LINES

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LINES
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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 has the greatest gradient
If you move along the line from left to right and
are climbing, it is a positive gradient.
This one has the smallest gradient
gradient
These are all positive gradients. The steeper
the line, the larger the gradient value.
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We compute the gradient by taking the ratio of
how much the line rises (goes up) and how much
the line runs (goes over)
We could compute the run by looking at the
difference between the x values.
run
(x2, y2)
If we took two points on the line, we could
compute the rise by looking at the difference
between the y values.
y2 - y1
rise
(x1, y1)
x2 - x1
So the gradient or gradient is the rise over the
run
This is the gradient formula
gradient is designated with an m
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This one has the greatest absolute value gradient
If you move along the line from left to right and
are descending, it is a negative gradient.
This one has the smallest 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).
Gradient
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Lets figure out the gradient of this line. We
know it should be a positive number.
(2, 4)
Choose two points on the line.
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(1, 2)
The rise over the run is 2 over 1 which is 2.
Lets compute it with the gradient formula.
2
(0, 0)
(-2, -4)
What if we'd chosen two different points on the
line?
It doesn't matter which two points we pick, we'll
always get a constant ratio of 2 for this line.
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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.
(-4, 3)
(2, 3)
(-1, 3)
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.
In general, the equation of a horizontal line is
y b, where b is the y coordinate of any point
on the line.
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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.
Let's choose the points (-2, 3) and (-2, - 2) and
compute the gradient.
(-2, 3)
(-2, 0)
(-2, -2)
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.
In general, the equation of a vertical line is x
a, where a is the x coordinate of any point on
the line.
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positive gradient
It is easy to remember undefined gradient because
you cant move along from left to right (it is
vertical)
negative gradient
undefined gradient
zero gradient (or no gradient)
It is easy to remember 0 gradient because the
line does not slope at all (it is horizontal)
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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.
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, -2)
Use the point (2, -2) and this general point in
the gradient formula subbing in the gradient we
found.
This is an equation for the line.
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.
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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.
get rid of brackets
get rid of fractions by multiplying by - 2
get the x and y terms on one side (with positive
x term)
general or standard form
constants on the other side
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.
x 0
(0, 1) is on the line
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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).
(x - x1)
Multiply both sides by x - x1
(x - x1)
rearranging a bit we have
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.
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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.
2
1
To find another point on the line, repeat this
process with your new point
(0,?3)
(0,?3)
1
2
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.
(1,?1)
(?1,?5)
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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.
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
(?1,?5)
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Let's take this equation and solve for y.
This form of the equation is called
gradient-intercept form because it contains the
gradient and the y intercept of the line.
gradient-intercept form
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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 gradient-intercept form
by solving for y.
-3x 4
-3x 4
Now plot the y intercept
-4
-4
Change in y
Change in x
From the y intercept, count the gradient
Now that you have 2 points you can draw the line
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Acknowledgement I wish to thank Shawna Haider
from Salt Lake Community College, Utah USA for
her hard work in creating this PowerPoint. www.sl
cc.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
Stephens School Carramar www.ststephens.wa.edu.
au