Chapter 5 the Straight Line

8/3/2019 Chapter 5 the Straight Line

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THE STRAIGHT LINES

Created By: Mohd Said B Tegoh


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Chapter 5

THE STRAIGHT LINE

5.1 Gradient of Straight Line

5.2 Gradient of Straight Line inCartesian Coordinates

5.3 Intercept

5.4 Equation of A Straight Line

5.5 Parallel Lines


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5.1 Gradient of Straight Line

A

Given a straight line that passes through points A and B

as shown.

A

B

O

OB is known as the vertical distance and OA is known as the

the horizontal distance.

Vertical Distance and Horizontal Distance


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P

Q

5 units

4u

n

its

DIAGRAM 1

vertical

horizontal

A Vertical Distance and Horizontal Distance


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Example 1


A

B

5 units

4u

n

its

Vertical distance

= _____________

Horizontal distance

= _____________

4 units

5 units


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Example 2


A(-4,-2)

Q(2,3)

Vertical distance

= _________________

Horizontal distance

= _________________

3 (-2)

2 (- 4)

3 (-2) = 5

2 (- 4) = 6


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Example 3


A(-5,-4)

Q(3,5)

Vertical distance

= _________________

Horizontal distance

= _________________

5 (-4)

3 (- 5)

5 (-4) = 9

3 (- 5) = 8


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B Determining The Gradient of A Straight Line

The gradient of a straight

line is theratio ofthevertical distance to the

horizontal distance between

two points onthat line.

P

Q

5 units

4u

n

its

Gradient= Vertical Distance

Horizontal Distance

DIAGRAM 1

For straight line PQ,

Gradient of PQ = 4

5

vertical

horizontal


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Find the gradient ofthe

line MN

M

N

DIAGRAM 2

=

63

2


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Notethatthe gradient of a

straight line is equals to tan ,where is the anglethat is

measured from the positive

direction ofthe x-axis to the

straight line.

Gradient of AB

= Vertical Distance

Horizontal Distance

= BC

AC

y

x

A

B

0

C

Tan = BC

AC

Thus,

Gradient of a straight line = Tan


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y

x

A

0

420

B

Example

(a)y

x

N0

600

M(b)

Gradient of straight line AB

= Tan 420

= 0. 9004

Gradient of straight line MN

= Tan 1200

= - Tan 600

= -1.732

1200


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The x-coordinate and y-coordinate


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xA

B

x

x

x

C

D

(-3,2)

(2,0)

(4,-3)

(0,-4)

Ex

(4,4)

F

x

(-7,-2)

The x-coordinate and

y-coordinate


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5.2Gradient of Straight Line inCartesian Coordinates

ACalculating The Gradient of A Straight LinePassing Through Two Points

A(x1,y1)

B(x2,y2)y2

y1

x1 x2x

y

y2 y1

x2 x1

0

Gradient of AB, m= y2 y1x2 x1

= y1 y2

x1 x2


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y

0 x

P(1,1)

Q(6,5)

Example 1


Based on the graph,

Gradient ofPQ, m

=

____________

=

____________

5 - 1

6 - 1

4

51

2

3

4

5

1 2 3 4 5 6


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y

0 x

P(6,1)

Q(1,5)

Example 2


Based on the graph,

Gradient ofPQ, m

=

____________

=

____________

= ____________

5 - 1

1 - 6

4

-5

- 45

1

2

3

4

5

1 2 3 4 5 6


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Example 3


POINT P POINT Q GRADIENT

(2,-5) (-1,3) 3-(-5) = -8

-1-2 3

(-1,0) (3,3)

(7,8) (5,2)

(-1,-5) (-8,-12)

3 O = 3

3 (-1) 4

2 8 = -6 = 3

5 7 -2

-12(-5) = -7 = 1

-8 (-1) -7


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5.3 Intercept

A The x-intercept and The y-intercept ofA Straight Line

y

x

N0

M(0,b)

(a,0)

The x-intercept of a straight line

is the x-coordinate at which that

line cuts the x-axisThe y-intercept of a straight line

is the y-coordinate at which that

line cuts the y-axis

y-intercept = b

x-intercept = a


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y

0 xP(6,0)

Q(0,5)

Example 1

AThe x-intercept and The y-intercept ofA Straight Line

y-intercept = 5

x-intercept = 6


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y

x

0

P(0,-4)

Q(5,0)

AThe x-intercept and The y-intercept ofA Straight Line

y-intercept = -4

x-intercept = 5

Example 2


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y

x

0

P

Q

B Calculating The Gradient of A Straight LineGiven The x-Intercept and y-Intercept

y-intercept = -4

x-intercept = 5-4

5Gradient

, m= - y-intercept

x-intercept

The gradient of a straightline can also be calculated

Ifwe knowthe x-intercept

and y-intercept ofthe line.

Gradient = --4

5

= 4

5

Example 1


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Based on the formula

CPerforming Calculations Involving Gradient,x-intercept and y-intercept

Gradient, m = - y-intercept

x-intercept

We can find the value oftheinterceptifwe knowthegradientand of

the intercepts.That is;

x-intercept = -y-intercept

m

y-intercept = - (m x x-intercept)

Performing Calculations Involving Gradient,x-intercept and y-intercept


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EXAMPLE 1

(a) If a straight linehas a ground of -1 and y-intercept of 3,2

find the x-intercept ofthe line.

SOLUTION

x-intercept = -y-intercept

m

y-intercept = 3, Gradient = -1

2

= -3

-1

2

= 3 x 2 = 6


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EXAMPLE 2

(b) A straight line passes throughthe points(-8,0) and y-intercept (0,p).Ifthe linehas a gradient of 5 , find the value of p.

4

SOLUTION

x-intercept = -8, Gradient = 5 , p = y-intercept

4

p = - 5 x (-8)4

= 10

p = - (m x x-intercept)


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5.4 Equation of A Straight Line

ADrawing The Graph of A Straight Line GivenAn Equation of The Form y = mx + c

To drawthe graph of anequation inthe form y = mx + c, it is sufficient

to find two points that satisfy theequation.

Notethatgraph that is obtained is a straight line.

Steps in drawing the graphwithequation y = mx + c:

find the x-intercept

find the y-intercept

plotthetwo points onthe Cartesian plane

using a ruler, draw a straight line joining thesetwo points

label the graph

Drawing The Graph of A Straight Line GivenAn Equation of The Form y = mx + c


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Example

A Drawing The Graph of A Straight Line GivenAn Equation of The Form y = mx + c

y = 2x + 4

1

2

3

4

-3 -2 -1 0 1

5

y

x

find the x-intercept of

y = 2x + 4.

Let y = 0

2x + 4 = 0

2x = -4

x = -2Thus, x-intercept = -2

find the y-intercept of

y = 2x + 4.

Let y = 0

y = 2(0) + 4

y = 4

Thus, y-intercept = 4

plottwo points onthe

CartesianPlane

draw a straight linethat

passes these points

y = 2x + 4


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B Writing The Equation of The Straight Line GivenThe Gradient and y-Intercept

Ifwe are giventhe gradient and y-intercept of a straight line, we

can find theequation ofthe line by substituting inthe given

values into the general equation y = mx + c.

ExampleFind theequation ofthe linewith gradient 4 and y-intercept = -2.

7

Gradient, m = 4 , y-intercept, c = -2

7

Solution

Substitute m = 4 and c = -2 into theequation y = mx + c.

7

y = 4

7x + ( )-2

= 4x - 2

7

Thus, theequation ofthe line is y= 4x - 2

7


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Gradient, m y-intercept Equation,

y = mx + c

3 -2

2

5

-8

-5 -3

y = x + ( )-23

y = 3x -2

y = x + ( )-825

y = 2x -8

5

y = x + ( )-3-5

y = -5x -3


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C Determining The Gradient and y-Intercept of AStraight Line from Its Equation

Given anequation ofthe form y = mx + c, we caneasilydetermine its gradient and y-intercept by looking atthe values

of m and c.

y = - a

bx +

If anequation ofthe form ax + by = c is given, we can determine

its gradient and y-intercept by simply writing theequation inthe

form y = mx + c.

ax + by = c

by = -ax + c

c

b


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C Determining The Gradient and y-Intercept of AStraight Line from Its Equation

Given anequation of a straight line is 4x + 2y = 7.Find

(a) the gradient,

(b) the x-intercept ofthe straight line.

SPM Clone

Solution4x + 2y = 7

2y = -4x + 7

y = -4x + 7

2 2

y = -2x + 7

2

a

The gradient ofthe line is -2

4x + 2y = 7

4x = 7

x= 7

4

b

The x-intercept ofthe line is 7

4

4x + 2(0) = 7

Let y = 0


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D (i) Finding The Equations of A Straight Line WhichIs Parallel to The x-Axis or y-Axis

y= a

x = b

y

y

x

x

0

0

In general, a linethat isparallel

to x-axis and has y-intercept

a has anequation ofthe form

y= a

a

b

In general, a linethat isparallel

to y-axis and hasx-interceptb has anequation ofthe form

y= b


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1

2

3

-3 -2 -1 0 1

5

y

x2 3

P4

y = 4


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1

2

3

-3 -2 -1 0 1

5

y

x3

P

4

x = 2

2


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Example 1The straight line MN is a parallel to the x-axis and passes throughthe point (5,7).

Writetheequation of line MN.

Solution

(5,7)y

x

0

7

5

y-intercept = 7

Thus, theequation of line MN is y = 7


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Example 2Find theequation ofthe straight linewhichhas a gradient of 1 and passes through

2

The point (-4,6)

SolutionSubstitutem = , x = -4 and y = 6 into theequation y = mx + c

y = mx + c

6 = ( ) + c-4

c = 8

Thus, theequation ofthe straight line is y = x + 8

D (ii) Finding The Equations of A Straight Line GivenThe Gradient and A Point On The Line


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Example 3Find theequation ofthe straight linethat passes throughthe points (-3,1) and

(2,6).

Solution

Substitutem = 1 x = 2 and y = 6 into theequation y = mx + c

y = mx + c

6 = 1 ( 2 ) + c

c = 4

Thus, theequation ofthe straight line is y = x + 4

D (iii) Finding The Equations of A Straight Line GivenTwo Points

Lettheequation of a straight line be y = mx + c.

Let (x1,y1) = (-3,1) and (x2,y2) = (2,6).

Gradient, m = 6 -1 = 1

2-(-3)


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1

2

3

-3 -2 -1 0 1

5

y

x2 3

4

5.5 Parallel Lines

P

Q

K

L

M

N

-1

GradientPQ = -3

-3

= 1

Gradient, m= - y-intercept

x-intercept

Gradient KL = -1

-1

= 1

Gradient MN = --1

1

= 1


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5.5 Parallel Lines

ADetermining Whether Two Straight Lines Are

Parallel Given The Equations

Example 1

0

2

-4

y

x

y = 3x + 2

y = 3x

y = 3x - 4

If two lines are parallel,then m1 = m2

Given two lines with

the equations l1 = m1x + c

and l2 = m2x + c;

If m1 = m2, then

two lines are parallel.

All thethree lines above are parallel and they

all have a gradient of 3.

2

3

4

3


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BFinding The Equation of A Straight Line Which PassesThrough A Given Point and Is Parallel to AnotherStraight Line.

Find theequation ofthe straight linethat is parallel to the straight line

2x + 3y = 27 and passes throughthe point (-1,6).

Lettheequation of a straight line be y = mx + c.

2x + 3y = 27

3y = -2x + 27

y = -2x + 9

3

Substitutem = -2 and the point (-1,6) into y = mx + c.

3y = mx + c

6 = -2 ( -1 ) + c

3

c = 16

3

Thus, theequation ofthe straight line is y = -2x + 16

3 3

Thus, m = -2

3

Example 1

Solution


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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.

SPM 2003:No.5In Diagram 1, the graph shows that PQ, QR, and RS are straight lines. P is on the

y-axis. OP is parallel to QR and PQ is parallel to RS.

y

x0

P

R

Q

DIAGRAM 1

S (8,-7)

y


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y

x0

P

R

QS (8,-7)y = 0

2x + 0 = 5

x = 5/2

2x + y =5

(a) x = 5

2

Solution:


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y

x0

P

R

Q

S (8,-7)

2x + y =5 (b) MRS = MPQ = -2

C = 9

y = -2x + 9y-intercept = 9

8 -7

= ( ) + Cy = mx +c

-2

Solution:


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9int

929

)8(272)(

2

5)(

!

!

!

!

!!

!

ercepty

xyC

CMMb

xa

PQRS

P1

P1K1

N1

P1


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Find

(a) the equation of the straight line QR,

(b) the equation of the straight line RT,

(c) the y-intercept of the line PT.

SPM TRIAL 2010:No.7Diagram 7 shows a trapezium PQRT drawn on a Cartesian plane. PQ is

parallel to RT and QR is parallel to x-axis. The equation of straight line PQ isy 2x 6 = 0. y

x0P

RQ

DIAGRAM 7

3

T

y 2x 6 = 0


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y

x0P

RQ

3

T

(a) the equation of the straight line QR,

Answer:

(a)

y 2(0) 6 = 0

y = 6

6

y = 6 P1


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y

x0P

RQ

3

T

(b) the equation of the straight line RT,:

y = 2x + 6

m = 2

6

2!! PQRT mm

(3,0)

0 = 2(3) + C

C = -6

y = 2x 6

P1

K1

N1

(c) y-intercept of line PT,

6 P1


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The diagram below shows the straight lines PQ

and SRT are parallel.

Find

(a) the gradient ofthe linePQ.

(b) theequation ofthe lineSRT.

(c) the x- intercept ofthe lineSRT.

Cloned SPM


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Find

(a) the gradient ofthe linePQ.Solution:

2

5

10

)2(3

212

!

!

)(a

Cloned SPM


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Solution:(b) theequation ofthe lineSRT.(c) the x- intercept ofthe lineSRT. )(b 2!! SRTPQ mm

c! )5(25

5!c

52 ! xy

)(c 052 !x

2

5!x

2

5int ! erceptx

Cloned SPM


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The diagram below shows thatthe straight line

EF and GH are parallel.

Find

(a) theequation of EF.

(b) the y - intercept and

x - intercept of EF.

Cloned SPM

Cl d SPM


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EF

and GH are parallel.Find

(a) theequation of EF.

Cloned SPM

Solution:)(a

5

7

)1(4

)5(2!

!

!EFGH

mm

c! )1(5

75

5

18!c

5

18

5

7! xy


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EF and GH are parallel.

(b) the y - intercept and

x - intercept of EF.

Cloned SPM

Solution:1875 ! xy

0187 !x

7

18!x

7

18int ! erceptx


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The diagram below shows STUV is a trapezium.

Giventhat gradient ofTU is -3, find

(a) the coordinates of pointT.

(b) theequation of straight lineTU.

(c) the value of p, iftheequation

of straight lineTU is 63

12 ! xy

Cloned SPM


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Giventhat gradient ofTU is -3, find

(a) the coordinates of pointT.

Cloned SPM

Solution:

)20,0(

20

182

360

2

T

p

p

p

!

!

!


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(b) theequation of straight lineTU.

Cloned SPM

Solution:

203

20

3

!

!

!

xy

c

mTU


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(c) the value of p, iftheequation

of straight lineTU is 63

12 ! xy

Cloned SPM

Solution:6

3

12 ! xy

3

3

62

6)0(

3

12

!@

!

!

!

p

y

y

y

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Chapter 5 the Straight Line