Presented by:

LEUNG Suk Yee

LEUNG Wing Yan

HUI Hon Yin

LED 3120B

PowerPoint Presentation

Target Audience : F.2 Students

Prerequisite knowledge :learnt about the distance between two points on the same horizontal line or on the same vertical line in a rectangle coordinate plane.

The use of slides : source of learning

LEUNG SUK YEE : Data Collection & Slide Production

LEUNG WING YAN : Data Collection & Slide Production

HUI Hon Yin:Slide Production & Layout Design

Hi, I am Joey. I am your guide in this presentation. Let explore the knowledge by click the below button.

GO!

1) Distance between any TWO points

2) Slope of a STRAIGHT line

3) PARALLEL lines

4) PERPENDICULAR lines

5) MC Questions

1.1 Distance between two points on the same horizontal line

y

x0

The Distance of AB:

AB = 4-1

= 3 unitsA(1,1) B(4,1)

1.2 Distance between two points on the same vertical line

y

x0

Distance of AC:

AC = 5-1

= 4 units

C(1,5)

A(1,1)

1.3 Distance between any two points y

x0

Then, do you know the distance between these two points ?

C(1,5)

B(4,1)

i) Draw a vertical line from A through C

ii) Draw a horizontal line from A through B

iii) Since ABC is right-angled triangle, do you remember that we can use the Pythagoras’ theorem to calculate the hypotenuse BC?

y

x0

C(1,5)

B(4,1)

A(1,1)

By the Pythagoras’ theorem,we have

BC 2 =AB 2 +AC 2

=3 2 +4 2

=9 +16

=25

∴BC =5

y

x0

C(1,5)

B(4,1)

A(1,1)

Why???

From the figure, we have

AB = x1 - x 2 and AC = y1 – y2

By Pythagoras’ theorem, we have

BC2 = AB2 + AC2

= (x1- x2 ) 2 + (y1 – y2 ) 2

1.4 Distance Formulay

x0

C(x2,y1)

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

) y - (y ) x- (x 221

221 BC

BC2 = (x1 - x2 ) 2 + (y1 - y2 ) 2

= (4-1) 2 + (2-6) 2

= 3 2 + (-4) 2

= 25

∴ BC=5

y

x0

C(1,5)

B(4,1)

C(-1,4)

A(-2,0)B(2,-1)

y

x

Find the perimeter of ABC correct to 1 decimal place.

Solution :

Example 1

Time Remain

5 sec

Time Remain

3 sec

Time Remain

2 sec

Time Remain

1 sec

Time Remain

4 sec

AB2 =(-2 -2) 2 + 0-(-1) 2 =17

BC 2=2-(-1) 2 + (-1- 4) 2 =34

AC2 = -2 - (-1) 2 + (0-4) 2 =17

Perimeter of ABC

=AB + BC +AC

=

=14.1(correct to 1d.p.)

173417

Lines AB and CD represent two steep roads. Which road is steeper?

Obviously road CD is steeper than road AB.

But, can you tell the steepness of the roads?

2.1 Definition of Slope

A

B

D

C

We can determine in this way

Difference in x-

coordinates

Difference in y-

coordinates

(x 1 , y 2)

(x 1 , y 1)

(x 2 ,y 2)

A(x , y ) and B (x , y ) are two points on a straight line.

Difference in y- coordinates =y1-y2

Difference in corresponding x- coordinates=x1-x2

Hence, Slope of a line = y1-y2 / x1-x2

..

.P(2,3)

Q (12,9)

R (7,9)

Example 2

Answer:

Slope of PQ

= 3-9 / 2-12 = -6 / -10 = 3/5

Slope of PR

= 3-9 / 2-7 = -6 / -5 = 6/5

Find the slope of the lines PQ & PR

Time Remain

5 sec

Time Remain

4 sec

Time Remain

3 sec

Time Remain

2 sec

Time Remain

1 sec

Example 3

Answer:

Slope of XY

= 1-10 / 10-4 = -9 / 6 = -3 / 2

Slope of XZ

= 1-7 / 10-2 = -6 / 8 = -3 / 4

Find the slope of the lines XY & XZ

.

.Z(2,7)

Y (4,10)

X (10,1)

Time Remain

5 sec

Time Remain

4 sec

Time Remain

3 sec

Time Remain

2 sec

Time Remain

1 sec

From the above two example, we can conclude that:

(1) when a straight line rises from left to right, the slope of the line is a positive .

(2) when a straight line falls from left to right, the slope of the line is negative.

Line with positive slope

Line with negative slope

2.2 The Characteristics of slope

There is a pair of parallel lines.

What’s the different between there slope?

Slopes of Parallel Lines

Slope of AB

= 1-3 / 1-5

= -2 / -4

=1 / 5

A(1,1)

B(5,3)

C(-1,1)

D(2,2.5)

Slope of CD

= 1-2.5 / -1- 2

= (-3 / 2) / -3

= 1 / 2

So, Slope of AB = Slope of CD =1 / 2

In general,

If two lines are parallel, then their slopes are

equal.Conversely,

If two lines have same slope,then they are parallel.

A(3, -1)

B(4,3)

C(-2,-2)

D(-3/2,0)

Show that two straight lines AB and CD are parallel.

Answer:

Slope of AB = (-1-3)/3-4

= -4/-1 = 4

Slope of CD = (-2-0)/-2-(-3/2)

= -2/(-1/2)

= 4

Example 4

Since slope of AB = slope of CD,so AB and CD are parallel lines.

Time Remain

5 sec

Time Remain

4 sec

Time Remain

3 sec

Time Remain

2 sec

Time Remain

1 sec

D(1,6)

B(6,5)

C(5,0)A(0,1)

Slopes of Perpendicular Lines

There is a pair of perpendicular lines.

What’s the relationship between there slope?

D(1,6)

B(6,5)

C(5,0)A(0,1)

Slope of AB

= 5-1 / 6-0

= 4 / 6

= 2/3

Slope of CD

= 6-0 / 1-5

= 6 / -4

= -3/2

Slope of AB x Slope of CD

= 2/3 x (-3/2)

=-1

In general,

If two lines are perpendicular,then the product of their slopesare equal to -1.

Conversely,

If the product of their slopes are equal to -1,then they are perpendicular.

Example 5 Show that two straight lines AB and CD are perpendicular.

A(0,3)

B(-2,0)

C(-6,4)

D(0,0)

Answer:

Slope of AB = (3-0)/0-(-2)

= 3/2

Slope of CD = (4-0)/(-6)-0

= -4/6

= -2/3

∵Slope of AB x slope of CD =3/2 x (-2/3) =-1 ∴AB and CD are perpendicular.

Time Remain

5 sec

Time Remain

4 sec

Time Remain

3 sec

Time Remain

2 sec

Time Remain

1 sec

1. Find the perimeter of a square if X(- 3,8) and Y(2,-4) are two adjacent vertices of the square.

A. 13

B. 52

C. 169

2. In the figure, which line has the greatest slope?

A. L1

B. L2

C. L3x

y

L3 L2

L1

.

. .

.

(4,3)

(3,n)

(2,5)

(1,0)

A

C

B

D

x

y

3. In the figure, AB is parallel to CD. What is the value of n?

C. 8

B. -1

A. -2

4. Given two points M(2,0) and N(5,1). If P(3,b) is a point such that MP=MN, find b..

A. 3 only

B. 2 or -2

C. 3 or -3

.

.

D

(a,b)

(-1,-2)

C

A

Bx

y

5. In the figure, AB is perpendicular to CD. (a,b) and (-1,-2) are two points on CD. If the slope of AB is –1/2, express a in terms of b.

A. a=b/2

B. a=-b/2

C. a=2b+

3

END

Produced by

Leung Suk Yee 99736862

Leung Wing Yan 99688683

Hui Hon Yin 99530432

RIGHTLet‘ s try another question by click the below button

RIGHTLet‘ s try another question by click the below button

RIGHTLet‘ s try another question by click the below button

RIGHTLet‘ s try another question by click the below button

RIGHTWell done, You have finish

all the question!

13 is only the length of one side of the square, not the perimeter of the square.

Wrong

Don’t give up!

Let’s try it again.

169 is the area of the square, not the perimeter of the square

Wrong

Don’t give up!

Let’s try it again.

L1 is parallel to the x-axis, there is no difference in y-coordinates, therefore slope of L1 isSlope of L1 = 0/x1-x2= 0

Wrong

Don’t give up!

Let’s try it again.

L2 falls from left to right, the slope must be negative.

Wrong

Don’t give up!

Let’s try it again.

You may have some problems in using the slope formula.Slope formula = y1-y2 / x1-x2

Are you sure that you have put x1 ,x2 ,y1 and y2 in the right place of the formula ?

Wrong

Don’t give up!

Let’s try it again.

In fact, 9 can be the square of both 3 and -3.

Wrong

Don’t give up!

Let’s try it again.

The situation is MP =MN , not MP = PN.

Wrong

Don’t give up!

Let’s try it again.

You may make some mistakes in the process of calculation. Please be more careful.

Wrong

Don’t give up!

Let’s try it again.

The slope formula is y1-y2 / x1-x2 , not x1-x2 / y1-y2

Wrong

Don’t give up!

Let’s try it again.