LINES
The gradient or gradient of a line is a number that tells us
how “steep” the line is and which direction it goes. This one
If you move along
the line from left
to right and are
climbing, it is a
positive gradient.
has the
greatest
gradient
These are all positive gradients.
The “steeper” the line, the larger
the gradient value.
We could compute the
run by looking at the
difference between the x
values.
rise
run
(x2, y2)
y2 - y1
(x1, y1)
x2 - x1
So the gradient or
gradient is the rise
over the run
y2  y1
m
x2  x1
gradient is designated with an m
We compute the
gradient by taking the
ratio of how much the
line rises (goes up) and
how much the line runs
(goes over)
If we took two points on
the line, we could
compute the rise by
looking at the difference
between the y values.
This is the
gradient formula
If you move along
the line from left to
right and are
descending, it is a
negative gradient.
This one has
the greatest
absolute value
gradient
These are all negative gradients.
The “steeper” the line, the larger
the absolute value of the
gradient. (basically this means if
you ignore the negative, the larger
the number, the steeper the line--but in the negative gradient
direction).
Let’s figure out
the gradient of
this line. We
know it should
be a positive
number.

y
(2, 4)
Choose two
points on the line.


1
(1, 2)
2









x

(0, 0)

(-2, -4)


What if we'd chosen two different points on
the line?
4   4 8
m
 2
2   2 4
The rise over the
run is 2 over 1
which is 2. Let’s
compute it with the
gradient formula.
20
m
2
1 0
It doesn't matter which two
points we pick, we'll always get
a constant ratio of 2 for this line.
If we look at any points on this line we see that they all have
a y coordinate of 3 and the x coordinate varies.
Let's choose the points (-4, 3) and
(2, 3) and compute the gradient.
y

(-4, 3) (-1, 3)
(2, 3)
















x

33
0
m
 0
2   4 6
This makes sense because as you go
from left to right on the line, you are
not rising or falling (so zero gradient).
The equation of this line is y = 3 since y is 3 everywhere
along the line.
In general, the equation of a horizontal line is y = b, where b
is the y coordinate of any point on the line.
If we look at any points on this line we see that they all have
a x coordinate of - 2 and the y coordinate varies.

(-2, 3)
y
Let's choose the points (-2, 3) and (-2, - 2)
and compute the gradient.



(-2, 0)






(-2, -2)






x

23
5
m

 undefined
 2   2 0
Dividing by 0 is undefined so we say
the gradient is undefined. You can't go
from left to right on the line since there
isn't a left and right.
The equation of this line is x = - 2 since x is - 2 everywhere
along the line.
In general, the equation of a vertical line is x = a, where a is
the x coordinate of any point on the line.
It is easy to remember 0 gradient
because the line does not slope at
all (it is horizontal)
undefined gradient
zero gradient (or no gradient)
It is easy to
remember
undefined
gradient because
you can’t move
along from left to
right (it is vertical)
We often have points on a line but want to find an equation of the
line. We'll see how to do this by looking at an example. Find the
equation of a line the contains the points (- 2, 4) and (2, - 2).

First let's plot
the points and
graph the line.
y
Now let's compute the
gradient---we know it
will be negative by
looking at the line.


(x, y)

Pick a general
point on the
line, (x, y).








24
6
3
(2, -2) m  2   2   4   2


x


Use the point (2, -2) and
This is an equation for the line.
this general point in the
3
3
y


2
gradient formula
 x2  y 2
 
2
subbing in the gradient
2
x2
we found.
Let's get it in a neater form. If we get rid of brackets and fractions
and get the x and y on one side (with positive x term) and constants
on the other side we'll have standard form.

 


 
If we get rid of brackets and fractions and get the x and y on one side
(with positive x term) and constants on the other side we'll have
standard form.
3

2
x  2  y   2
3
get rid of brackets
 x3 y2
2
3x  6  2 y  4
get rid of fractions by multiplying by - 2
get the x and y terms on one side
(with positive x term)
3x  2 y  6  4
general or
3x 3x 2 y2 y22 standard
constants on the other side
form
Choose any x and sub it in this equation and
solve for y and you will get a point (x, y) that
lies on the line.

y















x

x = 0 30  2 y  2
y 1
(0, 1) is on the line
Let's generalize what we did to get a formula for finding the equation of
a line. Let's call the specific point we know on the line (x1, y2).
Multiply both sides by x - x1
rearranging a bit we have:
y  y1 (x - x1)
(x - x1) m 
x  x1
y  y1  mx  x1 
This is called the point-gradient formula because it will find
the equation of a line when you have a specific point (x1, y2)
on the line and the gradient.
We can also use it when we know two points on the line
because we could find the gradient first and then use
one of the points for the specific point.
Example when you have a point and the gradient
A point on a line and the gradient of the line are given. Find two
additional points on the line.
1,5
+1 +2
To find another
point on the
line, repeat this
process with
your new point
Remember that gradient is the
change in y over the change in x.
The gradient is 2 which can be
made into the fraction
-1 0
(0,3)
m2
(0,3)
2
1
+1 +2
(1,1)(1,5) So this point is on the line also. You can see
that this point is found by changing (adding) 2
to the y value of the given point and changing
(adding) 1 to the x value.
A way to do the last problem using the equation of the line
A point on a line and the gradient of the line are given. Find two
additional points on the line.
1,5
If we find the equation of the line using
the point-gradient formula, we can
easily find additional points on the line
by subbing in various x values and
finding the y values.
x=0
20  y  3
y  3
-1 0
(1,5)
m2
y  y1  mx  x1 
y   5  2x   1
y  5  2x  1 y  5  2 x  2
2x  y  3
2x  y  3
Let's take this equation and solve for y.
y  2x  3
This form of the equation is
called gradient-intercept
form because it contains the
gradient and the y intercept
of the line.
-1 0
y  m x b
gradient-intercept form
Example of given an equation, find the gradient and y intercept
Find the gradient and y intercept of the given equation and graph it.
First let's get this in gradientintercept form by solving for y.
Now plot
the y
intercept
3x  4 y  4  0
-3x
+4
-3x +4
 4 y  3x  4
-4
-4
3
y  x 1
Change in x 4
Now that you
have 2 points
you can draw
the line
y intercept
From the
y intercept,
count the
gradient
gradient
Change in y
y  m x b
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
Shawna has kindly given permission for this resource to be downloaded
from www.mathxtc.com and for it to be modified to suit the Western
Australian Mathematics Curriculum.
Stephen Corcoran
Head of Mathematics
St Stephen’s School – Carramar
www.ststephens.wa.edu.au