Lecture 1 (Part A) Factorization

Analytical

Geometry

Grade 10 CAPS Mathematics

Video Series

Outcomes for this Video

In this DVD you will:

• Revise factorization.

LESSON 1.

• Revise simplification of algebraic fractions.

LESSON 2.

• Discuss when trinomials can be factorized.

LESSON 3.

2

In this video the focus will be on the distance

between two points, the gradient of a line / line

segment and the mid-point of a line segment.

This will be done via:

Informal investigations (Lesson 1)

Derivation of relevant formulae (Lesson 2)

Applications linked to polygons (Lesson 3)

Informal

Investigations


Video Series

Lesson 1

Outcomes for Lesson 1

In this lesson we will informally investigate the:

• Distance between two points

• Gradient of the line segment connecting the

two points

• Coordinates of the mid-point of the line

segment joining two given points

Line segment from origin to a point: Positive Gradient

Consider the point 3; 5 in the third quadrant.P

5 5Gradient 0

3 3

y

x

2 2 2 2 2 Pythagoras

with 5 and 3

OP OA AP BP OB

AP OB OA BP

2 2 2 2

Length of :

3 5

9 25 34

OP

OP OA AP

1 1 1 ;0 is midpoint of 1

2 2

1 1 0; 2 is midpoint of 2

2 2

M

M

D OA x

C OB y

1 1Mid-point of is 1 ; 2

2 2OP M

Line segment from origin to a point: Negative Gradient

Consider the point 4;5 in the second quadrant.P

5 5Gradient 0

4 4

y

x


with 5 and 4

OP OA AP BP OB

AP OB OA BP

2 2 2 2

Length of :

4 5

16 25 41

OP

OP OA AP

2;0 is midpoint of 2

1 1 0;2 is midpoint of 2

2 2

M

M

D OA x

C OB y

1Mid-point of is 2;2

2OP M

Line segment between two points: Zero Gradient

Consider the points 2;3 and 3;3 .P R

5PR

gradient is 0

1;3 is mid-point of

2M PR

0Gradient 0

5

y

x

Line segment between two points: Gradient Undefined

Consider the points 2;3 and 2; 2 .P R

5Gradient

0

y

x

5PR

gradient is undefined

12; is mid-point of

2M PR

Tutorial 1: Informal Investigations

PAUSE VIDEO

• Do Tutorial 1

• Then View Solutions

In each of the examples:

Complete an appropriate sketch in the Cartesian Plane;

Determine informally:

Gradient of segment joining the two given points.

Distance between the two points.

Mi

a

b

c d-point of the segment joining the two points.

1 0;0 and 5; 3

2 5;3 and 1; 2

3 5;4 and 5; 2

4 3; 3 and 4; 3

O P

P R

P R

P R

Tutorial 1: Problem 1: Suggested Solution

1 0;0 and 5; 3O P 3Gradient 0

5

y

x

2 2 2 25 3

25 9 34

OP OA AP

1 1 2 ;0 is midpoint of 2

2 2

1 1 0; 1 is midpoint of 1

2 2

M

M

D OA x

C OB y

1 1Mid-point of is 2 ; 1

2 2OP M

Line segment between any two points: Negative Gradient

Consider the points 2;3 and 5; 2 .P R

5 5Gradient 0

3 3

y

x


with 5 and 3

PR PA AR PB BR

PA BR AR PB

2 2 2 25 3

25 9 34

PR PA AR

1 3 ; 2 is midpoint of

2

13

2

1 5; is midpoint of

2

1

2

M

M

D AR

x

C BR

y

1 1

Mid-point of is 3 ;2 2

PR M


2 5;3 and 1; 2P R

5Gradient 0

4

y

x

2 2 2 24 5

16 25 41

PR RA AP

3; 2 is midpoint of 3

1 1 5; is midpoint of

2 2

M

M

C AR x

D PA y

1Mid-point of is 3;

2PR M


3 5;4 and 5; 2P R

6Gradient

0

y

x

gradient is undefined

6PR

5;1 is mid-point of M PR


4 3; 3 and 4; 3P R

0Gradient 0

7

y

x

gradient is 0

7PR 1

; 3 is mid-point of 2

M PR

Relevant Formulae


Video Series

Lesson 2


In this lesson we will derive formulae for the:

• Distance between two points

• Gradient of the segment connecting the two

points

• Coordinates of the mid-point of the line

segment joining the two points

Assume that ; and ;

are two points in the Cartesian Plane.

A A B BA x y B x y

Distance between two points

and A B A BAC y y BC x x

2 2 2

2 2

2 2

From Pythagoras:

A B A B

B A B A

AB BC AC

x x y y

x x y y

2 2 2 2

A B A B B A B AAB x x y y x x y y

Application of distance formula

Given the points 6; 2 , 2;4 and 10;10 .

1 Calculate , and by means of the distance formula.

2 Determine whether , and are co-linear lie on same line .

A B C

AB BC AC

A B C

2 2 2 2

A B A B B A B AAB x x y y x x y y

2 2

2 2

2 2

6 2 2 4 64 36 100 10

1 2 10 4 10 64 36 100 10

6 10 2 10 256 144 400 20

AB

BC

AC

2 the points will be co-linear.AB BC AC

Why are the three points co-linear?

We deduced that the points 6; 2 , 2;4 and 10;10

are co-linear because .

A B C

AB BC AC

Co-linearity is confirmed by the accurate sketch.

Shortest distances between two points is equal

to the length of the segment joining the points.

Co-linearity of three points can

be proved in different ways.

More about this later.

Assume that ; and ;


A A B BA x y B x y

Gradient of a line segment

Gradient of

Gradient of

tan

A B

A B

B A

B A

AB

BA

y

x

y y

x x

y y

x x

and A B A By y y x x x

or and B A B Ay y y x x x

Relationship between gradient and angle m

Gradient tany

mx

1 0 is acute.

2 0 is obtuse.

3 0 0

4 90

m

m

m

m

Parallel and Perpendicular Lines

1 1

2 2

Gradient of is tan

Gradient of is tan

But corr.

m

m

1 2 1 2m m

3

4

Gradient of is tan

Gradient of is tan 90

But tan 90 cot

b

a

a

b

1 2 3 4 1a b

m mb a

Using gradients to show when three points are co-linear.

Three points will be co-linear if they lie on the same line.

This will be the case if

Two with common end-point will suffice

AB AC BCm m m

Given the points 6; 2 , 2;4 and 10;10 .

1 Calculate , and .

2 Use these gradients to show that , and are co-linear.

AB BC AC

A B C

m m m

A B C

4 2 6

2 6 8

10 2 12 3

10 6 16 4

4 10 6 3

2 10 8

2 4 6

6 2 8

4

B A

A

A B

A B

A C

A C

C B

C

B

B A

C A

AC

C A

B

B C

BC

B C

y ym

x x

y ym

x

y y

x x

y y

x x

y

x

y ym

x x

y

x x

3

4

Assume that ; and ;


A A B BA x y B x y

Mid-point Formula

2

2

A D D B

A B D

A BD

AD DC

y y y y

y y y

y yy

2

2

E B A E

E A B

A BE

BE EC

x x x x

x x x

x xx

;2 2

A B A Bx x y yM

3;6 , ; and 8;5 are the vertices of .

2;4 and are the mid-points of and

respectively.

1 Determine the values of and .

2 The co-ordinates of .

A B a b C ABC

D E AB AC

a b

E

Application of mid-point formula

3 8 6 5 11 11

2 ; ;2 2 2 2

E E

3

1 2 3 4 72 2

6and 4 6 8 2

2 2

A BD

A BD

x x ax a a

y y by b b

Tutorial 2: Distance, Gradient and Mid-point formulae

PAUSE VIDEO

• Do Tutorial 2


1 Show by means of calculations that

with vertices 8; 4 , 6;2 and 14;10

is an isosceles triangle and determine the measure

of the two angles opposite the equal sides.

ABC

A B C

2 Show that by means of calculations that

with vertices 4;6 , 10;2 and 4; 6 is

a right-angled triangle and determine by means

of two methods which angle is equal to 90 .

ABC

A B C


with vertices 4;2 , 6;8 and 2; 4 will

satisfy the two conclusions of the mid-point theorem.

ABC

A B C



with vertices 8; 4 , 6;2 and 14;10

is an isosceles triangle and determine the measure

of the two angles opposite the equal sides.

ABC

A B C

2 2 2 2

2 2 2 2

2 2 2 2

8 6 4 2 14 6 2 58

6 14 2 10 20 8 4 29

14 8 10 4 6 14 2 58

AB

BC

AC

is isosceles with ABC AB AC

and 2 292

BCB C BD

2 29 29cos cos = =

2 58 58

BDB C

AB

1 29cos 45

58B C


2 Show that by means of calculations that

with vertices 4;6 , 10;2 and 4; 6 is

a right-angled triangle and determine by means

of two methods which angle is equal to 90 .

ABC

A B C

2 2 2

2 2 2

2 2 2

Use Pythagoras

6 4 2 13 52

8 12 4 13 208

14 8 2 65 260

AB AB

AC AC

BC BC

Method 1 :

2 2 2

90

BC AB AC

A

Use gradients

6 2 4 2

4 10 6 3

6 6 12 3

4 4 8 2

6 2 8 4

4 10 14 7

AB

AC

BC

m

m

m

Method 2 :

1 90AB AC

m m CA AB A



with vertices 4;2 , 6;8 and 2; 4 will

satisfy the two conclusions of the mid-point theorem.

ABC

A B C

and

2

a DE AC

ACb DE

Need to show that :

6 4 8 2; 1;5

2 2

6 2 8 4; 2;2

2 2

D D

E E

5 23

1 2

4 23

2 4

DE

AC

m

m

DE ACm m DE AC

2 2

2 2

1 3 10

2 6 40 2 10

DE

AC

22

ACAC DE DE

Polygon

Applications


Video Series

Lesson 3


In this lesson we will apply the distance, gradient and

mid-point formulae to confirm properties of:

• Kites

• Parallelograms

• Rectangles

• Rhombuses

• Squares

Definition of a Kite

Confirm definition by calculations:

Two pairs of consecutive sides equal. Definition :

2 2

2 2

2 2

2 2

4 2 20

2 4 20

0 10 10

6 8 10

AB

BC

AB BC

AD

CD

AD CD

Definition :

Some computation details left as an exercise.

Some diagonal properties of a Kite

Confirm properties below by calculations:

Diagonals are perpendicular line segments.

Long diagonal bisects the short diagonal.

a

b

Properties :

1

3 and 3

1

BD AC

BD AC

a m m

m m BD AC


16 Equation :

3 3

Equation : 3 22

5;7

Midpoint of is 5;7

xb AC y

BD y x

AC BD

AC M AM CM

M

Some angle properties of a Kite


Long diagonal bisects opposite angles.Property : 20

10

10

AB CB

AD CD

AM CM

Know :


1 1

1 1

10sin sin

20

10sin sin

20

45

AMABM

AB

CMCBM

BC

ABM CBM

1

Similarly:

10sin 18.435

10ADM CDM

Definition of a Rhombus


with two consecutive sides equal.mDefinition :


6 4 2 1

10 4 6 3

12 10 2 1

12 6 6 3

AB

DC

m

m

AB DC

10 4 63

6 4 2

12 6 63

12 10 2

AD

BC

m

m

AD BC

is a mABCD

2 22 6 40AD 2 26 2 40DC

Two consecutive sides are equal

Some properties of a Rhombus


Diagonals bisect each other at right angles.Properties :


Only need to show that 90 .

But we will also show that diagonals bisect.

AED

6 10 10 6

; 8;82 2

E E

2 2

2 2

2 2 2 2

2 2 2 2

Similarily: 4 2

DE

BE

DE BE

AE CE

1

1

1

DE DB

AE AC

DE AE

m m

m m

m m DE EA

You should be able to calculate

angles indicated by and

Tutorial 3: Polygon Applications

Use the definitions and show by means of calculations

that figures 1, 2 and 3 represent a parallelogram,

rectangle and square respectively.

Show by means of

a

b

Consider the three figures.

calculations that some additional

properties for these three quadrilaterals are satisfied.

PAUSE VIDEO

• Do Tutorial 3


Figure 1 Figure 2 Figure 3

Definition of a Parallelogram

m


Quadrilateral is a if both pairs

of opposite sides are parallel.

Definition :


4

3AB DCm m

AB DC

4AD BC

m m

AD BC

Some properties of a Parallelogram


Both pairs of opposite sides are equal.

Both diagonal bisect each other.

Properties :


2 17

10

AD BC

CD AB

4 12 8 8Midpoint of is ; 8;8

2 2

6 10 0 16Midpoint of is ; 8;8

2 2

Midpoints of diagonals coincide

DB

AC

Definition of a Rectangle


with one of angles equal to 90 .m Definition :


1

2AB DCm m

AB DC

2AD BC

m m

AD BC

is a mABCD12

2

Thus 1.

or 90

DC CB

DC CB

m m

m m

DC CB C

Some properties of a Rectangle


Diagonals are equal in length.Properties :


2 2

2 2

2 4 20 2 5

8 4 80 4 5

AD BC

AB CD

2 2 2

2 2 2

80 20 100

and

20 80 100

AC AB BC

DB AD AB

10AC DB You should be able to calculate all

unknown angles (left as exercise).

Definition of a Square


Rectangle with two consecutive sides equal.Definition :


2 2

2 2 2 2

Two consecutive sides equal because:

4 2 20 2 5

2 4 2 4 2 5

AB

BC AD

or AB BC AB AD

1

2 is a because

2

DC AB

AD BC

mm m DC AB

ABCD

m m AD BC

is a rectangle because 1

90

DC BCABCD m m

DC CB C

Some properties of a square


Diagonals equal in length

Diagonals bisect each other at right angles

Properties :


Diagonals bisect at 9;9M

20 20 40 2 10AC DB

10MD BM AM CM

10 8 2 1 12 6 6 and 3

6 12 6 3 10 8 2DB ACm m

1DB AC

m m DB AC

REMEMBER!

•Consult text-books for additional examples.

•Attempt as many as possible other similar

examples on your own.

•Compare your methods with those that were

discussed in the Video.

•Repeat this procedure until you are confident.

•Do not forget:

Practice makes perfect!

End of Video on Analytical Geometry

Lecture 1 (Part A) Factorization

Documents

Transcript of Lecture 1 (Part A) Factorization