gradient ppoint
TRANSCRIPT
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SEK. MEN. TEKNIK TUANKU JAAFAR
SUBJECT : MATHEMATICS
PREPARED BY :
PN. NORMALI ABDUL GHANI
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STRAIGHT LINE
GRADIENT
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WHICH PATH IS EASY TO RUN ON
AND WHICH IS DIFFICULT TO RUN.
WHY?
A B
C
D
In mathematics measuring the steepnessof a slope is known as the gradient (m)
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GRADIENT(m)
VERTICAL DISTANCE
HORIZONTALDISTANCE
Horizontal
Distance
VerticalDistance
GRADIENT = Vertical Distance
HorizontalDistance
B
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Find the vertical distance and the horizontaldistance for the straight line below:
Vertical distance=Horizontal distance
=
63
GRADIENT (m) = Vertical
DistanceHorizontal
Distancem = 6
3
m = 2
Use the formula to get thegradient
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Find the gradient of AB, BC and CD.
A
C
D
Straight lines have the samegradient
m AB = 2/2= 1
m BC = 5/5
= 1
m CD = 3/3
= 1
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P
Q
R
m PQ = m QR = m PR
Given the gradient of PQ is 5.Determine the gradient of
QR and PR.
Then m QR = m PR= 5
Therefore since PQR arecollinear points of a straight
line,
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Gradient of straight line on theCartesian Plane
Q(x2,y2)
P(x1,y1)
y2 y1
x2 x1
y2
y1
x1 x2
= y2 y1
x2 x1
y
x
GRADIENT (m) = VerticalDistance
Horizontal
Distance
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Gradient of two point on theCartesian Plane
Example 1:
A(2 , 2) and B(4 , 8)x1 y1 x2 y2
m = y2 - y1x2 - x1
m = 8 2
4 2
= 3
Or A(2 , 2) and B(4 , 8)x2 y2 x1 y1
m = 2
82 4
= 3
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Example 2
P(4, 0) and Q(-2,
12)
x1 y1 x2, y2
mPQ = 12 0
-2
4mPQ = 12
- 6
= - 2
m = x2 -x1
y2 -
y1m = x2 -
y1y2 -
x1m = y2 -
x1x2 -
y1
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Find thegradient:Q1. P(1, 1) and Q(3, 9)
Q2. R(2, -3) and S( - 4, 3)
m = 9 1
3 1
m = 4
m = 3 (-3)
- 4 2
m = -1
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Horizontal Distance
A
B
Vertical
Distance
Adjacent side
A
B
Opposite
side
GRADIENT= VerticalDistance
Horizontal
Distance
Tan = Oppositeside
Adjacent side
Gradient = Tan
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400
Gradient = Tan 400
m PQ =0.839
P
Q
Find the gradient of the straight linePQ if
PQR = 400
.
Gradient = Tan
R
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1. Find the gradient of the straight line ABand EF.
m AB =
m EF =
2. Find the gradient of the straight line ona
Cartesian plane.
A
B
F
E
(6,10)
(12,4)
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1. Find the gradient of a straight line thatjoins these two points
i. (2,0) and (4,8)ii. (-9,0) and (-12,15)
4. Diagram shows a straight lineOA subtends an angle withthe x- axis. Find
gradient b) tan
A ( 5, 3)y
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Answer for the worksheet:
Q1. m AB = 63
= 2
m EF = 36
= 0.5
Q2. m = 10 4
6 12
Or m = 4 1012 6
= - 1 = - 1
Q3. i) m = 8 0
4 2= 4
Or m = 0 8
2 4
= 4
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Q3. ii) m = 15 0
-12 (-9) = - 5
Or m = 0 15
-9 (-12)= - 5
Q4. a) m OA = 3 0
5 0
= 3
5
b) tan OA = m
OA= 3
5
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INTERCEPT
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Y-intercept
X-intercept
y
x
Gradient = - y-intercept
x-intercept
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6
3
y
x
Gradient = - 63
m = - 2
12
- 4
y
x
Gradient = - 12-4
m = 3
1 2
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Based on the graph given, determine
B(5,2)
y
xO
i. gradient of the straight line BC
ii. x-intercept of the straight line AD,
given the gradient of AD is 4
mBC = mBO
* B(5, 2) and O(0, 0)
x2 y2 x1 y1
mBC = 2 1 = 1
5 0 5
mAB = _ y-intercept
x-intercept
8
x-intercept = _ __8_
mAB
= _ _8_
- 4
= 2
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Conclusion GRADIENT
m
Vertical Distance
Horizontal Distance
y2 - y1x2 - x1
A straight line have
a same gradientfrom any two
points
Tan
Gradient = - y-interceptx-intercept
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EQUATION OF A STRAIGHT LINE
y = mx +
cm : gradientc : y - intercept
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EQUATION OF A STRAIGHT LINE
(0, c)
y
x0
Equation of a straight
line :
y = mx + c
Where m : gradient
c : y-intercept
Example :
y= 3x + 7
m= 3 and c = 7
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Draw the graph for each of thefollowing equation :
1. y = x + 5
0y
0x
5
- 5
y
x
2. y = 2x - 6
0y
0x
3
- 6
y
x
- 5
5
3
-6
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Determine whether the given point lies on
the given straight line below:
EXAMPLE:
i. (3, 4) ; y = 2x - 2
x = 3 and y = 4
y = 2x - 2
4 = 2(3) 2
4 = 4
LHS = RHS
ii. (-1, 3) ; y = 3x + 1
x = -1 and y = 3
y = 3x + 1
3 = 3(-1) + 1
3 = -2
LHS RHS
Yes No
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Determine whether the given point lies on
the given straight line below:
iii. (-2,-3) ; y = 3x +3
iv. (4, 0) ; y = 3x - 11
yes
No
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Determine the gradient and the y-intercept
of each of the following straight lines:
iii. 2y = 5x - 2 iv. 3x + 2y = 5
m= 5/2
c= -1
m= -3/2
c= 5/2
DETERMINE THE
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DETERMINE THEEQUATION
OF A STRAIGHT LINE
i. Given : m and c
Equation of a straight
line :
y = mx + cWhere m : gradient
c : y-intercept
Eg.
m=2 and c=8
y= 2x + 8
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(1,2)
m=3
ii. Given : one point and m
y= 3x + c : x=1, y=2
2= 3(1) + c
2= 3 + c
- 1 = c
y= 3x -1
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(1,2)
iii. Given : two points
m = 8 2
4 1
m = 2
y = 2 x + c
y= 2x + 0
(4,8)
y = 2 x + c , (1 , 2)
x = 1 , y = 2
2 = 2(1) + c
c = 0
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1. Find the equation of straight line whichpases through the point given and has agradient m.
a. (1, 3) and m =
1b. (-4, 5) and m =-6
2. Find the equation of straight line whichpases through the point given.
a. (5, 2) and (3,10)a. (-5, -3) and(-8,1)
PRACTICE
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Conclusion
EQUATIONOF A
STRAIGHTLINE
Given :
two points
Given :
m and c
y = mx + c
Where m :gradient
c: y-intercept
Given : one
point and m