1. •B(3,6) •1 •3 Investigation. •A(1,2)
Find the distance between two points A(1, 2) and B(3, 6) 6•
•B(3,6)
Form a triangle and use Pythagoras to find the distance between the points
(6 – 2 ) 4 2
2 •
•A(1,2)
x-length = 3 – 1 = 2 •1 •3
y-length = 6 – 2 = 4
(3 – 1 )
Length = √( = √20

2. Co-ordinate Geometry Chapter 7

3. Note 1 : Distance between two points
Let A (x1 , y1) and B (x2 , y2) be two points.
 B (x2 , y2) D
A (x1 , y1) 
y2 – y1
x2 – x1
We often need to find d, the distance between A and B. This is found using PYTHAGORAS

4. Example:
Find the distance between the points (–3,2) and (3,-6)
(x1, y1)
(x2, y2)
substitute into the formula
d = 10 units

5. Applications
Prove that the vertices A(1, 5), B(2, 9) and C(6, 10) are those of an isosceles triangle.
AB = √17
BC = √17
AC = √50
Because there is two sides with the same length the triangle is isosceles.

6. Theata Page 128
Exercise 16.2

7. Consider the two numbers 6 and 10.
Investigation
Consider the two numbers 6 and 10.
The number exactly halfway between them, their MIDPOINT, is 8.
The midpoint can easily be worked out by counting inwards from 6 and 10, but you can also find the midpoint by averaging the two numbers.
This averaging is a really useful process when the numbers are not as easy to work with as 6 and 10

8. This concept can now be used to find the midpoint of two points on an x-y graph.
Example.
Find the midpoint of (– 3, – 4) and (1, 2)
Step 1
Average the x’s
•(1,2)
Step 2
Average the y’s
(–1,–1) •
Step 3
M = (–1,–1)
(–3,–4) •

9. Note 2: Finding a Midpoint
Let A (x1 , y1) and B (x2 , y2) be two points.
 B (x2 , y2)
 M (? , ?)
 A (x1 , y1)
A general formula that finds the midpoint of any two points is:

10. Example:
Find the midpoint of the line segment joining
E = (10, -3) and F = (6, 0)
(6 , 0) (x1 , y1)
 (10,-3) (x2 , y2)
M = (8, -1.5)

11. Prove that the points A(-1,-2), B(1, 1),
C(8,-1) and D(6,-4) are the vertices of a parallelogram. (HINT: find the lengths of all four sides and the diagonals)
AB = 3.6 units BC = 7.3 units
DC = 3.6 units AD = 7.3 units
AC = 9.1 units BD = 7.1 units
Length of AB = length of CD and length of
BC = length of AD. The lengths of the diagonals are different, so therefore the shape is a parallelogram

12. Theata Page 127
Exercise 16.1

13. Starter
Find the midpoints of the line segments joining
(6, -2) and (1, 2)
(3, -2) and (1, -6)
midpoint = (3½, 0)
midpoint = (2, -4)

14. Note 3: Gradient (slope)
RISE
RUN
measures the steepness of a line
is defined as
is positive if the line leans to the right
is negative if the line leans to the left
is zero if the line is horizontal
is not defined if the line is vertical

15. Examples:
Leans left, so m is negative!
4 cm 8
2 cm 8 3 2
5 km
3 km

16. y2 C
(x2 – x1)
 B (x2 , y2 )
Let A (x1 , y1 ) and B (x2 , y2 ) be any two points
(y2 - y1) y1
 A (x1 , y1 )
The GRADIENT of AB is given by
m = x1 x2
rise is y2 – y1
run is x2 – x1

17. Example:
(x1, y1)
Find the gradient of the line joining points A(4, 3) and B(1, –3)
A(4,3) •
M = • C
B(1, –3)
(x2, y2)

18. Parallel and Perpendicular lines
Parallel lines have the same gradient
Perpendicular lines gradients are the negative reciprocals of each other
m1 x m2 = -1
Example: If line AB has a gradient of 2/3, and line CD is perpendicular to line AB, what is the gradient of CD?
Gradient of CD = -3/2
Points are collinear if they lie on the same line – their gradients are equal

19. Page 137
Exercise 7A

20. Note 4: Revision of Equations of Lines
The gradient/intercept form for the equation of a line is:
y = mx + c
gradient
y-intercept

21. Plot y = -2/3x + 2 using the gradient and intercept method
Example:
Plot y = -2/3x + 2 using the gradient and intercept method
y-intercept (c) = 2
Gradient (m) = -2 3
Fall is 2
Run is 3
Negative sign means it leans to LEFT
Plot the y-intercept
Plot the next point by making a triangle, whose fall is 2 and run is 3
Plot another point in the same manner
Connect the points with a straight line

22. Don’t forget arrows & label
Another option: up (+2) then left (-3) 
One option: down (-2) then right (+3)
Now join the dots 
y = -2/3x + 2
Don’t forget arrows & label

23. Exercises:
Plot the following graphs;
y = 2x – 3
Y = -⅖x + 2
y = x

24. The general form of the equation of a line is:
ax + by + c = 0
Equations can be rearranged from gradient-intercept form to the general form by performing operations on both sides of the equation. (a is usually positive)

25. Example:
Write the following equation in the general form
y = -2/3x + 2
3y = -2x + 6
2x + 3y – 6 = 0
Multiply equation by 3
Move everything to the LHS

26. Write the following equations in the general form:
Exercises:
Write the following equations in the general form:
y = 4x – 3
Y = -⅖x + 2
y = ⅙x – ⅚
4x – y - 3 = 0
2x + 5y - 10 = 0
x - 6y - 5 = 0

27. y – y1 = m (x – x1) Note 5: Finding Equations of Lines If you know
the gradient, m
and
any point, (x1 , y1)
then the equation of the line can be worked out using the formula
y – y1 = m (x – x1)

28. Example 1
Find the equation of the line passing through (– 3, 5) and with gradient 4
Step 1
Write m, x1 and y1
m = 4
x1 = –3
y1 = 5
Step 2
Put these values into the formula
y – y1 = m(x – x1)
y – 5 = 4(x – – 3)
Step 3
Remove brackets. Write in general form
y – 5 = 4(x + 3)
y – 5 = 4x + 12
y = 4x + 17
0 = 4x - y + 17

29. Find the equation of the line passing through (– 2, -13) and (3, 2)
Example 2
(x1, y1)
(x2, y2)
Step 1
Find m
m =
= 3
Step 2
Put these values into the formula
y – y1 = m(x – x1)
y – -13 = 3(x – – 2)
Step 3
Remove brackets. Write in general form
Y + 13 = 3(x + 2)
Y + 13 = 3x + 6
y = 3x - 7
0 = 3x - y - 7

30. Note 5a: Finding Equations of Lines Quickly
For slope the form of the line is Ax – By = …
For slope the form of the line is Ax + By = …
Example: Find the equation of the line which passes through (2,-5) with a gradient of ⅜
The equation is: 3x – 8y = 3(2) – 8(-5)
3x – 8y = 46

31. Exercise
7B and C

32. Starter
A car club has an annual hill climb competition. A cross section of part of the hill they race on is drawn below. The beginning and end points of the section, in metres from the start of the hill are given as co-ordinate pairs.

33. Find:
The gradient over this section of the hill climb
What is the distance of this section of the hill climb in metres to the nearest metre?
What are the co-ordinates of the half-way point of this section of the hill climb?
m = 1/5
153m
(215, 85)

34. STARTERS:
Graph the following lines:
4x + 12y = 24
y = - 2/3x + 4

35. Note 6: Perpendicular Bisector
The perpendicular bisector of AB is the set of all points which are the same distance from A and B.
The perpendicular bisector (or mediator) is a line which is perpendicular to AB and passes through the midpoint of AB.

36. Example: Find the equation of the perpendicular bisector between the points A(-2, 5) and B(4, 9)
Find the midpoint of AB:
Find the gradient of AB: M =
The perpendicular gradient of AB = - 3/2
Equation of perpendicular bisector:
M = (1, 7)
= ⅔
y – 7 = - 3/2 (x – 1)
2y – 14 = - 3 (x - 1)
2y – 14 = - 3x + 3
3x + 2y – 17 = 0

37. Page 146
Exercise 7D.1 and 7D.2

38. Starter
Find the equation of the mediator of AB for the triangle A(-2,3), B(4,0) C(-2,-3)
Find the midpoint of AB
Find the gradient of AB
Find the perpendicular gradient
Solution
Midpoint = (1, 1.5)
Gradient = -0.5
Perpendicular gradient = 2
Equation: y – 1.5 = 2( x – 1 )
= 2x – 2
y = 2x – 0.5 A
midpoint B C

39. Starter
3 darts are thrown at a dart board. The first land in the bulls eye, the 2nd lands 2cm to the left and 1 cm above the bulls eye, the 3rd lands 2cm to the right and 9cm above the bulls eye.
Calculate the distance between the points of the 2nd and 3rd darts
The equation of the line joining these two darts
Find the intersection of this line with its perpendicular bisector
Find the equation on this perpendicular bisector

40. Other Geometrical Terms
Equidistant – find the midpoint
Bisects – cuts into two equal parts
Perpendicular – at right angles
Vertex – corner of angle
Concurrent – pass through the same point
Collinear – lies on the same line

41. Note 7: Triangles – Altitude of Triangles
The perpendicular distance from a vertex to the opposite side of a triangle is called the altitude (or height)
Example: In the triangle A(-4,4), B(2,2) and C(-2,-1), calculate the length of the altitude of the triangle ABC through vertex C.
Find the intersection of CI and AB
Find equation of AB
Find equation of CI
Solve a and b simultaneously

42. Example: In the triangle A(-4,4), B(2,2) and C(-2,-1), calculate the length of the altitude of the triangle ABC through vertex C.
Find the intersection of CI and AB
Find equation of AB
Find equation of CI
Solve a and b simultaneously

43. Note 7: Triangles – Medians of Triangles
A line drawn from a vertex of a triangle to a midpoint of the opposite side is called the median

44. Note 7: Triangles – Medians of Triangles