co ordanate geometry intro and examples,

intro co ordanates

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AS-Level Maths:
Core 1
for Edexcel
C1.5 Coordinate
geometry
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2. © Boardworks Ltd 20052 of 33
C
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© Boardworks Ltd 20052 of 33
The distance between two points
The mid-point of a line segment
Calculating the gradient of a straight line
The equation of a straight line
Parallel and perpendicular lines
Examination-style questions
The distance between two points

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The Cartesian coordinate system
The Cartesian coordinate system is named after the French
mathematician René Descartes (1596 – 1650).
Points in the (x, y) plane are defined by their perpendicular
distance from the x- and y-axes relative to the origin, O.
The x-coordinate, or abscissa,
tells us the horizontal distance
from the y-axis to the point.
The y-coordinate, or ordinate,
tells us the vertical distance from
the x-axis to the point.
The coordinates of a point P are
written in the form P(x, y).

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The distance between two points
Given the coordinates of two points, A and B, we can find the
distance between them by adding a third point, C, to form a
right-angled triangle. We then use Pythagoras’ theorem.

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Generalization for the distance between two points
What is the distance between two general
points with coordinates A(x
1, y
1) and B(x
2, y
2)?
The horizontal distance between the points is . x
2
– x
1
The vertical distance between the points is . y
2 – y
1
Using Pythagoras’ Theorem, the square of the distance
between the points A(x
1
, y
1
) and B(x
2
, y
2
) is
The distance between the points A(x
1
, y
1
) and B(x
2
, y
2
) is
( ) ( )x x y y 
2 2
2 1 2 1
+
( ) ( )x x y y 
2 2
2 1 2 1
+

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Worked example
Given the coordinates of two points we can use the formula
to directly find the distance between them. For example:
What is the distance between the points
A(5, –1) and B(–4, 5)?
A(5, –1) B(–4, 5)
x
1 x
2y
1 y
2
( ) ( )x x y y
2 2
2 1 2 1
+
2 2 2 2
( 4 5) +(5 1) = ( 9) +6    
= 81+36
= 117
=3 13

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The distance between two points
The mid-point of a line segment
Calculating the gradient of a straight line
The equation of a straight line
Parallel and perpendicular lines
Examination-style questions
The mid-point of a line segment

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Finding the mid-point of a line segment

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Generalization for the mid-point of a line
In general, the coordinates of the mid-point of the line
segment joining (x
1
, y
1
) and (x
2
, y
2
) are given by:
,
1 2 1 2
+ +
2 2
x x y y 
 
 
is the mean of the
x-coordinates.
x x
1 2
+
2
is the mean of the
y-coordinates.
y y
1 2
+
2
(x
2, y
2)
(x
1
, y
1
)
x
y
0
,
1 2 1 2
+ +
2 2
x x y y 
 
 

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Finding the mid-point of a line segment
The mid-point of the line segment joining the point (–3, 4)
to the point P is (1, –2). Find the coordinates of the point P.
Let the coordinates of the points P be (a, b). We can then write
(1, –2)
Equating the x-coordinates:
–3 + a = 2
a = 5
Equating the y-coordinates:
4 + b = –4
b = –8
The coordinates of the point P are (5, –8)
,
3+ 4+
=
2 2
a b 
 
 
a3+
=1
2
b4+
= 2
2


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The distance between two points
The mid-point of a line segment
Calculating the gradient of a straight line
The equation of a straight line
Parallel and perpendicular lines
Examination-style questions
Calculating the gradient of a straight line

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Calculating gradients

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x
y
(x
1, y
1)
(x
2, y
2)
0
Finding the gradient from two given points
If we are given any two points (x
1
, y
1
) and (x
2
, y
2
) on a line we
can calculate the gradient of the line as follows:
the gradient =
change in y
change in x
x
2 – x
1
y
2 – y
1
Draw a right-angled triangle
between the two points on
the line as follows:
y y
m
x x


2 1
2 1
the gradient, =

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© Boardworks Ltd 200514 of 33
The distance between two points
The mid-point of a line segment
Calculating the gradient of a straight line
The equation of a straight line
Parallel and perpendicular lines
Examination-style questions
The equation of a straight line

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The equation of a straight line
The equation of a straight line can be written in several forms.
You are probably most familiar with the equation written in the
form y = mx + c.
The value of m tells us the gradient
of the line.
The value of c tells us where the
line cuts the y-axis.
This is called the y-intercept and it
has the coordinates (0, c).
For example, the line y = 3x + 4 has
a gradient of 3 and crosses the
y-axis at the point (0, 4).
x
y
1
m
c
0

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The equation of a straight line
A straight line can be defined by:
one point on the line and the gradient of the line
two points on the line
If the point we are given is the y-intercept and we are also
given the gradient of the line, we can write the equation of that
line directly using y = mx + c. For example:
Using y = mx + c with and c = –4 we can write the
equation of the line as
m
2
5
=
2
5
= 4y x
A line passes through the point (0, –4) and has a
gradient of . What is the equation of the line?
2
5

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Finding the equation of a line
Finding the equation of a line given a point on the line
and the gradient
Suppose we are given the gradient of a line but that the point
given is not the y-intercept. For example:
A line passes through the point (2, 5) and has a
gradient of 2. What is the equation of the line?
Let P(x, y) be any point on the line.
x
y
A(2, 5)
0
We can then write the gradient as
But the gradient is 2 so
y
x


5
2
y
x


5
=2
2
x – 2
y – 5
P(x, y)

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Finding the equation of a line
Rearranging:
y – 5 = 2(x – 2)
y – 5 = 2x – 4
y = 2x + 1
So, the equation of the line passing through the point (2, 5)
with a gradient of 2 is y = 2x + 1.
Now let’s look at this for the general case.
y
x


5
=2
2

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Finding the equation of a line
Suppose a line passes through A(x
1
, y
1
) with gradient m.
Let P(x, y) be any other point on the
line.
x
y
A(x
1
, y
1
)
0
This can be rearranged to give y – y
1
= m(x – x
1
).
The equation of a line through A(x
1, y
1) with gradient m is
y – y
1
= m(x – x
1
)
y y
x x


1
1
The gradient of AP =
So
y y
m
x x


1
1
=
x – x
1
y – y
1
P(x, y)
In general:

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Finding the equation of a line
Finding the equation of a line given two points on the
line
x
y
B(5, 4)
0
A(3, –2)
A line passes through the points A(3, –2) and
B(5, 4). What is the equation of the line?
Let P(x, y) be any other point on the line.
The gradient of AP, m
AP
=
( 2)
3
y
x
 

The gradient of AB, m
AB =
4 ( 2)
5 3
 

But AP and AB are parts of the
same line so their gradients must
be equal.
P(x, y)

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Finding the equation of a line
( 2)
3
y
x
 

4 ( 2)
5 3
 

=
+2
3
y
x
= 3
y + 2 = 3(x – 3)
y + 2 = 3x – 9
y = 3x – 11
So, the equation of the line passing through the points A(3, –2)
and B(5, 4) is y = 3x – 11.
Now let’s look at this for the general case.
Putting m
AP
equal to m
AB
gives the equation

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Finding the equation of a line
Suppose a straight line passes through the points A(x
1
, y
1
) and
B(x
2
, y
2
) with another point on the line P(x, y).
The equation of a line through A(x
1
, y
1
) and B(x
2
, y
2
) is
1 1
2 1 2 1
=
y y x x
y y x x
 
 
x
y
B(x
2
, y
2
)
0
A(x
1, y
1)
P(x, y)
The gradient of AP = the gradient of AB.
So
1 2 1
1 2 1
=
y y y y
x x x x
 
 
Or
1 1
2 1 2 1
=
y y x x
y y x x
 
 

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The equation of a straight line
One more way to give the equation of a straight line is in the
form
ax + by + c = 0.
This form is often used when the required equation contains
fractions. For example, the equation
3 1
4 2
=y x
can be rewritten without fractions as
4y – 3x + 2 = 0.
It is important to note that any straight line can be written in
the form ax + by + c = 0.
In particular, equations of the form x = c can be written in the
form ax + by + c = 0 but cannot be written in the form y = mx + c.

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The distance between two points
The mid-point of a line segment
Calculating the gradient of a straight line
The equation of a straight line
Parallel and perpendicular lines
Examination-style questions
Parallel and perpendicular lines

25. © Boardworks Ltd 200525 of 33
Parallel lines
If two lines have the same gradient they are parallel.
Show that the lines 3y + 6x = 2 and y = –2x + 7 are parallel.
We can show this by rearranging the first equation so that
it is in the form y = mx + c.
3y = –6x + 2
y =
–6x + 2
3
3y + 6x = 2
y = –2x +
2
/
3
The gradient, m, is –2 for both lines and so they are parallel.

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Exploring perpendicular lines

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Perpendicular lines
If the gradients of two lines have a product
of –1 then they are perpendicular.
Find the equation of the perpendicular bisector of the
line joining the points A(–2, 2) and B(4, –1).
The perpendicular bisector of the line AB has to pass through
the mid-point of AB.
In general, if the gradient of a line is m, then the gradient of
the line perpendicular to it is .
1
m

Let’s call the mid-point of AB point M, so
M is the point
2+4 2+( 1)
, =
2 2
  
 
 
1
2
(1, )

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Perpendicular lines
The gradient of the line joining A(–2, 2) and B(4, –1) is
The gradient of the perpendicular bisector of AB is therefore 2.
1
2
=2( 1)y x 
m
AB
=
1 2
=
4 ( 2)
 
 
3
=
6
 1
2

Using this and the fact that it passes through the point
we can use y – y
1
= m(x – x
1
) to write
1
2
(1, )
2 1=4( 1)y x 
2 1=4 4y x 
2 4 +3=0y x
So, the equation of the perpendicular bisector of the line
joining the points A(–2, 2) and B(4, –1) is 2y – 4x + 3 = 0.

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Sketching straight line graphs
Suppose we want to sketch the straight line with the equation
2y + 3x – 12 = 0.
It is sufficient to find two points on the line:
the y-intercept
the x-intercept
To find the y-intercept put x = 0
in the equation of the line:
2y – 12 = 0
y = 6
To find the x-intercept put y = 0
in the equation of the line:
3x – 12 = 0
x = 4
0 x
y
6
4

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The distance between two points
The mid-point of a line segment
Calculating the gradient of a straight line
The equation of a straight line
Parallel and perpendicular lines
Examination-style questions
Examination-style questions

31. © Boardworks Ltd 200531 of 33
Examination-style question
The line l
1
in the following diagram has equation
3x – 4y + 6 = 0
The line l
2 is perpendicular to the line l
1 and passes through
the point (2, 4).
The lines l
1 and l
2 cross the x-axis at the points A and B
respectively.
a)Find the equation of the line l
2.
b)Find the length of AB.
l
1
A0 x
y
l
2
B

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Examination-style question
a) Rearranging the equation of l
1
to the form y = mx + c gives
3x – 4y + 6 = 0
4y = 3x + 6
Using y – y
1 = m(x – x
1) with this gradient and the point (2, 4) we
can write the equation of l
2
as:
3y – 12 = –4x + 8
4x + 3y – 20 = 0
So the gradient of l
1 is .
3
4
y = x +
3
4
3
2
Since l
2
is perpendicular to l
1
its gradient is – .
4
3
4
3y – 4 = – (x – 2)

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Examination-style question
b) The point A lies on the line with equation 3x – 4y + 6 = 0.
When y = 0 we have 3x + 6 = 0
So A is the point (–2, 0).
x = –2
The point B lies on the line with equation 4x + 3y – 20 = 0.
When y = 0 we have 4x – 20 = 0
So B is the point (5, 0).
x = 5
 The length of AB is 5 – (–2) = 7