Gradient Ppoint

8/14/2019 Gradient Ppoint

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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

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Gradient Ppoint

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