Investigation.

Find the distance between two points A(1, 2) and B(3, 6)

x

y

•A(1,2)

•B(3,6)

•1 •3

2 •

6•Form a triangle and use Pythagoras to find the distance between the points

(3 – 1 )

(6 – 2 )

y-length = 6 – 2 = 4

x-length = 3 – 1 = 2

2

4

Length = √(22 + 42 = √20

Co-ordinate Geometry

Chapter 7

Let A (x1 , y1) and B (x2 , y2) be two points.

A (x1 , y1)

B (x2 , y2)

We often need to find d, the distance between A and B. This is found using PYTHAGORAS

y2 – y1

x2 – x1

D

Note 1Note 1 : Distance between two points

Example:

Find the distance between the points (–3,2) and (3,-6)

substitute into the formula

212

212 )y(y)x(xd

22 2)6(3)(3d

22 8)((6)d

100d

d = 10 units

(x1, y1)

(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.

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)

x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

(–3,–4) •

•(1,2)

Step 1Average the x’s

12

13

Step 2Average the y’s

12

24

Step 3 M = (–1,–1)

(–1,–1) •

A general formula that finds the midpoint of any two points is:

2

yy,

2xx

M 2121

Note 2Note 2: Finding a Midpoint

Let A (x1 , y1) and B (x2 , y2) be two points.

A (x1 , y1)

B (x2 , y2)

M (? , ?)

Example:

Find the midpoint of the line segment joining E = (10, -3) and F = (6, 0)

2

yy,

2xx

M 2121

(6 , 0) (x1 , y1)

(10,-3) (x2 , y2)

2-30

,2106

M

M = (8, -1.5)

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)

(3, -2) and (1, -6)

midpoint = (3½, 0)

midpoint = (2, -4)

Note 3: Gradient (slope)• measures the steepness of a line

• 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

• is defined as RunHorizontalRiseVertical

m

RISE

RUN

Examples:

4 cm

2 cm2

2

4m

Run

Risem

8

8

18

8m

2

3

3

2m

5 km

3 km

3

5m

Leans left, so m is negative!

x

y

Let A (x1 , y1 ) and B (x2 , y2 ) be any two points

B (x2 , y2 )

A (x1 , y1 )

C

The GRADIENT of AB is given by

x2 x1

y1

y2 (x2 – x1)

(y2 - y1)

rise is y2 – y1 run is x2 – x1

m =12

12

xxyy

x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

A(4,3) •

•B(1, –3)

Find the gradient of the line joining points A(4, 3) and B(1, –3)

M =

C

Example:

12

12

xxyy

4133-

23-6-

(x2, y2)

(x1, y1)

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

Parallel and Perpendicular lines

Page 137

Exercise 7A

The gradient/intercept form for the equation of a line is:

y = mx + c

gradient y-intercept

Note 4: Revision of Equations of Lines

Example:

Plot y = -2/3x + 2 using the gradient and intercept method

y-intercept (c) = 2Gradient (m) = -2

3

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

Fall is 2Run is 3

Negative sign means it leans to LEFT

Another option: up (+2) then left (-3)

y = - 2/3 x + 2

Don’t forget arrows & label

One option: down (-2) then right (+3)

Now join the dots

x

y

-5

-4

-3

-2

-1

1

2

3

4

5

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 gradient-intercept 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

Y = -⅖x + 2

y = ⅙x – ⅚

4x – y - 3 = 0

2x + 5y - 10 = 0

x - 6y - 5 = 0

Note 5: Finding Equations of Lines

If you know

• the gradient, mand

• any point, (x1 , y1)

then the equation of the line can be worked out using the formula

y – y1 = m (x – x1)

Example 1Find the equation of the line passing through (– 3, 5) and with gradient 4

Step 1

Write m, x1 and y1

m = 4 x1 = –3

Step 2

y1 = 5

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

Example 2Find the equation of the line passing through (– 2, -13) and (3, 2)

Step 1

Find m

m =

Step 2

Put these values into the formula

y – y1 = m(x – x1)

y – -13 = 3(x – – 2) Step

3Remove brackets. Write in general form

Y + 13 = 3(x + 2)

Y + 13 = 3x + 6

y = 3x - 7

(x1, y1)

(x2, y2)

= 3

0 = 3x - y - 7

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

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

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)

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:

Find the gradient of AB: M =

The perpendicular gradient of AB = - 3/2

Equation of perpendicular bisector:

M = (1, 7)

y – 7 = - 3/2 (x – 1)

2y – 14 = - 3 (x - 1)

2y – 14 = - 3x + 3

3x + 2y – 17 = 0

= ⅔

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

SolutionMidpoint = (1, 1.5)Gradient = -0.5Perpendicular gradient = 2Equation: y – 1.5 = 2( x – 1 )

= 2x – 2 y = 2x – 0.5

A

B

C

midpoint

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