Download File - Ms Burton`s Weebly

Investigation.
Find the distance between two points A(1, 2) and B(3, 6)
y
x-length = 3 – 1 = 2
6
•
•B(3,6)
(6 – 2 )
Form a triangle and use
Pythagoras to find the
distance between the
points
4
2
2 • •A(1,2)
y-length = 6 – 2 = 4
Length = √(22 + 42 = √20
•1
(3 – 1 )
•3
x
Co-ordinate Geometry
Chapter 7
Note 1 : Distance between two points
Let A (x1 , y1) and B (x2 , y2) be two points.
D
A (x1 , y1)

 B (x2 , y2)
y2 – y1
x2 – x1
We often need to find d, the distance between A and
B. This is found using PYTHAGORAS
D = (x 2 - x 1 )2 + (y 2 - y 1 )2
Example:
Find the distance between the points (–3,2) and (3,-6)
(x1, y1)
substitute into the formula
d  (x2  x1 )2  (y2  y1 )2
d  (3  3)2  ( 6  2)2
d  (6)2  ( 8)2
d  100
d = 10 units
(x2, y2)
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.
Theata Page 128
Exercise 16.2
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.
6+10
=8
2
This averaging is a really useful process when the
numbers are not as easy to work with as 6 and 10
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
y
31
 1
2
5.0
4
3
Step 2 Average the y’s
 4 2
 1
2
2
•(1,2)
1
x
-5.0
-4
-3
-2
(–1,–1)
Step 3 M = (–1,–1)
-1
•
1
-1
-2
-3
(–3,–4)
•
-4
-5.0
2
3
4
5.0
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:
 x1  x2 y1  y2 
M
,

2 
 2
Example:
Find the midpoint of the line segment joining
E = (10, -3) and F = (6, 0)
(6 , 0) (x1 , y1)

 x1  x2 y1  y2 
M
,

2 
 2
 6  10 0  -3 
M
,

2 
 2
M = (8, -1.5)
 (10,-3) (x
2
, y2)
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
Theata Page 127
Exercise 16.1
Starter
Find the midpoints of the line segments joining
(6, -2) and (1, 2)
midpoint = (3½, 0)
(3, -2) and (1, -6)
midpoint = (2, -4)
Note 3: Gradient (slope)
• measures the steepness of a line
• is defined as
m
Vertical Rise
Horizontal Run
• 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
RISE
RUN
Examples:
Leans left, so
m is negative!
4 cm
m
2 cm
8
4
2
2
8
Rise
m
Run
m
3
2
m
2
3
8
m    1
8
5
3
5 km
3 km
y
C
y2
Let A (x1 , y1 ) and B (x2 , y2 ) be any
two points
(y
2
y1
(x2 – x1)
 B (x2 , y2 )
- y1)
 A (x1 , y1 )
The GRADIENT of AB is given by
m=
y2  y1
x2  x1
x2
x1
rise is y2 – y1
run is x2 – x1
x
Example:
y
5.0
Find the gradient of the
line joining points A(4, 3)
and B(1, –3)
(x1, y1)
4
A(4,3)
3
•
2
1
M=
y2  y1
x2  x1
-3  3
14
-6
2
-3
-5.0
-4
-3
-2
-1
1
2
3
4
-1
-2
-3
-4
-5.0
•
B(1, –3)
(x2, y2)
C
5.0
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
Page 137
Exercise 7A
Note 4: Revision of Equations of Lines
The gradient/intercept form for the equation of
a line is:
y = mx + c
gradient
y-intercept
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
Another option:
up (+2) then left

(-3)
y
5
4
3
Now
join
the
dots
2
One option: down (-2)
then right (+3)

1






-1
-2
Don’t forget
arrows & label
-3
-4
-5



x


Exercises:
Plot the following graphs;
y = 2x – 3
Y = -⅖x + 2
y=x
The general form of the equation of a line is:
ax + by + c = 0
Equations can be rearranged from gradientintercept form to the general form by
performing operations on both sides of the
equation. (a is usually positive)
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
Exercises:
Write the following equations in the
general form:
y = 4x – 3
4x – y - 3 = 0
Y = -⅖x + 2
2x + 5y - 10 = 0
y = ⅙x – ⅚
x - 6y - 5 = 0
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)
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
Step 3
y – y1 = m(x – x1)
y – 5 = 4(x – – 3)
Remove brackets. Write in general form
y – 5 = 4(x + 3)
y = 4x + 17
y – 5 = 4x + 12
0 = 4x - y + 17
Example 2 Find the equation of the line passing
through (– 2, -13) and (3, 2)
Step 1
Step 2
(x1, y1)
(x2, y2)
Find m
m = 2- -13 = 3
3- -2
Put these values into the formula
y – y1 = m(x – x1)
Step 3
y – -13 = 3(x – – 2)
Remove brackets. Write in general form
Y + 13 = 3(x + 2)
y = 3x - 7
Y + 13 = 3x + 6
0 = 3x - y - 7
Note 5a: Finding Equations of Lines Quickly
For slope
A
B
the form of the line is Ax – By = …
A
For slope - the form of the line is Ax + By = …
B
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
Exercise
7B and C
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.
Find:
 The gradient over this section of the hill climb
m = 1/5
 What is the distance of this section of the hill
climb in metres to the nearest metre?
153m
 What are the co-ordinates of the half-way point
of this section of the hill climb?
(215, 85)
STARTERS:
Graph the following lines:
4x + 12y = 24
y = - 2/3x + 4
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.
Example: Find the equation of the perpendicular
bisector between the points A(-2, 5) and B(4, 9)
Find the midpoint of AB: M = æ -2 + 4 , 5 + 9 ö
ç
÷
è 2
Find the gradient of AB: M =
9-5
4 - -2
2 ø
=⅔
The perpendicular gradient of AB = - 3/2
Equation of perpendicular bisector:
y – 7 = - 3/2 (x – 1)
2y – 14 = - 3 (x - 1)
2y – 14 = - 3x + 3
3x + 2y – 17 = 0
M = (1, 7)
Page 146
Exercise 7D.1 and 7D.2
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
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
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
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
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
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
Note 7: Triangles – Medians of Triangles