Term 3 : Unit 3Linear Law

Name : ___________( ) Class: _______ Date : _______

3.1 Linear Law

3.2 Applications of Linear Law

Linear Law

In this lesson, you will learn how to

• convert a non-linear relation to linear form,

• use new variables X and Y to draw the graph of Y = mX + c,

• estimate the values of the gradient m and the Y-intercept, c,

• use the values of m and c to estimate unknown constants in the original

equation,

• and use the linear graph of Y = mX + c to obtain the estimated values of x and y.

3.1 Linear Law

Objectives

y(Y)

1x(X)

Calculate values of y for some values of x.

y(Y)

1x(X)

(0, 3)

(0.5, 4)

(1, 5)

(2, 7)

y(Y)

1x(X)

The gradient, 2.m

The intercept, 3.y c

Let and 1

.Y y Xx

x y

0 –

0.5 7

1 5

2 4

x X y Y

0 – – –

0.5 2 7 7

1 1 5 5

2 0.5 4 4

y(Y)

1x(X)

m = 2

y(Y)

1x(X)

m = 2

c = 3

Two variables x and y are related by the equation 2 1

3, 0. Draw the graph of against .y x yx x

Graph of Y against X.

The points lie on a straight line.

Linear Law

Example 1

xy(Y)

x (X)

Calculate values of xy(Y ) for some values of

x.

The gradient, 2.m

The intercept, 1.y c

Let and Y xy X x x Y

0 –1

1 1

4 3

9 5

x X y Y

0 0 – –1

1 1 1 1

4 2 0.75 3

9 3 0.56 5

Two variables x and y are related by the equation

2 1. Draw the graph of against .xy x xy x

Graph of Y against X.

The points lie on a straight line.

xy(Y)

x (X)

(0, – 1)

(2, 3)

(3, 5)

(1, 1)

xy(Y)

x (X)

xy(Y)

x (X)

m = 2

Linear Law

Example 2

yx (Y)

x (X)A(1, – 1)

B(4, 14)

The equation is 5 .Y X c 1 5 1 6c c

Let .Y mX c

Two variables x and y are related such that when

is plotted against , a straight line is obtained.

The line passes through the points A(1, –1) and B(4, 14). Express y in terms of x.

y

xx

14 1 155

4 1 3m

The equation is 5 6Y X

5 6y

xx

5 6y x x x

The line passes

through (1, – 1).

Linear Law

Example 3

lg y

lg x

lg y

lg x

( – 1, – 0.80)

( – 0.30, 0.25)

(0, 0.70)

(0.48, 1.41)

(a) Show that 3

lg lg lg5.2

y x

Two variables x and y are related by the equation

3

25 .y x3

25y x

The points lie on a straight

line.

3

2lg lg 5y x

3lg5 lg

2x

(b) Draw the graph of lg y against lg x.

x lg x lg y y

0.1 – 1 –0.80 0.16

0.5 – 0.30 0.25 1.77

1 0 0.70 5

3 0.48 1.41 26.0

lg y

lg x

( – 1, – 0.80)

( – 0.30, 0.25)

(0, 0.70)

(0.48, 1.41)

Linear Law

Example 4

Y (xy)

X 1x

A( – 1, – 1)

B(5, 2)

O

12The equation is .Y X c 1

21 1 c

Let .Y mX c

Two variables x and y are related in such a way that

when xy is plotted against , a straight line is

obtained. The line passes through the points A(–1, –1) and B(5, 2).

1

x

2 1 3 1

5 1 6 2m

1 12 2Y X 1 1

2 2xy

x

2

1 1

2 2y

x x

(a) Find an expression for y in terms of x.

12c

(b) Find the value of y when x = 2.

2

1 1

2 22 2y

1 1

8 4

1

8

Linear Law

Example 5

Linear Law

In this lesson, you will apply linear law to analyse experimental data.

3.2 Applications of Linear Law

Objectives

y(Y)

1x(X)0.5 1 1.5 2

-1

-0.5

0.5

1

1.5

2

2.5

3

2.38 1.20The gradient

1.58 1.00a

The intercept 0.80y b

Let and 1

.Y Xy

x x y

0.50 1.61

1.00 0.83

1.50 0.61

2.00 0.50

2.50 0.42

3.00 0.38

The table shows experimental values of two quantities x and y which are

known to be connected by an equation of the form 1

.a x by

Plot against and use the graph to estimate the values of a and b. 1

yx

x X y Y

0.50 0.71 1.61 0.62

1.00 1.00 0.83 1.20

1.50 1.22 0.61 1.64

2.00 1.41 0.50 2.00

2.50 1.58 0.42 2.38

3.00 1.73 0.38 2.63

y(Y)

1x(X)0.5 1 1.5 2

-1

-0.5

0.5

1

1.5

2

2.5

3

y(Y)

1x(X)0.5 1 1.5 2

-1

-0.5

0.5

1

1.5

2

2.5

3

y(Y)

1x(X)0.5 1 1.5 2

-1

-0.5

0.5

1

1.5

2

2.5

3

(1.58, 2.38)

(1.00, 1.20)

2.00a

Linear Law

Example 6

1.79 1.11gradient

0.90 0.48n

intercept lg 0.33y a

Let lg and lgY y X x

x y

3 13

4 20

5 29

6 39

7 49

8 61

The table shows experimental values of the variables x and y.

ny ax

It is known that x and y are related by the equation y = axn (a, n are constants)

(a) Express the equation in a form to draw a straight line graph.

lg lg ny axlg lgn x a

lg y(Y)

lg x(X)0.2 0.4 0.6 0.8 1

0.5

1

1.5

2

x X y Y

3 0.48 13 1.11

4 0.60 20 1.31

5 0.70 29 1.46

6 0.78 39 1.59

7 0.85 49 1.69

8 0.90 61 1.79

lg y(Y)

lg x(X)0.2 0.4 0.6 0.8 1

0.5

1

1.5

2

lg y(Y)

lg x(X)0.2 0.4 0.6 0.8 1

0.5

1

1.5

2

lg y(Y)

lg x(X)0.2 0.4 0.6 0.8 1

0.5

1

1.5

2(0.90, 1.79)

(0.48, 1.11)

1.62n

2.15a

(b) Draw the graph to estimate n and a.

1.622.15y x

Linear Law

Example 7

Linear Law

(c) Calculate the value of x when y = 66.

Using lg y = 1.62 lg x + 0.33,

lg 66 = 1.62 lg x + 0.33

1.82 = 1.62 lg x + 0.33

lg x = 0.92

x = 0.82

5.6 3.6gradient

3 2a

intercept 0.4y b

Let and Y Xy

xx

x y

1 1.6

2 7.2

3 16.8

4 30.4

The table shows experimental values of the variables x and y.

2y ax bx

It is known that x and y are related by the equation y = ax2 + bx (a, b are constants)

(a) Express the equation in a form to draw a straight line graph.

yax b

x

2a (b) Draw the graph to estimate a and b.

22 0.4y x x

x X y Y

1 1 1.6 1.6

2 2 7.2 3.6

3 3 16.8 5.6

4 4 30.4 7.6

yx (Y)

x(X)1 2 3 4

2

4

6

8

yx (Y)

x(X)1 2 3 4

2

4

6

8

yx (Y)

x(X)1 2 3 4

2

4

6

8

yx (Y)

x(X)1 2 3 4

2

4

6

8

(3, 5.6)

(2, 3.6)

Linear Law

Example 8