Coordinate Geometry - Carl Schurz High School

[Pages:43]8

Coordinate Geometry

Negative gradients: m 0

Chapter Contents

8:01 The distance between two points 8:02 The midpoint of an interval 8:03 The gradient of a line 8:04 Graphing straight lines 8:05 The gradient?intercept form of a straight

line: y = mx + c Investigation: What does y = mx + c tell us? 8:06 The equation of a straight line, given point and gradient

8:07 The equation of a straight line, given two points

8:08 Parallel and perpendicular lines 8:09 Graphing inequalities on the number plane

Fun Spot: Why did the banana go out with a fig? Mathemathical Terms, Diagnostic Test, Revision Assignment, Working Mathematically

Learning Outcomes

Students will be able to: ? Find the distance between two points. ? Find the midpoint of an interval. ? Find the gradient of an interval. ? Graph straight lines on the Cartesian plane. ? Use the gradient-intercept form of a straight line. ? Find the equation of a straight line given a point and the gradient, or two points on the line. ? Identify parallel and perpendicular lines. ? Graph linear inequalities on the Cartesian plane.

Areas of Interaction

Approaches to Learning (Knowledge Acquisition, Problem Solving, Communication, Logical Thinking, IT Skills, Reflection), Human Ingenuity

200

The French mathematician Ren? Descartes first introduced the number plane. He realised that using two sets of lines to form a square grid allowed the position of a point in the plane to be recorded using a pair of numbers or coordinates.

Coordinate geometry is a powerful mathematical technique that allows algebraic methods to be used in the solution of geometrical problems.

In this chapter, we will look at the basic ideas of: ? the distance between two points on the number plane ? the midpoint of an interval ? gradient (or slope) ? the relationship between a straight line and its equation.

We shall then see how these can be used to solve problems.

8:01 | The Distance Between Two Points

The number plane is the basis of coordinate geometry, an important branch of mathematics. In this chapter, we will look at some of the basic ideas of coordinate geometry and how they can be used to solve problems.

1 Which of the following is the correct statement of Pythagoras' theorem for the triangle shown?

A a2 = b2 + c2

B b2 = a2 + c2

C c2 = a2 + b2

c

b

For questions 2 to 4, use Pythagoras' theorem to find the value of d.

a

2

d cm

3 cm

3

5 cm

12 cm d cm

4

d m 2 m

4 cm

4 m

prep qu

8:01

iz

201 CHAPTER 8 COORDINATE GEOMETRY

5

A

B

?1 0 1 2 3 4 5 6 7 8 x

Distance AB = . . . units.

6y

2B 1 0 ?1 A ?2 ?3

Distance AB = . . . units.

7 Distance AB = . . . units.

y 4

3 2

A

B

1

-1-10 1 2 3 4 5 x

8

y

1

?1?10

1

23 A

4 5x

?2

?3

?4

B

Distance AB = . . . units.

9 Find the distance AB.

y 4 3A

B

2

4x

10 Find the distance AB.

y 4

A3

B

?1

3x

Pythagoras' theorem can be used to find the distance between two points on the number plane.

worked examples

1 Find the distance between the points (1, 2) and (4, 6). 2 If A is (-2, 2) and B is (4, 5) find the length of AB.

Solutions

1

y

7

A(4, 6)

6

5

4

4

3 2

3C

1 B(1, 2)

2

y

7

6

5

4

B(?2, 2) 3

2

1

A(4, 5)

3

6

C

?1 0 1 2 3 4 5 6 7 x ?1

?3 ?2 ?1 0 1 2 3 4 5 x ?1

45 is a surd. We simplify surds if they are perfect squares.

c2 = a2 + b2 AB2 = AC2 + BC2

= 42 + 32 = 16 + 9 = 25 AB = 25 the length of AB is 5 units.

c2 = a2 + b2 AB2 = AC2 + BC2

= 32 + 62 = 9 + 36 = 45 AB = 45 the length of AB is

45 unit.

By drawing a right-angled triangle we can use Pythagoras' theorem to find the

distance between any two points on the number plane.

y

y

y

B A

x

B

AC x

B BC

AC

AC x

202 INTERNATIONAL MATHEMATICS 4

Distance formula

A formula for finding the distance between two points, A(x1, y1) and B(x2, y2), can be found using Pythagoras' theorem. We wish to find the length of interval AB.

Now LM = x2 - x1 (since LM = MO - LO) AC = x2 - x1 (ACML is a rectangle)

and RS = y2 - y1 (since RS = RO - SO) BC = y2 - y1 (BCSR is a rectangle)

Now AB2 = AC2 + BC2 (Pythagoras' theorem) = (x2 - x1)2 + (y2 - y1)2

AB = (x2 ? x1)2 + (y2 ? y1)2

y y2 R

y2 - y1 y1 S

B(x2, y2) A(x1, y1) C

OL

M

0

x1 x2 - x1 x2 x

The distance AB between A(x1, y1) and B(x2, y2) is given by: d = (x2 ? x1)2 + (y2 ? y1)2

worked examples

1 Find the distance between the points (3, 8) and (5, 4).

Solutions

1 Distance = (x2 ? x1)2 + (y2 ? y1)2

2 Find the distance between the points (-2, 0) and (8, -5)

2 Distance = (x2 ? x1)2 + (y2 ? y1)2

(x1, y1) = (3, 8) and (x2, y2) = (5, 4)

(x1, y1) = (-2, 0) and (x2, y2) = (8, -5)

d = (5 ? 3)2 + (4 ? 8)2

d = (8 ? ?2)2 + (? 5 ? 0)2

= (2)2 + (?4)2

= (10)2 + (?5)2

= 4 + 16

= 100 + 25

= 20

Distance 4?47 (using a calculator to answer to 2 decimal places).

= 125

Distance 11?18 (using a calculator to answer to 2 decimal places).

? You should check that the formula will still give the same answer if the coordinates

are named in the reverse way. Hence, in example 1, if we call (x1, y1) = (5, 4) and (x2, y2) = (3, 8), we would produce the same answer.

203 CHAPTER 8 COORDINATE GEOMETRY

Exercise 8:01

Foundation Worksheet 8:01

Distance between points

1 Use Pythagoras' theorem to find the length

1 Use Pythagoras' theorem to find the length of each of the of the hypotenuse in each of the following.

following. (Leave your answer as a surd, where necessary.) a

x

b 53

y

a

b

y

y

12

7

10 9

B(7, 9)

10 9

8

8

2 Find the distance AB in each of the following.

ay

by

A

B

5

A

7 6

7

A(3, 6)

6

24 6 x 2

B

5

8

4

3

2 1

6 A(1, 1)

C(7, 1)

0 1 23 4 56 7 8 x

5

4

3

3 2

C(3, 3)

4

1

B(7, 3)

0 1 23 4 56 7 8 x

x

3 Find the length of AB in each of the following.

ay

B(5, 5)

b

y

A(-2, 2)

4

2

A

2 (1, 2) 24 x

-2

23x

-2 B(3, -2)

cy

4 3

B(4, 3)

2

1

5

?1 0 1 2 3 4 5 x

?1 ?2

3C

?3 A(1, ?2)

d

y

5

B(-3, 3) 4 3

2

2

C

51

-4 -3 -2 -1-10

-2

A(2, 1) 1 2 3x

e

y

A(-6, 2)

3 2

1

-6 5

-5

-4

-3

-2

-1-10

-2

C

12

-3 -4

1 2 3 4 5 6x B(6, -3)

2 Find the lengths BC and AC and use these to find the lengths of AB. (Leave your answers in

surd form.)

a

y

4

B(5, 4)

b

y

4

B(4, 4)

c

B(-7, 4)

y 4

3

3

3

2

1 A(0, 0)

C

-1-10 1 2 3 4 5 x

2

1 A(1, 1) C -1-10 1 2 3 4 5 x

C

2 A(-1, 2)1

-7 -6 -5 -4 -3-2 -1-10 1 x

d

y

5

A(4, 4)

4

3

2

1

-3 -2 -1-01 B(-3, -2-)2

1 2 3 4x C

e

y

B(-2, 3) 4

3

2

1

-3 -2 -1-01 -2

C -3

1 2 3 4x A(2, -3)

f

y

2

-4 -3 -2 -1-01

-2

-3

B(-4,

-4)

-4 -5

1 2 3x A(2, -1)

C

204 INTERNATIONAL MATHEMATICS 4

3 Use Pythagoras' theorem to find the length of interval AB in each of the following.

(Leave answers in surd form.)

a

y

6

A(3, 6)

b

y

5 A(4, 5)

c

y

5

5 4 3 2 1 B(1, 2)

-1-10 1 2 3 4 5 x

4 3 2 1

-1-10 1 2 3 4 5 x

-2

B(1, -2)

4 3 B(-2, 1)2 1

A(5, 3)

-2 -1-10 1 2 3 4 5 x -2

d

y

5

A(-2, 3) 4 3

2

1

-2 -1-10 -2

B(3, 1) 1 2 3 4x

e

y

5 A(1, 3) 4

3

2

1

-3 -2 -1-10 1 2 x B(-3, -2-)2

fy

3 2

1

A(8, 1)

-10 -2

1

2 3 4 5 6 7 8x B(2, -1)

-3

4 Use the formula d = (x2 ? x1)2 + (y2 ? y1)2 to find the distance between the points:

a (4, 2) and (7, 6)

b (0, 1) and (8, 7)

c (-6, 4) and (-2, 1)

d (-2, -4) and (4, 4)

e (-6, 2) and (6, 7)

f (4, 9) and (-1, -3)

g (3, 0) and (5, -4)

h (8, 2) and (7, 0)

i (6, -1) and (-2, 4)

j (-3, 2) and (-7, 3)

k (6, 2) and (1, 1)

l (4, 4) and (3, 3)

5 a Find the distance from the point (4, 2) to the origin. b Which of the points (-1, 2) or (3, 5) is closer to the point (3, 0)? c Find the distance from the point (-2, 4) to the point (3, -5). d Which of the points (7, 2) or (-4, -4) is further from (0, 0)?

Making a sketch will help.

6 a The vertices of a triangle are A(0, 0), B(3, 4) and C(-4, 5). Find the length of each side.

b ABCD is a parallelogram where A is the point (2, 3), B is (5, 5),

C is (4, 3) and D is (1, 1). Show that the opposite sides of the

parallelogram are equal.

c Find the length of the two diagonals of the parallelogram in part b. d EFGH is a quadrilateral, where E is the point (0, 1), F is (3, 2), G is (2, -1) and

H is (-1, -2). Prove that EFGH is a rhombus. (The sides of a rhombus are equal.) e (3, 2) is the centre of a circle. (6, 6) is a point on the circumference. What is the radius

of the circle? f Prove that the triangle ABC is isosceles if A is (-2, -1), B is (4, 1) and C is (2, -5).

(Isosceles triangles have two sides equal.) g A is the point (-13, 7) and B is (11, -3). M is halfway between A and B. How far is

M from B?

205 CHAPTER 8 COORDINATE GEOMETRY

pr

8:02 | The Midpoint of an Interval

ep quiz

8:02

1 4-----+--2---1---0--

2 ?----2---2-+-----4--

3 What is the average of 4 and 10?

4 What is the average of -2 and 4?

5 What number is halfway between 4 and 10?

1 2 3 4 5 6 7 8 9 10 1112 x

6 What number is halfway between -2 and 4?

?4 ?3 ?2 ?1 0 1 2 3 4 5 6 7 x

7 What number is

y

halfway between

6

1 and 5?

5

4

8 1-----+2-----5-- = ?

3 2

1

0

9 What number is

y

halfway between

4

-1 and 3?

3

2

10 ?-----1---2--+-----3- = ?

1 0

?1

?2

? The midpoint of an interval is the halfway position. If M is the midpoint of AB then it will be halfway between A and B.

y

8

B(10, 7)

7

6

5

M(p, q)

4

3 2

A(4, 3)

1

0 1 2 3 4 5 6 7 8 9 10 x

If M is the midpoint of AB then AM = MB.

Consider the x-coordinates.

Note: 7 is halfway between 4 and 10.

The average of 4 and 10 is 7.

Consider the y-coordinates.

Note: 5 is halfway between 3 and 7.

The average of 3 and 7 is 5.

4 + 10 -------2--------

=

7

3-----+2-----7-- = 5

Formula: p = x---1----+2-----x---2-

Formula: q = -y---1---+2-----y---2-

206 INTERNATIONAL MATHEMATICS 4

Midpoint formula

y y2

y1

B(x2, y2) M(p, q)

A(x1, y1)

M =

x1 + x2, 2

y1 + y2 2

Could you please say that in English,

Miss?

0

x1

x2 x

The midpoint, M, of interval AB, where A is (x1, y1) and B is (x2, y2), is given by:

M = x----1----+2----x----2-, -y--1----+2-----y---2- .

worked examples

1 Find the midpoint of the interval joining (2, 6) and (8, 10).

Solutions

1 Midpoint = -x--1----+2-----x---2-, -y---1---+2-----y---2-

= 2-----+2-----8--, 6-----+--2---1---0--

= (5, 8)

2 Find the midpoint of interval AB, if A is the point (-3, 5) and B is (4, -2).

2 Midpoint = x---1----+2-----x---2-, y----1---+2-----y---2-

= ?----3---2-+-----4--, -5----+--2---?---2--

=

(

1 2--

,

3 2--

)

or

(

1 2--

,

1

1 2--

)

Exercise 8:02

1 Use the graph to find the midpoint of each interval.

ay

5 (1, 4) 4

b

y

(-3, 1) 2

1

3

2

1

(5, 2)

-1 1 2 3 4 5 x

-3 -2 --11 -2 -3

1 2 3x (1, -3)

Foundation Worksheet 8:02

Midpoint

1 Read the midpoint of the interval

AB from the graph

2 Find the midpoint of the interval

that joins:

a (3, 4) to (10, 8)

b...

3 Find the midpoint of the interval

that joins: a (- 4, 6) to (- 3, - 5) b . . .

207 CHAPTER 8 COORDINATE GEOMETRY

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