Gradient
The gradient of a line is its slope. It is a very important feature of a line
because it tells you how fast things are changing. Look at the graphs
below to find the meaning of gradient in two different situations.
Sue’s journey
Distance
Sue’s line is steeper than
Felix’s so she is travelling at a
faster rate.
Felix’s journey
Time
Ivan’s savings
Dollars
Georgie’s line is going down
so she is losing money, while
Ivan’s line is going up so he is
gaining money.
Georgie’s savings
Time
Reviewing gradient
Some facts you already know about gradients are reviewed below.
y
Lines that go up from left to
right are said to have a positive
gradient, and lines that go
down have a negative gradient.
4
Positive gradient
3
2
1
–4
–3
–2
–1
0
1
2
3
4
5x
–1
–2
–3
Part 1
Midpoint, distance and gradient
Negative gradient
1
y
Lines with the same gradient
are parallel.
4
3
2
1
–4
–3
–2
–1
0
1
2
3
4
5x
–1
–2
–3
All these lines have a gradient of 3.
The gradient can be described using a number. This number not only
indicates whether the line is steep or gradual, but also whether it goes up
or down from left to right. In the past you have calculated this number
using the formula:
Gradient = ( + or )
2
rise
run
PAS5.2.3 Coordinate geometry
y
The rise is the distance you move up
the graph from one point to another.
B
4
rise = 2
3
2
The run is the distance you move
across from one point to the other.
A
run = 4
1
–1
0
1
2
3
4
5
6x
–1
Notice that this diagram is the same one you would use to the find the
distance between the two points A and B. In this section, the sides of the
right-angled triangle are used to find the gradient.
2
4
Gradient = +
=
This line has a positive
slope because it goes
up from left to right.
1
2
But which points on the line do you use? Can you just select any two
points on the line to find the gradient? Complete the following activity
to answer these questions.
Activity – Gradient
Try these.
1
A
6
y
5
4
B
3
C
2
1
–3
–2
–1
0
1
2
3
4
6x
5
–1
–2
–3
–4
–5
–6
Part 1
Midpoint, distance and gradient
D
3
a
Use the points A and B to find the gradient of the line shown in
the graph above. Remember to check its sign.
___________________________________________________
___________________________________________________
b
Use the points C and D to find the gradient of the line.
Remember to check its sign.
___________________________________________________
___________________________________________________
c
Comment on the answers you got in parts a and b.
___________________________________________________
___________________________________________________
d
Select a third pair of points on the line and check that you still
get the same result.
The two points are ____________________________________
Gradient = __________________________________________
___________________________________________________
Check your response by going to the suggested answers section.
Formula for gradient
You have discovered that to find the gradient of a straight line you can
select any two points on the line. But do you really need to draw the
graph? Can you find the gradient simply using the coordinates?
Complete the steps in the following activity to develop a formula for
calculating the gradient of a line passing through any two points.
4
PAS5.2.3 Coordinate geometry
Activity – Gradient
Try this.
2
The diagram below shows no scale on the axes. Two points have
been plotted on the line: A ( x1 , y1 ) and B ( x2 , y2 ) .
y
A ( x 1, y1)
B ( x 2, y2)
x
Use the coordinates of the points to write an algebraic expression
for:
the rise = _______________________________________________
the run = _______________________________________________
Combine these expressions to write a formula for the gradient:
Gradient =
rise
run
=
Check your response by going to the suggested answers section.
And so you can find the gradient of a line using two points on it, without
every having to draw a graph. The formula is:
Part 1
Midpoint, distance and gradient
5
But what about the sign of the gradient? Don’t you need the graph to
decide whether the gradient should be positive or negative?
The example below shows that the formula gives you the sign and the
number without needing to graph, provided you use it carefully,.
Follow through the steps in this example. Do your own working in the
margin if you wish.
Find the gradient of the line that passes through the points (2, 7)
and (6, 1).
Solution
The two solutions below show that you get the same gradient
no matter which point you decide will be ( x1 , y1 ) .
The solutions also show that the sign of the gradient is found by
the formula.
Method 1 – Using (2, 7) as the first point ( x1 , y1 )
Gradient =
=
y1 − y2
x1 − x2
7 −1
2−6
(2, 7)
(6, 1)
6
−4
3
=−
2
=
6
PAS5.2.3 Coordinate geometry
Method 2 – Using (6, 1) as the first point ( x1 , y1 )
Gradient =
y1 − y2
x1 − x2
=
1 − 7
6 − 2
(6, 1)
(2, 7)
−6
4
3
=−
2
=
Notice that the result was the same no matter which point you chose to
use as ( x1 , y1 ) . Also notice that the sign (+ or –) is automatically
determined by the formula. So the line joining (6, 1) and (2, 7) would
slope down because it has a negative gradient.
One final point about this example: gradients are left as improper
fractions because the numerator (top) tells you the rise and the
denominator (bottom) tells you the run. This is one of the rare times
when an improper fraction is a better answer than a mixed numeral.
Activity – Gradient
Try these.
3
Use the formula to calculate the gradient of the line that passes
through each pair of points.
a
(5, 2) and (0, 7)
___________________________________________________
___________________________________________________
___________________________________________________
Part 1
Midpoint, distance and gradient
7
b
(2, 3) and (–1, 2)
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
c
(–6, 10) and (–2, –4)
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
4
a
Use the formula to calculate the gradient of the line that passes
through (3, 5) and (3, –2).
___________________________________________________
___________________________________________________
___________________________________________________
b
Plot the points (3, 5) and (3, –2) and comment on why the
gradient is unusual.
8
AS5.2.3 Coordinate geometry
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
___________________________________________________
5
I used the points (3, 8) and (4, 5) and
got a gradient of 3.
But the correct answer is –3.
Please check my working and tell me
what I did wrong.
8−5
4−3
3
=
1
=3
Gradient =
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
6
Show that the lines AB and CD are parallel given the coordinates
below.
A (5, 2), B (8, –4), C (–10, 4) and D (–7, –2).
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
Part 1
Midpoint, distance and gradient
9
7
(Harder) The gradient of a line is
5
. If one point on the line is
2
(3,1), write the coordinates of another point on the line. (Graph
paper can be found in the additional resources section if needed.)
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
Check your response by going to the suggested answers section.
Gradient is a very important feature of lines and intervals. It tells you
how things are changing. You now have several methods for calculating
the gradient of lines. The formula is often the fastest method, but you
need to be very careful when using it.
10
PAS5.2.3 Coordinate geometry
Activity – Gradient
1
a
Rise = 2, run = 1
therefore gradient = 2.
b
Rise = 8, run = 4
8
4
=2
therefore gradient =
2
c
The two pairs of points gave the same gradient.
d
You could select any pair of points on the line and the gradient
would be the same.
If you cannot work with the algebraic coordinates, put in numbers
and work out what you would do to find the rise and run. Then go
back to working with the pronumerals.
The rise = distance from y2 up to y1
= y1 y2
The run = x1 x2
rise
run
y y
= 1 2
x1 x2
Gradient =
3
Part 1
27
50
5
=
5
= 1
a
Gradient =
b
Gradient =
3 2
2 1
1
=
3
Midpoint, distance and gradient
11
4
10 4
6 2
14
=
4
7
=
2
c
Gradient =
a
Gradient =
5 2
3 3
7
=
0
= error
You cannot divide by zero so you cannot find a value for the
gradient.
b
When you draw in some axes and plot the points, you find they
lie on a vertical line. Vertical lines are said to have no gradient
or an undefined gradient.
5
Lauren mixed the coordinates. She used the y-coordinate from (3, 8)
as her first y-value, but then used the x-coordinates from (4, 5) as the
first x-value.
The correct working is:
85
3 4
3
=
1
= 3
Gradient =
6
or
58
43
3
=
1
= 3
Gradient =
Parallel lines have the same gradient.
2 4
58
6
=
3
= 2
Gradient of AB =
4 2
10 7
6
=
3
= 2
Gradient of CD =
Since the gradients are the same, AB CD . (Remember,
means
‘is parallel to’.)
12
PAS5.2.3 Coordinate geometry
7
There are an infinite number of solutions to this because there are an
infinite number of points on the line. The best way to check your
own answer is to use it in the formula to make sure you get a
5
gradient of .
2
Two methods of finding some answers are shown below.
Method 1 – Using a graph
Plot the point (3, 1). Then draw in a right-angled triangle with a rise
of 5 and a run of 2. Remember to draw the triangle so the line would
have a positive slope.
y
2
(5, 6)
y
6
5
–1
0
1
2
–1
4
rise = 5
3
3
4
5
6x
rise = 5
–2
–3
2
(3, 1)
1
–1
(3, 1)
1
–4
run = 2
0
1
2
3
4
5
6x
(1, –4)
run = 2
–5
–6
–1
From these graphs you can see two answers are (5, 6) and (1, –4).
10 5
= giving
4 2
1
you the points (7, 11) and (–1, –9). Or you could use a rise of 2
2
You could also use a rise of 10 and a run of 4 because
and a run of 1. These equivalent fractions would lead to all the other
points on the line.
Method 2 – Using the formula
Let the missing point be ( x1 , y1 ) .
Gradient =
y1 y2
, therefore:
x1 x2
5 y1 1
=
2 x1 3
Looking at the numerators, if y1 1 = 5 then y1 = 6 .
Part 1
Midpoint, distance and gradient
13
Looking at the denominators: if x1 3 = 2 then x1 = 5 .
So one point is (5, 6).
But any fraction equivalent to
5
will also give the correct answer.
2
For example, you can use any of these equations to find a point.
10 y1 1
=
4 x1 3
14
20 y1 1
=
8
x1 3
1
2 = y1 1
x1 3
1
2
5 y1 1
=
2 x1 3
PAS5.2.3 Coordinate geometry