BAB 9 GARIS LURUS PT3

BAB 9 : GARIS LURUS

KECERUNAN m = Y2  Y1
X2  X1

Tulis kordinat titik-titik yang diberi dan seterusnya cari kecerunan garis yang diberi
dalam satah Cartes berikut.

y
1.

7
P( )

6

5 Kecerunan PQ
4 =
3
)
Q(
2
1

01 2 34 567 x

2.
y

2 P( )
1

-2 - 1 0 12 34 5 x Kecerunan PQ
-1 =

-2

-3
Q( )

-4

The Sttraight Line

3. y
4.
7
6 P( )

5 Kecerunan PQ
4 =
3
2 Kecerunan PQ
1 =

Q( )

01 2 34 567 x

P( ) y
4
3

2
1

-4 - 3 -2 -1 0 1 23 x
-1 Q( )

-2

The Sttraight Line

5. y
P( ) 4

Q( ) Kecerunan PQ
3 =
2
1 x

-4 - 3 -2 -1 0 1 2 3
-1

-2

6.
y

4
P( )

3

2

1

-4 - 3 -2 -1 0 1 23 x
-1 Q( )

-2

The Sttraight Line

MENENTUKAN KECERUNAN GARIS LURUS Latihan
1.) ( 2,1) dan (4,3)
Contoh

1) (1,2) dan (3,4)
42 2

m= = =1
31 2

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2) (-1,2) dan (3,4) 2.) (- 2,1) dan (4,3)

42 2 1
m= = =

3  (1) 4 2

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3) (1,-2) dan (3,4) 4.) (2,-1) dan (4,3)

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5) (-1,-2) dan (3,4) 6.) (-2,-1) dan (4,3)

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7) (-1,-2) dan (-3,-4) 8.) (-2,-1) dan (-4,-3)

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9) 1 ) dan ( 3 ,0 ) 13 1 1
(0, 10) (- , ) dan ( ,- )
22
22 2 2

The Sttraight Line

PINTASAN

● y-pintasan (x = 0)

●
x-pintasan (y = 0)

Pintasan-x y=0

Cari pintasan-x dany pintasan-y bagi garis lurus yang diberi dalam satah Cartes berikut.
1.

7

6

5
4 pintasan-x =
3 pintasan-y =
2
1

01 2 34 567 x

2.
y
2

1

-2 - 1 0 1 2 34 5 x pintasan-x =
-1 pintasan-y =
-2
-3

-4

The Sttraight Line

3. y x pintasan-x =
4. pintasan-y =
7
6 pintasan-x =
5 pintasan-y =
4
3
2
1

01 2 34 567

y
4
3

2
1

-4 - 3 -2 -1 0 1 2 3 x
-1

-2

The Sttraight Line

5. pintasan-x =
y pintasan-y =
4 x
3
2
1

-4 - 3 -2 -1 0 1 2 3
-1

-2

MENCARI PINTASAN-X MELALUI PERSAMAAN GARIS LURUS

Contoh Latihan

1) y  2x  4 1) y  3x  6
y  0  0  2x  4
 4  2x

2x  4

x   4  2
2

x- intercept = -2

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2) y  5x  2 2) y  2x  5

y  0  0  5x  2

2  5x

5x  2

x 2
5
5

x-intercept =
2

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The Sttraight Line

3) y  4x  5 4.) y  3x 1

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5) 2y  3x  6 6) 2y  2x  5

7) y   2 x  6 8) y   2 x  4
3 5

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9) 2 y  3 x 12 10) 3y  1 x  3
5 9

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11)  2y  x 12 12)  3y  3x  2

The Sttraight Line

MENCARI PINTASAN-X MELALUI PERSAMAAN GARIS LURUS

Pintasan-y x=0

Example Exersice
1) y  3x  8
1) y  2x  6
x = 0 , y = 2(0) + 6
y=6
y- intercept = 6

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2) y  2x  7 3.) y  3x  9

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4) y  2 x  6 5) y  1 x  6
3 2

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6) y   2 x  7 7) y   2 x 10
3 5

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Example

8) 2 y  3 x 12 9) 3y  1 x  21
5 9

3
X = 0, 2y = (0) + 12

5

2y = 12

y=6
y – intercept = 6

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10)  2y  x 12 11)  3y  3x  2

The Sttraight Line

……………………………………………………………………………………………………..

PERSAMAAN GARIS LURUS : y = mx + c
kecerunan pintasan-y

MENENTUKAN PERSAMAAN GARIS LURUS MELALUI TITIK DAN KECERUNAN

Example Exercise A
1) (2,7) and m = 3 1) (4,9) and m= -2

y = mx + c
7 = 3(2) + c

7=6+c
7–6=c

c=1

Equation of a straight line , y = 3x + 1

……………………………………………………………………………………………………….

2) (-3,5) and m = 4 2) (-1,7) and m = 5
y = mx + c
5 = 4(-3) + c
5 = -12 + c
C= 5 + 12
=17

Equation of a straight line, y = 4x +17

……………………………………………………………………………………………………….

1 3) (-5,11) and m = -2
3) (4,-9) and m =

2

y = mx + c

1
-9 = (4) + c

2

-9 = 2 + c
-9 – 2 = c

c = -11

1
equation of a straight line, y = x - 11

2

The Sttraight Line

……………………………………………………………………………………………………….

2 3
4) ( -3,-5) and m = 5) (3,-7) and m = -

3 4

2 4
6) (-1,-4) and m = 7) (-10,10) and m = -

5 5

…………………………………………………………………………………………….

1 4
8) (0,-3) and m = 9) (-6,-2) and m =

2 3

The Sttraight Line

PERSAMAAN GARIS LURUS JUGA BOLH DIPEROLEHI DENGAN
MENGGUNAKAN FORMULA

Example y –k = m (x –h)

1) (2,7) and m = 3 Exercise B
y  7  3(x  2) 1) (4,9) and m= -2
y  7  3x  6
y  3x  6  7
y  3x  1

……………………………………………………………………………………………………….

2) (-3,-5) and m = 4 3) (-1,7) and m = 5

……………………………………………………………………………………………………….

1 5) (5,-11) and m = -2
4) (4,-9) and m =

2

……………………………………………………………………………

The Sttraight Line

6) ( -4,-5) and m = 2 3
3 7) (4,-7) and m = -

y  (5)  2 (x  (4)) 4
3

y  5  2 (x  4)
3

3( y  5)  2(x  4)

3 y  15  2x  8

3 y  2x  8  15

3y  2x  7

…………………………………………………………………………………………………….

2 4
8 (-1,-4) and m = 9) (10,10) and m = -

5 5

…………………………………………………………………………………………….

1 4
10) (0,-3) and m = 11) (-5,-2) and m =

2 3

The Sttraight Line

PERSAMAAN GARIS LURUS – MELALUI 2 TITIK \ 2 KOORDINAT

Step 1 : Cari kecerunan
Step 2: Dapatkan pintasan-y
Step 3 : Tulis persamaan garis lurus dalam bentuk y = mx + c

Step 1 : Find the gradient
Step 2: Find the y- intercept
Step 3 : Write down the equation in a form of y = mx + c

Example Exercise
1. (2,1) and (4,3)
(-1,2) and (3,4)

42 2 1
m= = =

3  (1) 4 2
y = mx + c

1
4 = (3) + c

2
3
4= +c
2

3
C= 4 -

2
1
=2
2

15
Equation of a straight line, y = x +

22
Or 2y = x + 5

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2) (-1,2) and (3,4) 3) (-2,1) and (4,3)

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The Sttraight Line

4) (3,-5) and (-2,5) 5) (3,-7) and (-1,3)

6) (0,-5) and (2,1) 7) (-1,5) and (-3,-7)

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8) ( 1 ,4 ) and (3,-6) 9) ( 1 ,5 ) and (  1 ,3)
2 33

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The Sttraight Line

PERSMAAN GARIS LURUS MELALUI DUA GARIS YANG SELARI

Garis yang selari mempunyai kecerunan yang sama dan sebaliknya

Example Exercise
1) (1,2) and y = 3x – 4 1) (0,-3) and y = -2x + 5

m3
y  2  3(x 1)

y  2  3x  3

y  3x  3  2

y  3x 1

………………………………………………………………………………………………

2) (2,4) and y + x = 5 2) (0,6) and y = 5x -1

y = -x + 5

 m  1

y  4  1(x  2)

y 4  x  2

y  x  2 4

y  x 6

.............................................................................................................................................................

3) (8,-5) and 5y + 20x = -2 3) (2,3) and 2x -5y = 12

5y = -20x -2

 20x  2
y=

5

2
y = - 4x -

5

 m  4

y  (5)  4(x  8)

y  5  4x  32

y  4x  32  5

y  4x  37

………………………………………………………………………………………………

The Sttraight Line

4) (5,2) and y = x -3 5) (1,8) and 3x + y =4

………………………………………………………………………………………………

6) (3,4) and y = 2x -2 7) (0,-3) and 2x + y -5 = 0

………………………………………………………………………………………………

8) (-1,3) and y = 3x + 1 9) (4,0) and y = 3x - 11

………………………………………………………………………………………………

10) (4,-5) and y = -2x + 3 11) (3,4) and 4y – 2x = 3

The Sttraight Line

EXAMINATION FORMAT QUESTION EXAMPLE 1
(i) gradient AB= 4  2  2  1
A(0,4) 02 2
B(2,2) (ii) x-intercept for AB
 y 0
O the equation of AB 
C y  4  1(x  0)

Diagram 1 y  4  x

0  4  x

x4
x-intercept = 4

(iii) the equation of AC

M AC  M OB  20 1
20

y  4  1(x  0)

y4 x

y  x4

EXERCISE
1.Base on diagram 2
(i) gradient PQ

P(0,6)
Q(3,5)

(ii) x-intercept for PQ

RO

Diagram 2

(iii) the equation of PR

The Sttraight Line

A(2,6) 2 Base on diagram 3, find
(i) the gradient of AB

E (ii) the equation of ECF

B(-3,3) C(5,4)
F
(iii) x-intercept for ECF

Diagram 3

3 In the diagram 4, OP  6 units.
(i) find the gradient of PR

R Q(7,8)

OP (ii) find the y-intercept of PQ
Diagram 4

The Sttraight Line

B 4 The diagram 5, AB is a straight
line.
(i) state the y-intercept of AB

(ii) find the equation of of AB

O4 Diagram 5
-6

A

P(0,6) 5. In diagram 6, the gradient of PQ
is  3 . Line
OQ 2
Diagram 6 RS is parallel to the line PQ and
passes through a point (-5,3). Find
(i) the coordinate of Q.

ii) the equation of RS

The Sttraight Line

6. In diagram 7, line DE is parallel to the line GF.
y

D(-4,6)

G(h,4)

x

EO F
Diagram 7

Given the equation of straight line DG is x + 3y = 14 and the gradient of GF is -2,
Find
(a) the value of h,
(b) the coordinate F
( c) the equation of straight line DE

[ 5 marks]

7. In diagram 8, the gradient of the straight line MN is 2

y 3
● N(6, p)

M
3

x

-5 L Diagram 8

Find
(a) the value of p
(b) the equation of straight line LN

The Sttraight Line

8. In diagram 9, JKLM is a trapezium and JK is parallel to ML
y K (2,5)

L(4,3)

J(-4,2)

x

MO

Diagram 9

Find
(i) the x-intercept for the straight line JK
(ii) the equation of the straight line ML
(iii) the coordinate M

The Sttraight Line