CHAPTER 5The S t ra igh t L ine

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.

12

12

xx

yym

−−=

cmxy +=

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)

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

(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

Vertical distance = 10 m

Horizontal distance = 16 m

Therefore,

Solution:

6.110

16

distance horizontal

distance vertical

=

=

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

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

−−=

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

(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.

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

B

CAθ

θ

AB ofgradient distance horizontal

distance verticaltan

=

=θ

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

θ θ

θtan θtan

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)

θ

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

0

y

x

A B

Lines Gradient

AB 0

0

y

x

D

C

Lines Gradient

CD Undefined

0

y

x

F

E

Lines Gradient

EF Positive

0

y

x

H

G

Lines Gradient

GHGH NegativeNegative

0

y

x

A

D

HF

B

G

CE

Lines Gradient

AB 0

CD Undefined

EF Positive

GHGH NegativeNegative

5.3 Intercepts

• Another way finding m, the gradient:

x-intercept

y-intercept

intercept-

intercept-

x

ym −=

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

−−=

−−

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?