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CHAPTER 5 The Straight Line

FORM 4 CHAPTER 5

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Page 1: FORM 4 CHAPTER 5

CHAPTER 5The

Straight Line

Page 2: FORM 4 CHAPTER 5

Learning Objectives

5.1 Understand the concept of gradient of a straight line.

5.2 Understand the concept of gradient of a straight line in Cartesian coordinates.

5.3 Understand the concept of intercept.5.4 Understand and use equation of a straight

line.5.5 Understand and use the concept of parallel

lines.

Page 3: FORM 4 CHAPTER 5

12

12

xx

yym

cmxy

Page 4: FORM 4 CHAPTER 5

5.1 GRADIENT OF A STRAIGHT LINE

(A) Determine the vertical and horizontal distancesvertical and horizontal distances between two given points on a straight line

E

F

G

Example of application: AN ESCALATOR.

EG - horizontal distance(how far a person goes)

GF - vertical distances(height changed)

Page 5: FORM 4 CHAPTER 5

Example 1State the horizontal and vertical distances for the following case.

10 m

16 m

Solution:The horizontal distance = 16 mThe vertical distance = 10 m

Page 6: FORM 4 CHAPTER 5

(B)Determine the ratioratio of the vertical distance to the horizontal distance

Let us look at the ratio of the vertical distance to the horizontal distances of the slope as shown in figure.

10 m

16 m

Page 7: FORM 4 CHAPTER 5

Vertical distance = 10 mHorizontal distance = 16 m

Therefore,Solution:

6.110

16

distance horizontal

distance vertical

Page 8: FORM 4 CHAPTER 5

5.2 GRADIENT OF THE STRAIGHT LINE IN CARTESIAN COORDINATES

• Coordinate T = (X2,Y1)

• horizontal distance = PT= Difference in x-coordinates

= x2 – x1

• Vertical distance = RT= Difference in y-coordinates

= y2 – y1

y

x0

P(x1,y1)

R(x2,y2)

T(x2,y1)

y2 – y1

x2 – x1

Page 9: FORM 4 CHAPTER 5

REMEMBER!!!For a line passing through two points (x1,y1) and (x2,y2),

where m is the gradient of a straight line

12

12

distance horizontal

distance vertical ofgradient

xx

yyPT

RT

PR

Solution:

12

12

xx

yym

Page 10: FORM 4 CHAPTER 5

Example 2• Determine the gradient of the straight line

passing through the following pairs of pointsi) P(0,7) , Q(6,10)ii)L(6,1) , N(9,7)Solution:

2

1units 6

units 306

710Gradient

PQ

2units 3

units 669

17Gradient

LN

Page 11: FORM 4 CHAPTER 5

(C) Determine the relationship between the value of the gradient and the

(i)Steepness(ii)Direction of inclination of a straight line

• What does gradient represents??Steepness of a line with respect to the x-axis.

Page 12: FORM 4 CHAPTER 5

• a right-angled triangle. Line AB is a slope, making an angle with the horizontal line AC

B

CA

AB ofgradient distance horizontal

distance verticaltan

Page 13: FORM 4 CHAPTER 5

When gradient of AB is positive:

When gradient of AB is negative:

• inclined upwards • acute angle• is positive

• inclined downwards • obtuse angle. • is negative

y

x

y

x0 0

B

A

B

A

tan tan

Page 14: FORM 4 CHAPTER 5

Activity: Determine the gradient of the given lines in figure and measure the angle between the line and the x-axis (measured in anti-clocwise direction)

Line Gradient Sign

MN

PQ

RS

UV

y

x

N(3,3)V(1,4)

R(3,-1)

P(2,-4)U(-1,-4)

M(-2,-2)

0

S(-3,1)

Q(-2,4)

Page 15: FORM 4 CHAPTER 5

REMEMBER!!!The value of the gradient of a line:

• IncreasesIncreases as the steepness increases

• Is positivepositive if it makes an acute angle

• Is negativenegative if it makes an obtuse angle

Page 16: FORM 4 CHAPTER 5

0

y

x

A B

Lines Gradient

AB 0

Page 17: FORM 4 CHAPTER 5

0

y

x

D

C

Lines Gradient

CD Undefined

Page 18: FORM 4 CHAPTER 5

0

y

x

F

E

Lines Gradient

EF Positive

Page 19: FORM 4 CHAPTER 5

0

y

x

H

G

Lines Gradient

GHGH NegativeNegative

Page 20: FORM 4 CHAPTER 5

0

y

x

A

D

HF

B

G

CE

Lines Gradient

AB 0

CD Undefined

EF Positive

GHGH NegativeNegative

Page 21: FORM 4 CHAPTER 5

5.3 Intercepts

• Another way finding m, the gradient:

x-intercept

y-intercept

intercept-

intercept-

x

ym

Page 22: FORM 4 CHAPTER 5

5.4 Equation of a straight line

• Slope intercept form y = mx + c

• Point-slope formgiven 1 point and gradient:

given 2 point:

)( 11 xxmyy

12

12

1

1

xx

yy

xx

yy

Page 23: FORM 4 CHAPTER 5

5.5 Parallel lines• When the gradient of two straight lines

are equal, it can be concluded that the two straight lines are parallel.

Solution:

2x-y=6y y=2x-6 gradient is 2.

2y=4x+3 gradient is 2.

Since their gradient is same hence they are parallel.

2

32xy

Example: Is the line 2x-y=6 parallel to line 2y=4x+3?