1. CHAPTER 5
The Straight Line

2. 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.

3. y2 − y1
m=
x2 − x1
y = mx + c

4. 5.1 graDient OF a
straigHt Line
(A) Determine the vertical and horizontal distances
between two given points on a straight line
F
E G
Example of application: AN ESCALATOR.
EG - horizontal distance(how far a person goes)
GF - vertical distances(height changed)

5. Example 1
State the horizontal and vertical
distances for the following case.
10 m
16 m
Solution:
The horizontal distance = 16 m
The vertical distance = 10 m

6. (B)Determine the ratio of the vertical
distance to the horizontal distance
10 m
16 m
Let us look at the ratio of the vertical distance
to the horizontal distances of the slope as
shown in figure.

7. Vertical distance = 10 m
Horizontal distance = 16 m
Therefore,
Solution:
vertical distance 16
=
horizontal distance 10
= 1.6

8. 5.2 GRADIENT OF THE STRAIGHT LINE IN
CARTESIAN COORDINATES
y • Coordinate T = (X2,Y1)
• horizontal distance
R(x2,y2)
= PT
= Difference in x-coordinates
y 2 – y1
= x2 – x 1
x2 – x 1 • Vertical distance
P(x1,y1) T(x2,y1)
= RT
x = Difference in y-coordinates
0
= y2 – y 1

9. Solution:
vertical distance
gradient of PR =
horizontal distance
RT
=
PT
y 2 − y1
=
x2 − x1
REMEMBER!!!
For a line passing through two points (x1,y1) and (x2,y2),
y2 − y1
m=
x2 − x1
where m is the gradient of a straight line

10. Example 2
• Determine the gradient of the straight line
passing through the following pairs of points
i) P(0,7) , Q(6,10)
ii)L(6,1) , N(9,7)
Solution:
10 − 7 7 −1
Gradient PQ = Gradient LN =
6−0 9−6
3 units 6 units
= =
6 units 3 units
1 =2
=
2

11. (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.

12. B
• a right-angled triangle. Line
AB is a slope, making an
angle θ with the horizontal
line AC
θ
A C
vertical distance
tan θ =
horizontal distance
= gradient of AB

13. y y
B
B
θ θ
x x
0 0 A
A
When gradient of AB is When gradient of AB is
positive: negative:
• inclined upwards • inclined downwards
• acute angle • obtuse angle.
• tan θ is positive • tan θ is negative

14. 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)
y
Line Gradient Sign θ
V(1,4) N(3,3)
Q(-2,4)
MN
S(-3,1)
0
x PQ
M(-2,-2)
R(3,-1) RS
U(-1,-4)
P(2,-4)
UV

15. REMEMBER!!!
The value of the gradient of a line:
• Increases as the steepness increases
• Is positive if it makes an acute angle
• Is negative if it makes an obtuse angle

16. Lines Gradient
y
AB 0
A B
0 x

17. Lines Gradient
y
D
CD Undefined
C
0 x

18. Lines Gradient
y
F
E
EF Positive
0 x

19. Lines Gradient
y
H
G
0 x
GH Negative

20. Lines Gradient
y
AB
D
H
F
0
A B
G
CD Undefined
E
EF
C
Positive
0 x
GH Negative

21. 5.3 Intercepts
y-intercept
x-intercept
• Another way finding m, the gradient:
y - intercept
m=−
x - intercept

22. 5.4 Equation of a straight line
• Slope intercept form
y = mx + c
• Point-slope form
given 1 point and gradient:
y − y1 = m( x − x1 )
given 2 point:
y − y1 y 2 − y1
=
x − x1 x2 − x1

23. 5.5 Parallel lines
• When the gradient of two straight lines
are equal, it can be concluded that the
two straight lines are parallel.
Example:
Is the line 2x-y=6 parallel to line 2y=4x+3?
Solution:
2x-y=6y→ → is 2.
y=2x-6 gradient
3
2y=4x+3 → + 2 →
y = 2x gradient is 2.
Since their gradient is same hence they are parallel.