INTRODUCTION TO COORDINATE GEOMETRY
910YEARS
The Improving Mathematics Education in Schools (TIMES) Project NUMBER AND ALGEBRA Module 29
A guide for teachers - Years 910 June 2011
Introduction to Coordinate Geometry
(Number and Algebra : Module 29)
For teachers of Primary and Secondary Mathematics
510
Cover design, Layout design and Typesetting by Claire Ho
The Improving Mathematics Education in Schools (TIMES)
Project 20092011 was funded by the Australian Government
Department of Education, Employment and Workplace
Relations.
The views expressed here are those of the author and do not
necessarily represent the views of the Australian Government
Department of Education, Employment and Workplace Relations.
The University of Melbourne on behalf of the International
Centre of Excellence for Education in Mathematics (ICEEM),
the education division of the Australian Mathematical Sciences
Institute (AMSI), 2010 (except where otherwise indicated). This
work is licensed under the Creative Commons Attribution
NonCommercialNoDerivs 3.0 Unported License. 2011.
http://creativecommons.org/licenses/byncnd/3.0/
Peter Brown
Michael Evans
David Hunt
Janine McIntosh
Bill Pender
Jacqui Ramagge
INTRODUCTION TO COORDINATE GEOMETRY
910YEARS
The Improving Mathematics Education in Schools (TIMES) Project NUMBER AND ALGEBRA Module 29
A guide for teachers - Years 910 June 2011
{1} A guide for teachers
ASSUMED KNOWLEDGE
Fluency with the arithmetic of the rational numbers
Knowledge of ratios
Congruent and similar triangles
Basic algebraic notation
Fluency with algebraic expressions and equations
Basic plotting points in the Cartesian plane including plotting points from a table of values.
MOTIVATION
Coordinate geometry is one of the most important and exciting ideas of mathematics.
In particular it is central to the mathematics students meet at school. It provides a
connection between algebra and geometry through graphs of lines and curves. This
enables geometric problems to be solved algebraically and provides geometric insights
into algebra.
The invention of calculus was an extremely important development in mathematics that
enabled mathematicians and physicists to model the real world in ways that was previously
impossible. It brought together nearly all of algebra and geometry using the coordinate
plane. The invention of calculus depended on the development of coordinate geometry.
CONTENT
It is expected that students have met plotting points on the plane and have plotted points
from tables of values of both linear and non linear functions.
The number plane (Cartesian plane) is divided into four quadrants by two perpendicular
axes called the x-axis (horizontal line) and the y-axis (vertical line). These axes intersect at
a point called the origin. The position of any point in the plane can be represented by an
ordered pair of numbers (x, y). These ordered pairs are called the coordinates of the point.
INTRODUCTION TO COORDINATE GEOMETRY
{2}The Improving Mathematics Education in Schools (TIMES) Project
The point with coordinates (4, 2) has been
plotted on the Cartesian plane shown.
The coordinates of the origin are (0, 0).
Once the coordinates of two points are
known the distance between the two points
and midpoint of the interval joining the points
can be found.
DISTANCE BETWEEN TWO POINTS
Distances are always positive, or zero if the points coincide. The distance from A to B is
the same as the distance from B to A. We first find the distance between two points that
are either vertically or horizontally aligned.
EXAMPLE
Find the distance between the following pairs of points.
a A(1, 2) and B(4, 2) b A(1, 2) and B(1, 3)
SOLUTION
a The distance AB = 4 1 = 3
0x
y
A(1, 2)
B(4, 2)Note: The distance AB is obtained from
the difference of the xcoordinates of
the two points.
b The distance AB = 3 (2) = 5
0x
yB(1, 3)
A(1, 2)
Note: The distance AB is obtained from
the difference of the ycoordinates of the
two points.
0 1
1
11
2
2
3
3
4
4
2
2 (4, 2)
x-coordinate
y-coordinate
3
4
3 4x
y
{3} A guide for teachers
The example above considered the special cases when the line interval AB is either
horizontal or vertical. Pythagoras theorem is used to calculate the distance between two
points when the line interval between them is neither vertical nor horizontal.
The distance between the points A(1, 2) and B(4, 6) is calculated below.
0x
y
B(4, 6)
C(4, 2)A(1, 2)
AC = 4 1 = 3 and BC = 6 2 = 4.
By Pythagoras theorem,
AB2 = 32 + 42 = 25
And so AB = 5
The general case
We can obtain a formula for the length of any interval. Suppose that P(x1, y1) and Q(x2, y2)
are two points.
0x
y
P(x1, y
1) X(x2, y1)
Q(x2, y
2)
x2 x
1
y2 y
1
Form the rightangled triangle PQX, where X is the point (x2, y1),
PX = x2 x1 or x1 x2 and QX = y2 y1 or y1 y2
depending on the positions of P and Q.
By Pythagoras theorem:
PQ2 = PX2 + QX2
= (x2 x1)2 + (y2 y1)2
Therefore PQ = QP = (x2 x1)2 + (y
2 y
1)2
Note that (x2 x1)2 is the same as (x1 x1)2 and therefore it doesnt matter whether we go
from P to Q or from Q to P the result is the same.
{4}The Improving Mathematics Education in Schools (TIMES) Project
EXAMPLE
Find the distance between the points A(4, 3) and B(5, 7).
SOLUTION
In this case, x1 = 4, x2 = 5, y1 = 3 and y2 = 7.
AB2 = (x2 x1)2 + (y2 y1)2
= (5 (4))2 + (7 (3))2
= 92 + 102
= 181
Thus, AB = 181
Note that we could have chosen x1 = 5, x2 = 4, y1 = 7 and y2 = 3 and still obtained the
same result. As long as (x1, y1) refers to one point and (x2, y2) the other point, it does not
matter which one is which.
EXERCISE 1
Show that the distance between the points A(a, b) and B(c, d) is the same as the
distance between
the points P(a, d) and Q(c, b)
the points U(b, a) and V(d, c)
Illustrate both of these.
EXERCISE 2
The distance between the points (1, a) and (4, 8) is 5. Find the possible values of a and use
a diagram to illustrate.
THE MIDPOINT OF AN INTERVAL
The coordinates of the midpoint of a line interval can be found using averages as we will see.
We first deal with the situation where the points are horizontally or vertically aligned.
EXAMPLE
Find the coordinates of the midpoint of the line interval AB, given:
a A(1, 2) and B(7, 2) b A(1, 2) and B(1, 3)
{5} A guide for teachers
SOLUTION
a AB is a horizontal line interval, the
0x
y
A(1, 2)
B(7, 2)
midpoint is at (4, 2), since 4 is halfway
between 1 and 7.
Note: 4 is the average of 1 and 7, that is, 4 = 1 + 72 .
b The midpoint of AB has coordinates 1, 12 .
0x
yB(1, 3)
A(1, 2)
Note that 12 is the average of 3 and 2.
When the interval is not parallel to one of the axes we take the average of the
xcoordinate and the ycoordinate. This is proved below.
0x
y
S
T
x
M(x, y)
2
1 5
y
8
A(1, 2)
B(5, 8)
Let M be the midpoint of the line AB. Triangles AMS and MBT are congruent triangles
(AAS), and so AS=MT and MS=BT.
Hence the xcoordinate of M is the average of 1 and 5.
x = 5 + 12 = 3
The y coordinate of M is the average of 2 and 8.
x = 2 + 82 = 5
Thus the coordinates of the midpoint M are (3, 5).
The general case
We can find a formula for the midpoint of any interval. Suppose that P(x1, y1) and Q(x2, y2)
are two points and let M(x, y) be the midpoint.
{6}The Improving Mathematics Education in Schools (TIMES) Project
P(x1, y
1)
Q(x2, y
2)
0x
y
S
T
x
M(x, y)y
y2
y1
x1
x2
Triangles PMS and MQT are congruent triangles (AAS), and so PS=MT and MS=QT.
Hence the xcoordinate of M is the average of x1 and x2, and ycoordinate of M is the
average of y1 and y2. Therefore
x = x
1 + x
2
2 and y = y
1 + y
2
2
Midpoint of an interval
The midpoint of an interval with endpoints P(x1, y1) and Q(x2, y2) is x
1 + x
2
2 , y
1 + y
2
2 .
Take the average of the xcoordinates and the average of the ycoordinates.
EXAMPLE
Find the coordinates of the midpoint of the line interval joining the points (6, 8) and (3, 2).
SOLUTION
The midpoint has coordinates, 6 + (3)2 , 8 + 2
2 = 32, 5
EXAMPLE
If C(3, 6) is the midpoint of line interval AB and A has coordinates (1, 1), find the
coordinates of B.
SOLUTION
Let the coordinates of B be (x1, y1).
x
1 + (1)
2 = 3 and y
1 + 12 = 6
x1 1 = 6 y1 + 1 = 12
so x1 = 7 so y1 = 11.
Thus B has coordinates (7, 11).
{7} A guide for teachers
EXERCISE 3
A square has vert
