................................
Key point
The equation of all vertical straight lines take the form
x = . . ..
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) using
the 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 into
the equation giving:
y − b = m (x − a)
y − 7 = 2 (x − 5)
y − 7 = 2x − 10
y = 2x − 3
expand the brackets
get rid of - 7 from the left by + 7
( - 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.
..........................................
© H ERIOT-WATT U NIVERSITY
TOPIC 1. THE STRAIGHT LINE
15
Top tip
If we know the gradient and the y-intercept use:
y = mx + c
to 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
Q15: Find the equation of the line in the diagram in the form y = mx + c.
Go online
..........................................
Q16: Find the equation of the line in the diagram.
..........................................
© H ERIOT-WATT U NIVERSITY
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TOPIC 1. THE STRAIGHT LINE
Q17: Find the equation of the line in the diagram.
..........................................
Q18: Find the equation of the line which passes through C(-3,8) and D(2,-2).
..........................................
Equation of a straight line exercise
Q19: Find the equation of the line in the diagram in the form y = mx + c.
Go online
The equation of the line is y = ?
..........................................
Q20: Find the equation of the line in the diagram in the form y = mx + c.
© H ERIOT-WATT U NIVERSITY
TOPIC 1. THE STRAIGHT LINE
The equation of the line is y = ?
..........................................
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?
..........................................
Q22:
a) Find the equation of the line passing through A(-3,1) and B(2,6) in the form
y = mx + c.
b) Find the equation of the line passing through C(-2,1) and D(4,-2) in the form
y = 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 form
y = 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) in
the form
y = mx + c.
..........................................
1.3
Identifying the gradient and the y-intercept of a straight
line
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;
© H ERIOT-WATT U NIVERSITY
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18
TOPIC 1. THE STRAIGHT LINE
•
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.
..........................................
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).
..........................................
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.
..........................................
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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TOPIC 1. THE STRAIGHT LINE
19
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.
..........................................
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 · 5
We can see that m = − 2 so the gradient is -2.
..........................................
Key point
The equation of a straight line must be written in the form
y = mx + c before we can identify the gradient and the
y-intercept.
Identifying the y-intercept and gradient of a straight line practice
Q23: Identify the gradient of the line with equation y = 5x − 1.
..........................................
Q24: Identify the coordinates of the y-intercept of the line with equation y = 5x − 8.
..........................................
Q25: Identify the gradient of the line with equation y = − 3x + 2.
..........................................
Q26: Identify the coordinates of the y-intercept of the line with equation y = − 3x.
..........................................
Q27: Identify the gradient of the line with equation y + x = 7.
..........................................
© H ERIOT-WATT U NIVERSITY
Go online
20
TOPIC 1. THE STRAIGHT LINE
Q28: Identify the coordinates of the y-intercept of the line with equation
y − 4x + 6 = 0.
..........................................
Q29: Identify the gradient of the line with equation 3y − 12x = 2.
..........................................
Q30: Identify the coordinates of the y-intercept of the line with equation y = 2.
..........................................
Identifying the y-intercept and gradient of a straight line exercise
Identifying the gradient
Go online
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
..........................................
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
..........................................
© H ERIOT-WATT U NIVERSITY
TOPIC 1. THE STRAIGHT LINE
1.4
21
Finding equations of parallel lines
We already know that parallel lines have the same gradient so it is only the y-intercept
which changes.
Equation of parallel lines
Q33:
Go online
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 value
for m but increase or decrease the value of c.
..........................................
Now try this again by drawing graphs of parallel lines using the following equations of
straight 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.
..........................................
Examples
1.
Problem:
Identify the equation of the line which is parallel to y = 2x + 3 and passes through the
point (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.
..........................................
© H ERIOT-WATT U NIVERSITY
22
TOPIC 1. THE STRAIGHT LINE
2.
Problem:
Identify the equation of the line which is parallel to y = − x + 3 and passes through
the point (2,5).
Solution:
This time the parallel line has gradient -1 so m parallel = − 1.
We don’t know the y-intercept but we do have a point on the parallel line (2,5) so by
using 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.
..........................................
3.
Problem:
Identify the equation of the line which is parallel to y − 3x + 4 = 0 and passes through
the 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 m parallel = 3.
We don’t know the y-intercept but we do have a point on the parallel line (-6,4) so by
using m = 3, a = − 6 and b = 4 we can find its equation by substituting into
y − 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.
..........................................
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.
..........................................
© H ERIOT-WATT U NIVERSITY
TOPIC 1. THE STRAIGHT LINE
23
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 in
this case (4, . . .).
So the equation of the parallel line is x = 4.
..........................................
Finding equations of parallel lines practice
Q34: Identify the equation of the line which is parallel to y = 4x + 1 which passes
through (0,-3).
Go online
..........................................
Q35: Identify the equation of the line which is parallel to y = − 0 · 5x − 2 which passes
through (2,6).
..........................................
Q36: Identify the equation of the line which is parallel to y + 3x − 1 = 0 which passes
through (1,-4).
..........................................
Q37: Identify the equation of the line which is parallel to 4y − 4x + 15 = 0 which
passes through (-2,-1).
..........................................
Q38: Identify the equation of the line which is parallel to y = − 4 and passes through
the point (0,3).
..........................................
Q39: Identify the equation of the line which is parallel to x = 1 and passes through the
point (2,5).
..........................................
Finding equations of parallel lines exercise
Q40: Identify the line which passes through (0,4) and is parallel to the line with equation
y = 3x + 1 in the form y = mx + c.
..........................................
Q41: Identify the line which passes through (1,9) and is parallel to the line with equation
y = − 3x + 4 in the form y = mx + c.
..........................................
© H ERIOT-WATT U NIVERSITY
Go online
24
TOPIC 1. THE STRAIGHT LINE
Q42: Identify the line which passes through (-5,2) and is parallel to the line with equation
y − 2x + 1 = 0 in the form y = mx + c.
..........................................
Q43: Identify the line which passes through (3,3) and is parallel to the line with equation
5y − 10x = 12 in the form y = mx + c.
..........................................
Q44: Identify the line which passes through (2,1) and is parallel to the line with equation
y = − 4.
..........................................
Q45: Identify the line which passes through (4,3) and is parallel to the line with equation
x = 1.
..........................................
© H ERIOT-WATT U NIVERSITY
TOPIC 1. THE STRAIGHT LINE
1.5
Learning points
•
The formula for the gradient, m, of a straight line requires two points on the line
y1
(x1 , y1 ) and (x2 , y2 ). The formula is m = xy22 −
− 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 line
must be in the form y = mx + c. If the equation is not in that form it must be
re-arranged.
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TOPIC 1. THE STRAIGHT LINE
1.6
End of topic test
End of topic 13 test
Gradient of a straight line
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Q46:
What is the gradient of the line which passes through the points A( -5,4) and B(2, -2) ?
..........................................
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).
..........................................
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 M N ? (answer yes or no)
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TOPIC 1. THE STRAIGHT LINE
Equation of a straight line
Q49: Find the equation of the line in the diagram in the form y = mx + c.
The equation of the line is y = ?
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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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Equation of a straight line
Q51: Find the equation of the straight line passing through A(-3,-6) and B(3,0) in the
form y = mx + c.
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TOPIC 1. THE STRAIGHT LINE
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 the
line with equation y = 7x − 2 in the form y = mx + c.
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GLOSSARY
Glossary
parallel lines
two lines are parallel when they are always the same distance apart and have the
same gradient
x-intercept
the x-intercept is the x coordinate of the point where the line cuts the x-axis
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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 − y1
x2 − x1
=
1−4
9−3
=
-3
6
=
-1
2
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 − y1
x2 − x1
=
2−4
4−2
=
−2
2
= − 22 = −1
Determining the gradient of a straight line exercise (page 5)
Q4:
mAB =
Q5:
1
Q6:
mEF =
-11 /
3
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 − y1
x2 − x1
=
8−1
3−3
=
7
0
= undef ined 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 − y1
x2 − x1
=
4−4
(−2) − 2
=
0
−4
= 0 EF is a horizontal line.
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ANSWERS: TOPIC 13
31
Determining the gradient of horizontal and vertical straight lines exercise (page
9)
Q9:
a) undefined
b) 0
Q10: P Q
Identifying parallel lines practice (page 11)
Q11: a) Parallel
y2 − y1
x2 − x1 =
y2 − y1
7−1
x2 − x1 = 4 − 0
Q12: mGH =
5 − ( - 2)
7
7 − ( - 1) = 8
= 64 = 32
mKL =
since the gradients of GH and KL are not equal, the lines are not parallel.
Identifying parallel lines exercise (page 11)
Q13:
a) 3 /2
b) 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 =
3−0
0−3
= −1
Remember we could use the coordinates of any two points that lie on the line to calculate
the 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
So the equation is y = 0x + (−1). This can be simplified to y = − 1.
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= 0
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ANSWERS: TOPIC 13
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
undef ined
2−0
(−2) − (−2)
=
All points on the line have an x coordinate of -2 i.e. (-2,. . .). So the equation is x = − 2.
Q18:
mCD =
( - 2) − 8
2 − ( - 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 =
substitute into the formula:
y − b = m (x − a)
− 3 and b = 8 we can
y − 8 =
- 2 (x − ( - 3))
y − 8 =
- 2 (x + 3)
y − 8 =
- 2x − 6
get rid of - 8 on the right by + 8
y =
- 2x + 2
( - 6) + 8 = 2
deal with double negative
expand the brackets
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 and
where 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 =
1
2x
− 5
Q20:
Hint:
•
To find the equation of a line, you must first identify the gradient of the line and
where it crosses the y-axis.
Steps:
•
What is the gradient of the line?−3
•
What are the coordinates of the y-intercept? (0,-2)
Answer: y = −3x − 2
Q21:
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ANSWERS: TOPIC 13
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 · 5x
c) 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)
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ANSWERS: TOPIC 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 than
the example lines shown here.
m = 1
The 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
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ANSWERS: TOPIC 13
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:
•
To find the equation of a line, you must first identify the gradient of the line and
where it crosses the y-axis.
Steps:
•
What is the gradient of the line? 2 /3
•
What is the y-intercept? -1
Answer: y =
2/ x
3
− 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
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