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GRADIENT AND EQUATION

OF A STRAIGHT LINE

Bumpy Road

When you ride a car, bicycle, or other type of vehicle, you must

have passed a flat road, a road that goes down, and a road

that goes up. Roads that go up or down usually have a certain

slope that has been taken into account the level of slope, so it

is safe and comfortable for vehicles to pass. The uphill road

also has a slope. if it is too steep, the vehicle will experience,

difficulty crossing it.

KEY VOCABULARY

Straight line equation

Graphs

Gradient

cut point

BASIC COMPETENCE

8As4. Construct tables of values and use all four quadrants

to plot the graphs of linear functions, where y is given

explicitly in terms of x; recognise that equations of the form

y = mx + c correspond to straight-line graphs.

WHY STUDY THIS CHAPTER

Learning about gradient will open doors to the careers in

mathematics and physics. The formulae used can

provide exact calculation to problems in product design.

In addition, construction engineers, especially those

involved in land surveying, use gradient to determine the

stabillity or elevation of a building area.

Diagnose Readiness

Quick Check Quick Review

Relation Example 1

Known set A = {Jakarta, Bangkok, Known P = {2, 4, 6} and Q = {2, 3, 4}.

Tokyo, Manila} and set B = The set of sequential pairs from P to

{Indonesia, Japan, Thailand, Q that expresses "multiples of" is . . .

Philippines, Malaysia}. The relation -> The set of sequential pairs from P

from A to B can be expressed by... to Q that express "multiples of" are:

Known K = {2, 3, 4, 5} and L = {3, 4, {(2, 2), (4, 2), (4, 4), (6, 2), (6, 3)}.

5, 6, 8, 10, 12}. If a set of sequential

pairs is specified {(2, 4), (3, 6), (4, Example 2

8), (5, 10)}, then the relation from set

K to set L is . . If A = {1, 3, 5} and B = {2.4} then A x B

Take a look at the arrow chart

below! is . . . .

-> A x B = {(1, 2), (1, 4), (3, 2), (3, 4),

(5, 2), (5, 4)}

Example 3

→On the f:x 2x−2 mapping, the

→shadow of 2 is . . .

-> f:x 2x − 2

The relation from A to B is . . . . y = f(x) = 2x − 2

Function The shadow of 2 is f(2) = 2.2 − 2 = 4

→The function f:x Example 4

ax - 5 is known. If →Function f:x

→f(−2) = 3 then the value of a is . . . . 3x + 5 If f(a) = −1 then a

Known function f:x 2x(x−3). The value →= . . . .

of f(5) is . . . . -> f:x 3x + 5

A function f(x) = mx − 3 maps 2 to 1. The f(x) = 3x + 5

→map of 4 is . . . . f(a) = 3a + 5 = −1

In the f:x ax + b mapping, if f(2) = 3a = −1 − 5

3 and f(3) = 5, then the values of f(1) 3a = −6

and a − b= . . . . a = −6 : 3

a = −2

EQUATIONLION

FESTRAIGHT

The equation of a straight line is a mathematical

equation that gives the relation between the

coordinate points lying on that straight line. It can

be written in different forms and tells the slope, x-

intercept, and y-intercept of the line.

In general, there are two kinds of straight-

line equations, so the way to determine the

gradient is also different, depending on the

shape of the line equation.

The equation of the line y = mx + c

The gradient is the coefficient of the

variable x itself, i.e. m

Equation of line ax + by + c = 0

If you know the equation of the line ax + by

+ c = 0, then the first step you have to do is

change the equation of the line to the form

y = mx + c

EXAMIPdLEentify Equation Of

Straight Line

GRAD

IENT

Gradient is a value that shows

the slope of a straight line

What does it stand for?

Equation Of a Straight Line

y = mx + b

Slope or y value when x = 0

Gradient (see Y Intercept)

y = how far up

x = how far along

m = Slope or Gradient (how

steep the line is)

b = value of y when x=0

HOW DO YOU FIND "M" AND "B"?

b is easy: just see where

the line crosses the Y axis.

m (the Slope) needs some

calculation:

EXAMPLE

The gradient of the equation

y = 2x + 1 is...

DISCUSS

What are you trying to find ?

How to draw a graph of the equation of a straight line

y = 2x + 1 ?

EXPLORE

y = mx + b

SOLVE

b = 1 (value of y when x = 0)

CONFIRM

With that equation we can now choose any value for x and find the

matching value for y. For example, when x is 1:

y=2x1+1=3

Check for yourself that x=1 and y=3 is actually on the line.

POSITIVE SALNO

DPENEGATIVE

Positive Slope Negative Slope

The line rises from left to right The line falls from left to right

As an example, let's look at the As an example, let's look at the equation:

equation: 2x + y = 7 -> y = -2x + 7

-x + y = 3 -> y = x + 3 Our slope here is the coefficient of x,

Our coefficient for x here is 1: which is (-2):

y = 1x + 3 y = -2x + 7

So the slope is +1 So the slope is -2

Slope of zero undefined Slope

This happens when we encounter a when we encounter a vertical line, the

horizontal line. slope is said to be "undefined"

As an example, let's look at the Let's take a look at:

equation: x = -2

y = 5 -> y = 0x + 5 m our slope is undefined. We can show this

Our coefficient for x here is 0: using the slope formula. We can pick any

two points such as (-2,4) and (-2,7).

y = 0x + 5

So the slope is 0

THE EQUATION OF A

STRAIGHT LIN

E THROUGH

TWO GIVEN POINTS

Slope Formula:

1 23

Pick or generate Label one point as Plug into the slope

any two points (x1,y1) and the formula

other as (x2, y2)

on the line

EXAMPLE

Find the slope of the line that passes through the given points

(0,4) and (-2,-2).

1 The first point as point 1, meaning it is: A(x1,y1)= A(0,4) and the

second point as point 2, meaning it is: B(x2, y2) = B(-2,-2).

2

3

-> The slope is 3

WREITQIUNAGT

ILOINNSEAR

MAIN IDEA EXAMPLE

The equation of the Calculate the equation of the straight line

line through the two passing through the points A(1,2) and

points can be written B(-1,3) in the form ax + by + c = 0

in the form: DISCUSS

What are you trying to find ?

Where m is the slope Identify the x coordinates and y coordinates of

of the line having the the given points.

value EXPLORE

x1 =

x2 =

y1 =

y2 =

SOLVE

Calculate the value of the gradient of the line.

Now substitute the values in the formula.

CONFIRM

Use the x

coordinates and the

y coordinates to

graph the equation.

Name Class

Exercise

1

The gradient of the line perpendicular to the line a is

Answer:

2

Find the slope of the line that passes through the given points

(1,2) and (3,4).

Answer:

3

The equation of the line that passes through the point (2, -5) and is

parallel to the line4y - 3x = -4 is...

Answer: