Gradient Ppoint

SEK. MEN. TEKNIK TUANKU JAAFAR SUBJECT : MATHEMATICS

PREPARED BY :

PN. NORMALI ABDUL GHANI

STRAIGHT LINE

GRADIENT

WHICH PATH IS EASY TO RUN ON AND WHICH IS DIFFICULT TO RUN. D WHY?

C A

B

In mathematics measuring the steepness of a slope is known as the gradient (m)

GRADIENT (m)

B

 

VERTICAL DISTANCE HORIZONTAL DISTANCE

Vertical Distance Horizontal Distance

A

GRADIENT = Vertical Distance Horizontal Distance

Find the vertical distance and the horizontal distance for the straight line below: Vertical distance 6 = Horizontal distance 3 = Use the formula to get the gradient GRADIENT (m) = Vertical Distance Distance

m =Horizontal 6 3 m= 2

Find the gradient of straight line I, II and III

I

II

III

GRADIENT = Vertical Distance Horizontal Distance mI=4=1 4

m II = 6 = 3 2

m III = 3 = 1 6

Find the gradient of AB, BC and CD. D m AB = 2/2 =1

C

m BC = 5/5 =1

B A

m CD = 3/3 =1

Straight lines have the same gradient

R

Given the gradient of PQ is 5. Determine the gradient of QR and PR. Q P

Therefore since PQR are collinear points of a straight line,

m PQ = m QR = m PR Then m QR = m PR =5

y

Gradient of straight line on the Cartesian Plane Q(x2,y2)

y2

y2 – y1

y1

P(x1, y1) x1

x2 – x1 x2

GRADIENT (m) = Vertical Distance Horizontal Distance

x = y2 – y1 x2 – x1

Gradient of two point on the Cartesian Plane m

=

Example 1:  A(2 , 2) and B(4 , 8) x1 y1 x2 y2 

m =

=

8–2 4–2 3

y 2 - y1 x2 - x1 Or

A(2 , 2) and B(4 , 8) x2 y2 x1 y1 

8

m =

2– 2–4

=

3



Example 2 P(4, 0) and Q(-2, 12) x1 y 1 x2, y2

mPQ = 12 – 0 -2 – 4 mPQ =

12 -6

= -2

m = x2 x1 y2 y1 m = x2 y1 y2 x1 m = y2 x1 x2 y1

-

Find the gradient: Q1. P(1, 1) and Q(3, 9) m=

9–1 3–1

m= 4 Q2. R(2, -3) and S( - 4, 3) m= (-3)

3– -4–2

m = -1

B

B

Vertical

Opposite

Distance

side

Ө

A Horizontal Distance GRADIENT = Vertical Distance

A

Adjacent side TanӨ side

Horizontal Distance

Gradient = Tan Ө

= Opposite Adjacent side

Find the gradient of the straight line PQ if ∠PQR = 400. P

400

R

Q

Gradient = Tan Ө Gradient = Tan 400 m PQ 0.839

=

1. Find the gradient of the straight line AB and EF. B F m AB =

A

E

m EF =

2. Find the gradient of the straight line on a Cartesian plane. (6,10)

(12,4)

1. Find the gradient of a straight line that joins these two points i. (2,0) and (4,8) ii. (-9,0) and (-12,15)

4. Diagram shows a straight line OA subtends an angle θ with y the x- axis. Find • gradient

b) tan θ

A ( 5, 3)

θ

x

Answer for the worksheet: Q1.

m AB = 6 3 =2

Q2. m = 10 – 4

m EF = 3 6 = 0.5 Or

12 – 6

= 6- –1 12 Q3. i) m = 8 – 0 = 44– 2

m = 4 – 10 = -1

Or

m=0–8 2–4 = 4

Q3. ii) m = 15 – 0 -12 – (-9)

Or

m = 0 – 15 -9 – (-12)

= -5

= -5

Q4. a) m OA = 3 – 0 5–0 = 3 5

b) tan OA = m OA = 3 5

INTERCEPT

y

Yintercept X-intercept x

Gradient = - y-intercept x-intercept

1

2

y

y

12 6

3

Gradient = - 6 3 m =-2

x

-4

x

Gradient = - 12 -4 m =3

Based on the graph given, determine i. gradient of the straight line BC y 8

mBC = mBO

A

* B(5, 2) and O(0, 0) x2 y2

x1 y1

B(5,2) D C

O

x

mBC = 2 – 1 = 1 5–0 5 ii. x-intercept of the straight line AD, given the gradient of AD is – 4

mAB = _ y-intercept

x-intercept = _ __8_

x-intercept

mAB

= _ _8_ -4

= 2

Conclusio n

GRADIENT m

Vertical Distance Horizontal Distance

A straight line have a same gradient from any two points

y2 - y1 x 2 - x1 Tan Ɵ

Gradient = - y-intercept x-intercept

EQUATION OF A STRAIGHT LINE

y = mx + m : gradient c c : y - intercept

EQUATION OF A STRAIGHT LINE y

Equation of a straight line : y = mx + c

(0, c)

Where m : gradient 0

x

c : y-intercept Example : y= 3x + 7 m= 3 and c = 7

Draw the graph for each of the following equation : 1. y = x + 5 x y

2. y = 2x - 6

0 -5 5 0

x y

0 3 -6 0

y

y

5 3

-5

x

x

-6

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

ii. (-1, 3) ; y = 3x + 1 x = -1 and y = 3 y = 3x

+1

4 = 2(3) – 2

3 = 3(-1) + 1

4 =

3 =

4

-2

LHS = RHS

LHS ≠ RHS

Yes

No

Determine whether the given point lies on the given straight line below:

iii. (-2,-3) ; y = 3x +3

yes

iv. (4, 0) ; y = 3x - 11

No

Determine the gradient and the y-intercept of each of the following straight lines: EXAMPLE:

i. y = 5x + 3 y = mx + c m = 5 and c = 3

ii. 3y = 6x + 4 y = mx + c y = 6x + 4 3

3

y = 2x + 4/3 m= 2 and c = 4/3

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

m= -3/2

c= -1

c= 5/2

DETERMINE THE EQUATION OF A STRAIGHT LINE i. Given : m and c Equation of a straight line : y = mx + c Where m : gradient c : y-intercept Eg. m=2 and c=8 y= 2x + 8

ii. Given : one point and m m=3 (1,2) y= 3x + c : x=1, y=2 2= 3(1) + c 2= 3 + c -1=c y= 3x -1

iii. Given : two points (4,8) (1,2) m=8–2 4–1

y = 2 x + c , (1 , 2) x=1 ,y=2

m=2

2 = 2(1) + c

y=2x+c

c=0

y= 2x + 0

PRACTICE 1. Find the equation of straight line which pases through the point given and has a gradient m. a. (1, 3) and m = 1 b. (-4, 5) and m = -6 2. Find the equation of straight line which pases through the point given. a. (5, 2) and (3, 10) a. (-5, -3) and (-8,1)

Conclusio n

y = mx + c Where m : gradient c : yintercept

EQUATION OF A STRAIGHT LINE

Given : m and c

Given : one point and m Given : two points