51955900 form-4-chapter-5
TRANSCRIPT
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CHAPTER 5The
Straight Line
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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.
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12
12
xx
yym
cmxy
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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)
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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
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(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
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Vertical distance = 10 mHorizontal distance = 16 m
Therefore,Solution:
6.110
16
distance horizontal
distance vertical
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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
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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
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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
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(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.
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• 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
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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
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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)
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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
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0
y
x
A B
Lines Gradient
AB 0
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0
y
x
D
C
Lines Gradient
CD Undefined
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0
y
x
F
E
Lines Gradient
EF Positive
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0
y
x
H
G
Lines Gradient
GHGH NegativeNegative
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0
y
x
A
D
HF
B
G
CE
Lines Gradient
AB 0
CD Undefined
EF Positive
GHGH NegativeNegative
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5.3 Intercepts
• Another way finding m, the gradient:
x-intercept
y-intercept
intercept-
intercept-
x
ym
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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
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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?