1) Nasirah Binti Che Daud D20081032296

2) Wan Masturah Binti Wan Mad Mohtar D20081032356

3) Ayuni Amalina Binti Mukhtar D20081032369

4) Nur Ain Bt Ahmad Fikri D20081032311 

5) Nur Syazwani Bt Wan Aziz D20061026742

Coordinate Plane

• A basic concept for coordinate geometry.

• It describes a two-dimensional plane in terms of two perpendicular axes: x and y.

• The x-axis-horizontal direction

• the y-axis-vertical direction of the plane.

y-axis

X-axis

• Points are indicated by their positions along the x and y-axes in the form (a,b)

• L coordinates is (–3, 1.5)

Equation Of A Line

• An equation of a line can be written

y = mx + b

where

m is the slope

b is the y-intercept

• Slant of a line is called the slope/gradient.

• Slope is the ratio of the change in the y-

value over the change in the x-value.

Slopes = Change in y value

Change in x value

- The rate at which line rises (or falls) vertically for every unit across to the right.

y

x0

m

Q(x2,y2)

P(x1, y1)(y2-y1)

(x2-x1)

Gradient line

P(x1,y1) ,Q (x2,y2):

12

12

x-x

y-y m

Where.. 12 xx

y

x0

P(2,2)

Q(6,5)

If θ < 90, m is positive

θ

Gradient line of PQ :

12

12

x-x

y-y mPQ

26

25

4

3m PQ

Example 1:• Given two points, P = (0, –1) and Q = (4,1), on the

line we can calculate the slope of the line.• Slopes= Change in y value

Change in x value

= 1-(-1)

4 - 0

= 2

4

= 1

2

y

0x

M (2,7)

N (5,2)

If θ > 90, m is negative

θ

Gradient line of MN :

52

27mMN

3

5

3

5mMN

• Consider the two points, R(–2, 3) and S(0, –1) on the line. What would be the slope of the line?

Slopes = Change in y value

Change in x value

= -1-3

0-(-2)

= -4

2

= -2

Example 2:

y

X0

P Q

-If the line PQ is parallel with the x-axis,

θ = 0° OR θ = 180° ….Hence, m = 0

m = 0

How about parallel & perpendicular line ??What are their gradient ??....

1.) Two lines are parallel if and only if both have the same gradient ; m1=m2

2.) Two lines with m1 and m2 gradient

perpendicular if and only if m2m1= -1

Slopes Of Parallel Lines

y

x0

P(2,2)

Q(6,5)

Gradient line of PQ :

12

12

x-x

y-y mPQ

26

25

4

3m PQ

R (4,2)

S (8,5)

Gradient line of RS :

12

12

x-x

y-y mRS

48

25

4

3m RS

mPQ // mRS

Question…

1.) Does the straight-line AB and CD below parallel??....

A(1,9) , B(5,8) , C(5,2) ,D(1,3)

12

12

ABmxx

yy

12

12

CDmxx

yy

51

23

15

98

4

1

4

1

mAB = mCD…….hence, AB // CD

Slopes Of Perpendicular Lines

• Two lines are perpendicular if the product of their slopes (m) is –1

• The line y= ½ x-1

perpendicular to

y= -2x-1

Because:

½ x (-2) =-1

y

x0

P(2,5)

Gradient line of PQ :

12

12

x-x

y-y mPQ

42

25

2

3m PQ

Q (6,3)

R (9,5)

Gradient line of RS :

12

12

x-x

y-y mQR

69

35

3

2m RS

mPQ mRS

Y-intercept

• The y-intercept is where the line intercepts (meets) the y-axis.

• The midpoint of a segment divides the segment into two segments of equal length.

• The midpoint between the two points (x1,y1) and (x2,y2) is

2,

2

2121 yyxx

Example:

• The midpoint of the points A(1,4) and B(5,6) is

)5,3(2

10,

2

6

2

64,

2

51

Distance Formula

• The distance between the two points (x1,y1) and (x2,y2) is   

http://www.toonups.com/animated-clip-art-fairy-tale-item.1668.c

Dividing Point with Ratio

Formula for inside point :

nm

nymy

12

nm

nxmx

12

x = y =

Inside dividing ratio

11, yxA

),( yxP

),( 22 yxB

Given points A(1,2) and B(19,23). If P (x,y) dividing inside AB with ratio 1:2, find the value of x and y.

A(1,2)

y

xA(1,2)

P(x,y)

B(19,23)

m

n

Solution:

y=

=

=

=

nm

nymy

12

21

)2)(2()23)(1(

3

27

9

x =nm

nxmx

12

3

219

21

)1)(2()19)(1(

3

21 7

)9,7(P

Outside dividing ratioy

x),( 11 yxA

),( 22 yxB

),( yxP

Outside dividing point

nm

nymy

12

nm

nxmx

12

x=

Formula :

y =

Given that points A(-5,-6) and B(-1,-2). Get the coordinates that

dividing outside AB with the ratio of 5:3

1

5

2

1

x

x2

6

2

1

y

y

3

5

n

m

nm

nymy

nm

nxmxP

1212

,

35

)6(3)2(5,

35

)5(3)1(5P

2

1810,

2

155P )4,5(

STRAIGHT-LINE EQUATION

Equation Type

EQUATION TYPE

• Gradient Type

• Interception Type

• General Type

GRADIENT TYPE

y = mx + c

y

x

P(x,y)

m = gradient

c = y-interception

c

INTERCEPTION TYPE

x + ya b = 1

a = x-interception

b = y-interception

x

y

P(0,b)

Q(a,0)

GENERAL TYPE

ax + by + c = 0

a, b, c are constant

METHOD TO FIND STRAIGHT-LINE EQUATION

•Give Two Points

or

•One Point, One Gradient

GIVEN TWO POINTS

P(1,-4)

Q(3,4)

First, find gradient (m):

m =

m = =

m = 4

Then, find c

Substitute q(3,4) in equation y = mx + c

(4) = (4)(3) + c

4 = 12 + c

C= -8

the equation

y = 4x-8

y2 – y1

x2 – x1

4-(-4)3-1

82

y = 4x - 8

ONE POINT, ONE GRADIENT

P(2,8)

Substitute m and point p(2,8)

In equation y = mx + c

(8) = (3)(2) + C

8 = 6 + C

Hence c = 2

So the equation=

y = 3x + 2 y = 3x + 2

m = 3

CHANGING EQUATION TYPE

y = 2x + 10Change this equation into General Type

y = 2x + 10

y – 2x – 10 = 0 or 2x – y + 10 = 0

Then, change 2x – y + 10 = 0 into Interception type

2x – y + 10 = 0

2x – y = -10

2x – y = -10

-10 -10 -10

2x + y = 1 or x + y = 1

10 10 5 10

Brain-Storming Corner

Given a line with two given point. Find the equation of the straight line

(-3, 2)

(2, 7)

y

x

Gradient ;

7 – 2 = 5

2 – (-3) 5

= 1

Find c by substitute (2,7) in equation y = mx + c

7 = (1)(2) + c

c = 5

Equation

y = mx + c

y = x + 5

Change the equation into general type and interception type

y = x + 5

y – x – 5 = 0 or x – y + 5 = 0

Then, change x – y + 5 = 0 into Interception type

x – y + 5 = 0

x – y = -5

x – y = -5

-5 -5 -5

x + y = 1

5 5

Subtopics….

• The nearest point to the straight line

• The distance between two straight lines that parallel to each other

• Intersection of the straight line• Area of rectangle• Area of triangle

P

Q

ax + by +c = 0

d

(h,k)

The nearest point to straight line

Perpendicular distance :

-The shortest distance betweenthem / the length of a perpendicularline segment from the line to thepoint

y

x

The nearest point to straight line

22d

ba

cbkah

d

P

Q

ax + by +c = 0

(h,k)

The nearest distance from point (h,k) to straight line ax +by +c =0 is ;

0 if22

ba

cbkah Hence, the point is on the other side.

Example :• Find the point distance and location of (2,1) and (-3,2)

towards straight line 2y-3x-1=0

Solution :

From straight line 2y-3x-1= 0 ,

a = -3 b = 2 c = -1

Point (2,1), hence h=2, k=1

d1 22 )2()3(

1)1(2)2(3

13

5

22 )2()3(

1)2(2)3(3

13

12

Point (-3,2), hence, h=-3, k=2

d2

d1 d2 13

5

13

12

2y-3x-1=0

y

x

(2,1)(-3,2)

d1d2

0

The distance between 2 straight-lines that parallel

to each other

L1

L2

METHOD

1) Find the coordinate on one of the line2) Find the point perpendicular distance from the other line

Example :• Find the distance between the parallel lines 5x+12y+1=0 and

5x+12y+8=0

Solution:

Take 5x+12y +1=0

When x=0, y=

Coordinate is (0, )

12

1

12

1

The distance to line 5x+12y+8=0 is :

a= 5 b= 12 c=8 h= 0 k=

13

7

14425

8121

120 d

12

1,0

22d

ba

cbkah

12

1

Intersection of straight-line

The coordinate for two straight lines intersection can be found by

solving both equation stimultaneously

7

4 y

Solution :

2x-3y=6 …………(1)

4x+y =16 …………(2)

(1) x 2

4x+y =16 …………(2)

4x-6y=12 …………(3)

(2) - (3)

7y= 4

7

27 x

2x-3y=6

4x+y=16

y

x0

P

QR

S

Choose A Quizzes

SELECT THE DIFFICULTY LEVEL

EASY

EXPERT

y

5

4

3

2

1 x

-5 -4 -3 -2 -1 1-1

2 3 4 5

-2

-3

-4

-5

QUESTION 1

QUESTION 2

QUESTION 3

QUESTION 4

QUESTION 5

QUESTION 6C D

F

B

E

A

y

5

4

3

2

1 x

-5 -4 -3 -2 -1 1-1

2 3 4 5

-2

-3

-4

-5

QUESTION 1

3 UNITS LEFT,

5 UNITS UP

C D

F

B

E

A

QUESTION 2

QUESTION 3

QUESTION 4

QUESTION 5

QUESTION 6

y

5

4

3

2

1 x

-5 -4 -3 -2 -1 1-1

2 3 4 5

-2

-3

-4

-5

QUESTION 1

QUESTION 2

QUESTION 3

QUESTION 4

QUESTION 5

QUESTION 6

C D

F

B

E

A

2 UNITS RIGHT,

4 UNITS DOWN

y

5

4

3

2

1 x

-5 -4 -3 -2 -1 1-1

2 3 4 5

-2

-3

-4

-5

QUESTION 1

QUESTION 2

QUESTION 3

QUESTION 4

QUESTION 5

QUESTION 6

C D

F

B

E

A

5 UNITS RIGHT,

0 UNITS UP/DOWN

y

5

4

3

2

1 x

-5 -4 -3 -2 -1 1-1

2 3 4 5

-2

-3

-4

-5

QUESTION 1

QUESTION 2

QUESTION 3

QUESTION 4

QUESTION 5

QUESTION 6

C D

F

B

E

A

5 UNITS LEFT,

3 UNITS DOWN

y

5

4

3

2

1 x

-5 -4 -3 -2 -1 1-1

2 3 4 5

-2

-3

-4

-5

QUESTION 1

QUESTION 2

QUESTION 3

QUESTION 4

QUESTION 5

QUESTION 6

C D

F

B

E

A

0 UNITS LEFT/RIGHT

4 UNITS UP

y

5

4

3

2

1 x

-5 -4 -3 -2 -1 1-1

2 3 4 5

-2

-3

-4

-5

QUESTION 1

QUESTION 2

QUESTION 3

QUESTION 4

QUESTION 5

QUESTION 6C D

F

B

E

A

0 UNITS RIGHT/LEFT

0 UNITS UP/DOWN

Positive slopeFor example: Given two points, P = (0, –1) and Q = (4,1), on the line we can calculate the slope of the line.

• y-intercept

• Gradient

• Equation

1/2 4

Y = 1/2x - 1

Y = 2x + 4

-1 2

Q4Q1 Q3Q2 Q5 Q6

Negative slopeFor example: Consider the two points, R(–2, 3) and S(0, –1) on the line. What would be the slope of the line?

• y-intercept

• Gradient

• Equation

-2 3

Y = 3x - 2

Y = -2x - 1

-1 -2

Q4Q1 Q3Q2 Q5 Q6

• In coordinate geometry, two lines are parallel if their slopes (m) are equal.

• For example: The line y=1/2x+1 is parallel to the line y=1/2x-1. Their slopes are both the same.

Slopes Of Parallel Lines

Q4Q1 Q3Q2 Q5 Q6

• In the coordinate plane, two lines are perpendicular if the product of their slopes (m) is –1.

• For example: The line Y=1/2X-1 is perpendicular to the line y = –2x– 1. The product of the two slopes is 1/2 x (-2) = -1

Slopes Of Perpendicular Lines

Q4Q1 Q3Q2 Q5 Q6

• To find a point that is halfway between two given points, get the average of the x-values and the average of the y-values.

• The midpoint between the two points (x1,y1) and (x2,y2) is:

   

• For example:

The midpoint of the points A(1,4) and B(5,6) is

Mid Point Formula

Q4Q1 Q3Q2 Q5 Q6

• For example: To find the distance between A(1,1) and B(3,4), we form a right angled triangle with AB as the hypotenuse. The length of AC = 3 – 1 = 2. The length of BC = 4 – 1 = 3.

Applying Pythagorean Theorem:

• AB2 = 22 + 32

AB2 = 13AB = /13

Distance Formula

Q4Q1 Q3Q2 Q5 Q6

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