[Worksheet] Chapter 5 - The Straight Line

8/21/2019 [Worksheet] Chapter 5 - The Straight Line

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The Straight Line

THE STRAIGHT LINE

GRADIENT m =12

12

X X

Y Y

−

−

Example Exercises

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

m =13

24

−

− =

2

2= 1

------------------------------------------------------------------------------------------------------------

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

m=)1(3

24

−−

−=

4

2=

2

1

------------------------------------------------------------------------------------------------------------

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

------------------------------------------------------------------------------------------------------------

5) (-1,-2) and (3, 4) 6) (-2,-1) and (4, 3)

------------------------------------------------------------------------------------------------------------

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

------------------------------------------------------------------------------------------------------------

9) (0,2

1) and ( 0,

2

3) 10) (-

2

1,2

3) and (

2

1,-

2

1)


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The Straight Line

INTERCEPT

X-intercept y = 0

Example Exercise

1) 42 += x y 1) 63 += x y

4200 +=⇒= x y

22

4

42

24

−=−

=

−=

=−

x

x

x

x- intercept = -2---------------------------------------------------------------------------------------------------------------------

2) 25 −= x y 2) 52 +−= x y

2500 −=⇒= x y

5

225

52

=

=

=

x

x

x

x-intercept =2

5

---------------------------------------------------------------------------------------------------------------------

3) 54 +−= x y 4.) 13 +−= x y

---------------------------------------------------------------------------------------------------------------------

5) 632 −= x y 6) 522 +−= x y

● y-intercept (x = 0)

●x-intercept (y = 0)


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The Straight Line

7) 63

2+−= x y 8) 4

5

2−−= x y

---------------------------------------------------------------------------------------------------------------------

9) 125

32 += x y 10) 3

9

13 −= x y

--------------------------------------------------------------------------------------------------------------------

11) 122 +=− x y 12) 233 −=− x y

y –intercept x = 0

Example Exersice

1) 62 += x y 1) 83 += x y

x = 0 , y = 2(0) + 6

y = 6y- intercept = 6

---------------------------------------------------------------------------------------------------------------------

2) 72 −= x y 3.) 93 −= x y

---------------------------------------------------------------------------------------------------------------------


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The Straight Line

4) 63

2+= x y 5) 6

2

1−= x y

---------------------------------------------------------------------------------------------------------------------

6) 73

2+−= x y 7) 10

5

2−−= x y

---------------------------------------------------------------------------------------------------------------------

Example

8) 125

32 += x y 9) 21

9

13 −= x y

X = 0, 2y =5

3(0) + 12

2y = 12y = 6

y – intercept = 6--------------------------------------------------------------------------------------------------------------------

10) 122 +=− x y 11) 233 −=− x y

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


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The Straight Line

EQUATION OF A STRAIGHT LINE : y = mx + c

Gradient y-intercept

THE EQUATION OF A STRAIGHT LINE PASSING THROUGH THE POINTS AND

GRADIENT

Example Exercise A1) (2,7) and m = 3 1) (4,9) and m= -2

y = mx + c

7 = 3(2) + c7 = 6 + c7 – 6 = c

c = 1

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

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

5 = 4(-3) + c5 = -12 + c

C= 5 + 12=17

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

3) (4,-9) and m =2

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

y = mx + c

-9 =2

1(4) + c

-9 = 2 + c-9 – 2 = c

c = -11

equation of a straight line, y = 2

1x - 11

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

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

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

4

3


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The Straight Line

6) (-1,-4) and m =5

2 7) (-10,10) and m = -

5

4

…………………………………………………………………………………………….

8) (0,-3) and m =2

1 9) (-6,-2) and m =

3

4

The straight line can also be obtained by using the formula

y –k = m (x –h)

Example Exercise B

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

13

763

637)2(37

+=

+−=

−=−

−=−

x y

x y

x y x y

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

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

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


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4) (4,-9) and m =2

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

……………………………………………………………………………

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

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

4

3

723

15823

82153

)4(2)5(3

)4(3

25

))4((32)5(

−=

−+=

+=+

+=+

+=+

−−=−−

x y

x y

x y

x y

x y

x y

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

8 (-1,-4) and m =5

2 9) (10,10) and m = -

5

4

…………………………………………………………………………………………….

10) (0,-3) and m =2

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

3

4


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The Straight Line

EQUATION OF A STRAIGHT LINE – THROUGH TWO POINTS

Step 1 : Find the gradientStep 2: Find the y- intercept

Step 3 : Write down the equation in a form of y = mx + c

Example Exercise

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

m=)1(3

24

−−

−=

4

2=

2

1

y = mx + c

4 =2

1(3) + c

4 =2

3+ c

C= 4 -2

3

= 22

1

Equation of a straight line, y =2

1x +

2

5

Or 2y = x + 5---------------------------------------------------------------------------------------------------------------------

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

---------------------------------------------------------------------------------------------------------------------

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


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The Straight Line

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

---------------------------------------------------------------------------------------------------------------------

8) ( 4,2

1) and (3,-6) 9) ( 5,

3

1) and ( 3,

3

1−− )

---------------------------------------------------------------------------------------------------------------------

EQUATION OF A STRAIGHT LINE PASSING THROUGH PARALLEL LINES

Lines which are parallel have the same gradient and vice versa.

Example Exercise

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

3=⇒ m

13

233

332

)1(32

−=

+−=

−=−

−=−

x y

x y

x y

x y

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

y = -x + 5

1−=⇒ m

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

6

42

24

)2(14

+−=

++−=

+−=−

−−=−

x y

x y

x y

x y


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3) (8,-5) and 5y + 20x = -2 3) (2,3) and 2x -5y = 125y = -20x -2

y = 5

220 −− x

y = - 4x -5

2

4−=⇒ m

374

5324

3245

)8(4)5(

−=

−−=

−=+

−=−−

x y

x y

x y

x y

………………………………………………………………………………………………

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


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The Straight Line

EXAMINATION FORMAT QUESTION

A(0,4)

B(2,2)

CO

Diagram 1

R O

P 0 6

Q(3,5)

Diagram 2

EXAMPLE 1

(i)

gradient AB= 12

2

20

24−=

−=

−

−

(ii) x-intercept for AB

0=⇒ y

the equation of AB ⇒

4

40

4

)0(14

=

−=−

−=−

−−=−

x

x

x y

x y

x-intercept = 4

(iii)

the equation of AC

102

02=

−

−== OB AC M M

4

4

)0(14

+=

=−

−=−

x y

x y

x y

EXERCISE1.Base on diagram 2

(i) gradient PQ

(ii) x-intercept for PQ

(iii) the equation of PR


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B(-3,3)C(5,4)

F

E

A(2,6)

Diagram 3

R

O P

Q(7,8)

3 In the diagram 4, 6=OP units.

(i) find the gradient of PR

(ii) find the y-intercept of PQ

Diagram 4

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

(ii) the equation of ECF

(iii) x-intercept for ECF


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The Straight Line

A

-6

O

B

4

4 The diagram 5, AB is a straight

line.

(i) state the y-intercept of AB

(ii) find the equation of of AB

Diagram 5

P(0,6)

QO

5. In diagram 6, the gradient of PQ

is2

3− . Line

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

Diagram 6


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The Straight Line

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

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 is3

2

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

8. In diagram 9, JKLM is a trapezium and JK is parallel to ML

Find

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

y

x

D(-4,6)

G(h,4)

E O FDiagram 7

y

x

3

-5

● N(6, p)

Diagram 8

M

L

y K (2,5)

J(-4,2)

L(4,3)

M Ox

Diagram 9


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The Straight Line

SPM PAST YEAR QUESTIONS (PAPER 2)

1. SPM 2003

In Diagram 2, the graph shows that PQ, QR, and RS are straight lines. P is on the y-axis. OP isParallel to QR and PQ is parallel to RS.

The equation of PQ is 2x + y = 5(a) State the equation of the straight line QR(b) Find the equation of the straight line RS and hence, state its y-intercept.

[ 5 marks ]

2. SPM 2004

In Diagram 3, OPQR is parallelogram and O is the origin.

Find

(a) the equation of the straight line PQ,(b) the y-intercept of the straight line QR

[ 5 marks ]

O Q x

P●

● R

● (8, -7)

y

y

● R(4,12)

●Q

● P (3, -6)

0 x

Diagram 3

Diagram 2


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The Straight Line

3. SPM 2005

In Diagram 2, O is the origin, point R lies on the x-axis and point P pies on the y-axis.

Straight line PU is parallel to the x-axis and straight line PR is parallel to straight line ST . Theequation of straight line PR is x + 2y = 14.

(a) State the equation of the straight line PU

(b) Find the equation of the straight line ST and hence , state its x-intercept

[ 5 marks]

4. SPM 2006( June)

In Diagram 3, O is the origin and PQRS is a trapezium. PS is parallel to QR. Straight line RS is

parallel to x-axis. Point Q and S lies on x-axis

Find

(a) Equation of a straight line QR(b) x-intercept of the straight line QR

[ 5 marks ]

●

R(4, -11)

yP (-1, 10)

●

xQ● ●S

Diagram 3

y

x●O R

P● ●U

x + 2y -14

S ●

●T(2, -5)

Diagram 2


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The Straight Line

5. SPM 2006(Nov)

Diagram 5 shows a straight line PQ and a straight line RS drawn on a Cartesian plane. PQ is

parallel to RS.

Finda) the equation of the straight line PQ,

b) the x-intercept of the straight line PQ[ 5 marks ]

6. SPM 2007( June)

In Diagram 2, O is the origin . Point P and point R lie on the y-axis. PS is parallel to the x-axis

and straight line SR is parallel to straight line PQ .

Find(a) the equation of a straight line SR

(b) the x-intercept of the straight line SR[ 5 marks ]

Diagram 5

y

R (0, 4) ●

x

Q●O

P (0, 4)●

●

S (0, 4)

Diagram 2

y

xQ -4,0

S 8 , 3

R

P

O


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7. SPM 2007(Nov)

In Diagram 2, O is the origin .Straight line KL is parallel to straight line MN. The equation ofstraight line KL is 2x + y = 4. The points L and N lie on y-axis .

Findc) the equation of the straight line MN,

d) the x-intercept of the straight line MN[ 5 marks ]

8. SPM 2008( June)

In Diagram 5, PQRS is parallelogram and O is the origin . It is given that the equation of the

straight line PQ is 2y = x + 18.

Find(a) the equation of a straight line RS,

(b) the x-intercept of the straight line RS.[ 5 marks ]

y

M (-3, 7) ●

x

K●

●L

N●

O

Diagram 2

Diagram 5

y

x

R 7, 2

S

Q

P

O


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9. SPM 2008( Nov)

Diagram 6 show a trapezium PQRS drawn on a Cartesian plane. SR is parallel to PQ .

Find(a) the equation of a straight line SR

(b) the y-intercept of the straight line SR[ 5 marks ]

Diagram 6

y

xQ 11,1

S 8 , 9

RP (3 , 5)

O


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The Straight Line

ANSWERS

Gradient

Exercise1) 1

2)3

1

3) 3 4) 25)

2

3

6)2

5

7) 1 8) 19)

3

1−

10) 2

Intercept

x-intercept

1. x = -2

2. x = 2

5

3. x = 4

5

4. x = 3

1

5. x = 2

6. x = 2

5

7. x = 9 8. x = 10 9. x = -20 10. x = -27 11. x = -1212.

3

2

y-intercept

1. y = 8 2. y = -7 3. y = -9 4 y = 6 5. y = -6 6. y = 7

7. y = -10 9. y = -7 10. y = -611. y =

3

2

Equation of a straight line – given the point and gradient

Exercise

1. y = -2x + 17 2. y = 5x +12 3. y = -2x + 14. y =

3

2x – 3 5. y = -

4

3x -19

6. y =5

2x -

5

18 7. y = -

5

4x + 2 8. y =

2

1x – 3 9. y =

3

4x - 1

The equation of a straight line passing through

two points

1. y = x – 12. y =

4

1x +

4

9

3. 3y = x + 5 4. y = -2x – 1 5. 2y = -5x + 1

6. y = 3x – 5 7. y = 6x + 11 8. y = -4x + 6 9. y = 12x + 1


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The Straight Line

Equation of a straight line through parallel lines Exercise

1. 32 −−= x y 2. y = 5x + 6 3. y = 2x + 11 4. y = x – 3 5. y = -3x +11

6. y = 2x -2 7. y = -2x – 3 8. y = 3x + 6 9. y = 3x – 12 10. y = -2x + 3

11. 2y = x + 5

SPM Format Question

1. (i) -3

1 (ii) x = 18

2. (i)53 (ii) y =

53 x + 1 (iii) x = -

35

3. (i)3

4 (ii) y = 8x – 48

4. (i) -6 (ii) y =2

3x – 6

5. (i) Q = (4,0) (ii) 2y = -3x - 9

6. (i) x = 2 (ii) ( 4,0 ) (iii) y = -2x + 2

7. (a) p = 7 (b) y = 2x – 7

8. (a)2

1 (b) 2y = x + 2 (c) (-2,0)


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The Straight Line

Past Years SPM Questions

1. SPM 2003 Q5

(a) x =2

5 (b) y-intercept =9

2. SPM 2004 Q6

(a) y = 3x – 15 (b) y-intercept =20

3. SPM 2005 Q5(a) y = 7 (b) 2y = -x - 8

x = 8

4. SPM 2006 ( June) Q7

(a) y = -2x -3 (b) x = -23

5. SPM 2006 ( Nov) Q10

(a) y = -2x + 11 (b) x –intercept =2

11

6. SPM 2007 ( June) Q7

(a) 34

3−= x y (b) x-intercept = 4

7. SPM 2007 ( Nov) Q5

(a) y = -2x + 1 (b) x –intercept =2

1

8. SPM 2008 ( June) Q7

(a)2

3

2

1−= x y (b) x-intercept = 3

9. SPM 2008 ( Nov) Q5

(a) 132

1+−= y (b) y –intercept = 13

[Worksheet] Chapter 5 - The Straight Line

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