chapter 2: the straight line and applications. · chapter 2: the straight line and applications. ©...

CHAPTER 2: THE STRAIGHT LINE AND APPLICATIONS.

© John Wiley and Sons 2013 www.wiley.com/college/Bradley © John Wiley and Sons 2013

Essential Mathematics for Economics and Business, 4th Edition

http://www.wiley.com/college/Bradley

www.wiley.com/college/Bradley © John Wiley and Sons 2013

Calculate and plot a demand schedule (slide show)Worked Example 2.6, Figure 2.19: Slides 2 – 11 Plot a demand function from the points of intersection with the axis Worked Example 2.6, Figure 2.19: Slides 12, 13, 14, 15 Calculate and plot a supply schedule Worked Example 2.7, Figure 2.22: Slide 16, 17 Fixed Cost, Variable Cost and Total Cost: Calculate and plot a TC function (slide show), Worked Example 2.9, Figure 2.25: Slides 18, 19, 20, 21 Calculate and Plot a Total Revenue function (TR) Worked Example 2.10, Figure 2.26: Slide 22 Linear Profit function. Worked Example 2.10b. Figure 2.26: Slides 23, 24, 25, 26


Demand Function P =100 - 0.5Q

Label the horizontal and vertical axis Calculate the demand schedule from the equation given above, for positive values of P and Q Plot the points Draw the graph Label the graph

Table 2.3 Demand schedule Q P = 100 - 0.5Q 0 P = 100 -0.5(0) = 100 40 P = 100 -0.5(40) = 80 80 P = 100 -0.5(80) = 60 120 P = 100 -0.5(120) = 40 160 P = 100 -0.5(160) = 20 200 P = 100 -0.5(200) = 0 Intercept (vertical) = 100: Slope = - 0.5: Horizontal intercept = 200

Demand Function P =100 - 0.5Q :

Calculate the demand schedule (Method A)

P on

the

verti

cal (

y) a

xis

Q on the horizontal (x) axis

Table 2.3 Demand schedule Quantity (x) Price (y) 0 100 40 80 80 60 120 40 160 20 200 0 Intercept (vertical) = 100 Slope = - 0.5: Horizontal intercept = 200


Plot the demand schedule

P =100 - 0.5Q

Quantity Q

Price P

0

20

40

60

80

100

120

0

40

80

120

160

200




P =100 - 0.5Q

Quantity Q

Price P

0

20

40

60

80

100

120

0 40

80

120

160

200




120

160

200

P =100 - 0.5Q

Quantity Q

Price P

0

20

40

60

80

100

120

0 40

80




P =100 - 0.5Q

Quantity Q

Price P

0

20

40

60

80

100

120

0 40

80

120

160

200




P =100 - 0.5Q

Quantity Q

Price P

0

20

40

60

80

100

120

0 40

80

120

160

200

240




Quantity

P =100 - 0.5Q

Q

Price P

0

20

40

60

80

100

120

0 40

80

120

160

200

240



Join the points and label the graph

P =100 - 0.5Q

Quantity Q

Price P

: Demand

P = 100 - 0.5 Q

D

0

20

40

60

80

100

120

0 40

80

120

160

200

240

Figure 2.17



Alternatively, plot the vertical intercept(Method B)

Q Quantity

Price

P

0

20

40

60

80

100

120

40

80

120

160

200

240

Vertical intercept = 100



and plot the horizontal intercept

Quantity Q 0

20

40

60

80

100

120

0

40

80

120

160

200

Horizontal intercept = 200

Vertical intercept = 100



Join the horizontal and vertical intercepts

Figure 2.19

Join the horizontal and vertical intercepts

Quantity Q

Price P

D : Demand

0

20

40

60

80

100

120

0 40

80

120

160

200

240

The General Linear Demand function such as

0

20

40

60

80

100

120

0

40

80

120

160

200

240

Figure 2.19 Demand function, P = 100 - 0.5 Q

a = 100

P = 100 - 0.5

Q

P

Slope = - 0.5

a b

= 200

P = a - bQ P =100 - 0.5Q

Supply Function P = 10 + 0.5Q Calculate and plot the supply schedule P = 10 + 0.5Q

Table 2.4 Supply schedule Q: Quantity P: (P = 10 +0.5Q) 0 10: (P =10 +0.5(0)) 20 20: (P =

10+0.5(20)) 40 30 60 40 80 50 100 60 Intercept (vertical) = 10 Slope = 0.5 Horizontal intercept = - 20

P = 10 + 0.5Q

0

10

20

30

40

50

60

70

-20 0 20 40 60 80 100Figure 2.22

c = 10

P = 10 + 0.5 Q

Slope = 0.5

Q

P

S

Supply Function P = 10 + 0.5Q Calculate and plot the supply schedule P = 10 + 0.5Q

Table 2.4 Supply schedule Q: Quantity 0 20 40 60 80

100 P: Price per unit 10 20 30 40 50

60 Intercept (vertical) = 10 Slope = 0.5 Horizontal intercept = - 20 The supply function may be plotted (i) from the table of points (ii) Or by simply joining the intercepts

P = 10 + 0.5Q

0

10

20

30

40

50

60

70

-20 0 20 40 60 80 100

Figure 2.22

c = 10

P = 10 + 0.5Q

Slope = 0.5

Q

P

Total Cost Function:Derive the equation :

Example: Given the following: Fixed Costs (FC) = £10 Variable Costs, VC = £2 for one unit Variable costs, VC = £2Q for Q units Hence, Total Cost for Q units is

TC = FC + VC TC = 10 + 2Q. Plot Costs on the vertical axis Q on the horizontal axis

TC = 10 + 2Q

C o

n th

e ve

rtica

l axi

s

Q on the horizontal axis

Total Cost Function: TC = 10 + 2Q

Plot Fixed Costs

This is a horizontal line, which cuts the vertical axis at

Costs = 10

Plot the horizontal line: FC = 10

Cost

C

Quantity Q 0

2

4

6

8

10

12

14

16

18

20

22

24

0 1 2 3 4 5 6

FC = 10


Table 2.5: TC = FC +VC TC = 10 + 2Q • Calculate and plot the total cost schedule Points Q VC = 2Q TC = 10 +2Q (Q, TC) 0 0 TC = 10 + 0 = 10 (0, 10) 1 2(1) = 2 TC = 10 + 2 = 12 (1, 12) 2 2(2) = 4 TC = 10 + 4 = 14 (2, 14) 3 2(3) = 6 TC = 10 + 6 = 16 (3, 16) 4 2(4) = 8 TC = 10 + 8 = 18 (4, 18) 5 2(5) = 10 TC = 10 + 10 = 20 (5, 20) 6 2(6) = 12 TC = 10 + 12 = 22 (6, 22)

Plot the points

Cost

C

Quantity Q 0

2

4

6

8

10

12

14

16

18

20

22

24

0 1 2 3 4 5 6

FC = 10


Join the points: Label the graph

0

2

4

6

8

10

12

14

16

18

20

22

24

0 1 2 3 4 5 6

Figure 2.24

Table 2.5: TC = FC +VC TC = 10 + 2Q Calculate and plot the total cost schedule Q VC = 2Q TC = FC +VC (Q, TC) 0 0 TC = 10 + 0 = 10 (0, 10) 1 2(1) = 2 TC = 10 + 2 = 12 (1, 12) 2 2(2) = 4 TC = 10 + 4 = 14 (2, 14) 3 2(3) = 6 TC = 10 + 6 = 16 (3, 16) 4 2(4) = 8 TC = 10 + 8 = 18 (4, 18) 5 2(5) = 10 TC = 10 + 10 = 20 (5, 20) 6 2(6) = 12 TC = 10 + 12 = 22 (6, 22)

FC = 10

TC = 10 + 2 Q

Total Cost TC

Quantity Q

Total Revenue: TR = 3.5Q: Calculate and plot the TR

Table 2.6 Total revenue Price is fixed at P = 3.5 Q TR = PQ =3.5 Q

(Q, TR) 0 TR = 3.5 (0) = 0 (0, 0) 2 TR = 3.5 (2) = 7 (2, 7) 4 TR = 3.5 (4) = 14 (4, 14) 6 TR = 3.5 (6) = 21 (6, 21)

TR = 3.5Q

0

7

14

21

28

0 2 4 6

TR = 3.5 Q

Total revenue,

TR

Quantity, Q

Figure 2.25: Linear total revenue function

Profit from chicken snack boxes Price = £3.5 per box : FC = £800, Cost = £1.5 per box.

Total Revenue from the sale of Q boxes TR = Price per box × number of boxes = Price × Quantity = 3.5Q Total cost of producing Q boxes TC = Fixed Cost + Cost per box × number of boxes TC = FC + VC TC = 800 + 1.5Q Profit = Total Revenue – Total Cost Π = TR – TC Π = (3.5Q) - (800 + 1.5Q) Π = 3.5Q- 800 -1.5Q = 2Q - 800

TR = 3.5Q

TC = 800+1.5Q

Profit = 2Q-800

Profit from chicken snack boxes TR = 3.5Q: FC = 800, VC = 1.5 per unit.

TR = PQ =3.5 Q

TC = 800 + 1.5Q Profit = 2Q - 800

Figure 2.26: Linear profit function and TR, TC

-1000

-500

0

500

1000

1500

2000

2500

3000

0 100 200 300 400 500 600 700 800

TR

TC

Profit

Break-even quantity of chicken snack boxes TR = 3.5Q: FC = 800, VC = 1.5 per unit.

TR = 3.5 Q TC = 800 + 1.5Q Break-even when TR = TC 3.5Q = 800 + 1.5Q 2Q = 800 Q = 400 At the break-even Q TR = TC = 1400


-1000

-500

0

500

1000

1500

2000

2500

3000

0 100 200 300 400 500 600 700 800

TR

TC

Profit

Profit from chicken snack boxes TR = 3.5Q: FC = 800, VC = 1.5 per unit.

TR = 3.5 Q TC = 800 + 1.5Q Profit = TR – TC = 2Q - 800 When profit = 0 2Q – 800 = 0 Q = 400 But break-even Q = 400 Hence TR = TC = 1400 when Profit = 0


-1000

-500

0

500

1000

1500

2000

2500

3000

0 100 200 300 400 500 600 700 800

TR

TC

Profit

CHAPTER 2 : BUDGET AND COST CONSTRAINTS.





Worked Example 2.22, Figure 2.39:How to plot a budget

constraint.

Worked Example 2.23, Figure 2.40:The effect of a price change

in good X (on the horizontal).

Worked Example 2.23, Figure 2.41:

The effect of a price change in good Y (on the vertical).

Worked Example 2.23, Figure 2.42: The effect of a change in

the budget limit

Plot an Isocost constraint

Effect of change in the price of labour.



How to plot any Linear Budget Constraint

Rearrange the equation in the form y = mx + c (see above) Plot y on the vertical axis, against x on the horizontal axis Calculate and plot the vertical and horizontal intercepts Join the points and label the graph

xPP

PMyMyPxP

Y

X

YYX

−=→=+

M P X

M P Y Quantity of good Y ,

Quantity of good X , 0

10

20

30

40

0 30 60 90

y

x



Plot the Budget Constraint:

Given: PX =£2: PY = £6: M = 180: plot the constraint. xPX + yPY = M x (2) + y(6) = 180 (units of X) (price per unit) + (units of Y )(price per unit) = budget limit

This is the equation of the budget constraint.

To plot, rearrange the equation into the form y = mx + c:

Hence, 2x + 6y = 180 is rearranged as: y = 30 - 0.33x

In this form, it is easy to read off intercepts

Vertical intercept = 30 (value of y when x = 0):

Horizontal intercept = 90 (value of x when y = 0)

MPyPx YX =+



Plot the Budget Constraint: 2x + 6y = 180 where PX =£2: PY = £6: M = 180: x(2) + y(6) = 180

Plot the horizontal intercept: x = 90 Plot the vertical intercept: y = 30 Join these points

0

10

20

30

40

0 30 60 90

Quantity of good Y , y

Quantity of good X , x

y = 30 - 0.33 x

M P X

M P Y

Slope = − P P

X

Y

Figure 2.39



xP yP MX Y+ =

Another Example : Plot the Budget Constraint given PX =£3: PY = £6: M = 180:

Substitute the prices and budget limit into the general equation: x(3) + y(6) = 180. Rearrange the equation into the form y = mx + c: y = 30 - 0.5x Hence, vertical intercept = 30; horizontal intercept = 60.

Quantity of good Y, y

Quantity of good X, x

Budget constraint: M = 180

30

60



Adjust the equation of the Budget Constraint y= 30 -0.5x when the price of good X decreases. PX decreases from 3 to 1.5

The original budget constraint, PX =£3: PY = £6: M = 180,

Hence, the equation 3x + 6y = 180 (or y = 30 - 0.5x)

To find the equation of the new budget constraint write PX = 1.5 in the original budget constraint (nothing else

changes) Hence, the equation of the new budget constraint is: (1.5)x + 6y =

180. Rearrange this equation (for plotting later) to the form y = mx + c:

Hence y = 30 - 0.25x is the equation of the budget constraint when P has decreased from £3 to £1.5 The intercept (30) is the same: slope has changed from -0.5 to -0.25



Original Budget Constraint: y=30 - 0.5x Plot the new budget constraint when price of good X decreases from £3 to £1.5

When PX decreases from 3 to 1.5 Equation of new budget constraint becomes: y = 30 - 0.25x Plot the vertical intercept = 30 (this is the same as the original) Plot the horizontal intercept = 120 (this is different from the

original)

0

10

20

30

0 20 40 60 80 100 120 x

y



Plot the original budget constraint: y = 30 - 0.5x and the new budget constraint when PX decreases from £3 to £1.5.

New budget constraint is y = 30 – 0.25x For the new budget constraint: y = 30 – 0.25x: (i) vertical intercept = 30 (ii) horizontal intercept = 120. Join the vertical intercept = 30 and the horizontal intercept = 120

0

10

20

30

0 20 40 60 80 100 120 x

y

Original Constraint

Constraint with PX decreases from £3 to £1.5

Figure 2.40



Summary: Change in the equation and graph of the Budget Constraint: y = 30 - 0.5x when the price of good X decreases from £3 to £1.5

When PX decreases from 3 to 1.5 the new constraint is y = 30 – 0.25x The new budget constraint pivots out from the unchanged vertical intercept (see Figure 2.40) (note: as X decreases in price, more units of X are affordable, so x increases)

Figure 2.40 ∆ P X and its effect on the budget constraint

0

10

20

30

40

0 20 40 60 80 100 120

Original: y = 30 - 0.5x

New: y = 30 - 0.25x

x

y



Adjust the equation of the original Budget Constraint: y = 30 - 0.5x, when the price of good Y decreases from £6 to £3 The equation of original budget constraint, where PX =£3: PY = £6: and M = 180 is 3x + 6y = 180 (or y = 30 - 0.5x) Find the equation of the new constraint when PY changes from 6 to 3.

To obtain the equation of the new budget constraint Replace 6 (the value of PY ) by 3 in the original equation (nothing else

changes) The equation of new budget constraint is: 3x + (3) y = 180 Rearrange the equation into the form y = mx + c (for plotting): Hence, the equation of the new budget constraint is y = 60 – x.

Horizontal intercept is the same: slope has changed from -0.5 to -1



Adjust the graph of the Budget Constraint: y = 30 - 0.5x when the price of good Y decreases When PY changes from 6 to 3 The equation of the budget constraint becomes: y = 60 - x Read off the intercepts; Plot the vertical intercept = 60 Plot the horizontal intercept = 60 Join the intercepts

0

10

20

30

40

50

60

70

0 20 40 60

x



Compare the graphs of the original budget constraint and new budget constraint following a decrease in the price of good Y

The new constraint pivots upwards from the unchanged horizontal intercept (see Figure 2.41) Comment: When Y decreases in price, more units of Y are

affordable

Figure 2.41 ∆ P Y and its effect on the budget constraint

0

10

20

30

40

50

60

70

0 20 40 60

y = 30 - 0.5 x

y = 60 - x

x

y



Adjust the equation of the Budget Constraint: y = 30 - 0.5x when the budget limit increases

M changes from 180 to 240 The original budget constraint, where PX =£3: PY = £6: M =

180, was given by the equation was: 3x + 6y = 180 ( or y = 30 - 0.5x) To obtain the equation of the new budget constraint replace M by 240 (nothing else changes) The equation of the new budget constraint becomes: 3x + 6y =

240: Rearrange this equation into the form y = mx + c (for plotting

later): The equation of the budget constraint is y = 40 - 0.5x.

Slope is the same: vertical intercept has changed from 30

to 40



Change in the graph of the Budget Constraint: y = 30 - 0.5x when the budget limit increases

M changes from 180 to 240 Equation of budget constraint becomes: y = 40 - 0.5x Plot the vertical intercept = 40 Plot the horizontal intercept = 80

0

10

20

30

40

50

0 20 40 60 80 x

y



Change in the graph of the Budget Constraint: y = 30 - 0.5x when the budget limit increases

The adjusted budget constraint is: y = 40 - 0.5x Join the vertical intercept = 40 and the horizontal intercept = 80

0

10

20

30

40

50

0 20 40 60 80

x

y

New budget limit = 240

Original budget limit = 180



Summary: Change in the graph of the Budget Constraint: y = 30 - 0.5x when the budget limit increases

When the budget limit increases, the constraint moves upwards, parallel to the original constraint Comment: When the budget limit increases, more units of both X

and Y are affordable

Figure 2.42 ∆ Y and its effect on the Budget constraint

0

10

20

30

40

50

0 20 40 60 80

x

y

Budget = 240

Budget = 180



Plot the Isocost line:4K + 5L = 8000 as K = f(L). Method: Rearrange the equation to K = f(L): Hence K = 2000 - 1.25L Plot K on the vertical axis, L on the horizontal axis. Calculate and plot the horizontal and vertical intercepts. A and B are simply two extra points. It is a safeguard (against arithmetic errors) to plot at least one extra point

when plotting lines Join the points

0

500

1000

1500

2000

2500

0 400 800 1200 1600

K = 2000 - 1.25L

L

K

A

B

C r

C w



Isocost line:4K + 5L = 8000, hence K = 2000 - 1.25L Labour increases from £5 to £8 per hour: The equation of the new isocost line is K = 2000 -2L Plot the horizontal and vertical intercepts : draw the isocost line The horizontal intercept moves towards the origin, along the

horizontal axis Comment: when the price of labour increases, fewer units are

affordable Figure 2.48

0

500

1000

1500

2000

2500

0

200

400

600

800

1000

1200

1400

1600

Effect of change in price of labour on the isocost line

K = 2000 - 1.25L

L

K

K = 2000 – 2L


CHAPTER 2 : PROPERTIES OF STRAIGHT LINES.





• Measure Slope and Intercept Figures 2.4. • Different lines with (i) same slopes, (ii) same intercept.. • How to draw a line, given slope and intercept. Worked Example

2.1 Figure 2.6 • What is the equation of a line ? • Write down the equation of a line, given its slope and intercept,

Worked Example 2.2, • Given the equation of a line, write down the slope and intercept • Plot a line by joining the horizontal and vertical intercepts. • Equations of horizontal and vertical lines, Figure 2.10



Measuring Intercept The point at which a line crosses the vertical axis is referred as the ‘Intercept’

intercept = 2

intercept = 0

intercept = - 3 -4

-3

-2

-1

1

2

3

-4 -3 -2 -1 0 1 2 3 4



-6

-5

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

Measuring Slope

Slope

=change in height

change in distance xy

∆∆

=

2.156

−=−

Line CD

slope =

Figure 2.4

D

C A

B Line AB

slope = 4 2 5.0

42=

-6

5

y increases by 0.5 when x increases by 1.

y decreases by 1.2 when x increases by 1.



Slope alone does not define a line Intercept alone does not define a line

Lines with same intercept but different slopes are different lines

Lines with same slope but different intercepts are different lines

Figure 2.2

x -4 -3 -2 -1 0y -1 0 1 2y -2 -1 0 1y -3 -2 -1 0y -5 -4 -3

-4

-3

2

-5-4-3-2-1012345

-4 -3 -2 -1 0 1 2 3 4

-3-2-101234567

-1 0 1 2 3 4 5 6Inte

rcep

t =

2



0

1

2

3

4

5

-1 0 1 2 3 4

A line is uniquely defined by both slope and intercept

In mathematics, the slope of a line is referred to by the letter m

In mathematics, the vertical intercept is referred to by the letter c

Intercept, c = 2

slope, m = 1



Draw the line, given slope, m =1: intercept, c = 2 Worked Example 2.1(b)

1. Plot a point at intercept = 2

2. From the intercept draw a line with slope = 1 by

(a) moving horizontally forward by one unit and

(b) vertically upwards by one unit

3. Extend this line indefinitely in either direction, as required

The graph of the line whose intercept = 2, slope = 1

0

1

2

3

4

5

6

-1 0 1 2 3 4

(0, 2)

(1, 3)

x Figure 2.6



What does the equation of a line mean?

Consider the equation y = x From the equation, calculate the values of y for x = 0, 1, 2, 3, 4, 5, 6. The calculated points are given in the following table. x 0 1 2 3 4 5 6 y 0 1 2 3 4 5 6 Plot the points as follows



x 0 1 2 3 4 5 6 y 0 1 2 3 4 5 6

0

1

2

3

4

5

6

0 1 2 3 4 5 6 x

y

(0, 0)

Plot the point x = 0, y = 0



x 0 1 2 3 4 5 6 y 0 1 2 3 4 5 6

0

1

2

3

4

5

6

0 1 2 3 4 5 6

(0, 0)

(1, 1)


y

x



x 0 1 2 3 4 5 6 y 0 1 2 3 4 5 6

0

1

2

3

4

5

6

0 1 2 3 4 5 6

(2, 2)

(1, 1)

(0, 0)


y

x



x 0 1 2 3 4 5 6 y 0 1 2 3 4 5 6

0

1

2

3

4

5

6

0 1 2 3 4 5 6

(2, 2)

(1, 1)

(0, 0)

(3, 3)

Plot the point x = 3, y = 3 y

x



x 0 1 2 3 4 5 6 y 0 1 2 3 4 5 6

0

1

2

3

4

5

6

0 1 2 3 4 5 6

y

(2, 2)

(1, 1)

(0, 0)

(3, 3)

(4, 4)


x



x 0 1 2 3 4 5 6 y 0 1 2 3 4 5 6

0

1

2

3

4

5

6

0 1 2 3 4 5 6

(2, 2)

(1, 1)

(0, 0)

(3, 3)

(4, 4)

(5, 5)


x



x 0 1 2 3 4 5 6 y 0 1 2 3 4 5 6

0

1

2

3

4

5

6

0 1 2 3 4 5 6

(2, 2)

(1, 1)

(0, 0)

(3, 3)

(4, 4)

(5, 5)

(6, 6)


x



x 0 1 2 3 4 5 6 y 0 1 2 3 4 5 6

0

1

2

3

4

5

6

0 1 2 3 4 5 6

y

(2, 2)

(1, 1)

(0, 0)

(3, 3)

(4, 4)

(5, 5)

(6, 6)

Join the plotted points

x



x 0 1 2 3 4 5 6 y 0 1 2 3 4 5 6

0

1

2

3

4

5

6

0 1 2 3 4 5 6

(2, 2)

(1, 1)

(0, 0)

(3, 3)

(4, 4)

(5, 5)

(6, 6)

The y co-ordinate = x co-ordinate, for every point on the line:

Figure 2.9 The 45o line, through the origin

y

x



x 0 1 2 3 4 5 6 y 0 1 2 3 4 5 6

0

1

2

3

4

5

6

0 1 2 3 4 5 6

(2, 2)

(1, 1)

(0, 0)

(3, 3)

(4, 4)

(5, 5)

(6, 6)

y = x is the equation of the line.

Similar to Figure 2.9

y

x



Deduce the equation of the line, given slope, m = 1; intercept, c = 2

1. Start by plotting at least 2 points. Then observe the relationship between

the x and y coordinate 2. Plot the intercept: x = 0, y = 2 (c

=2) 3. Since slope = 1, move forward

1unit then up 1 unit. See Figure 2.6 This gives the point (x = 1, y = 3) 4. Map out further points, (2, 4) etc. in

this way 5. Observe that value of the y co-ordinate

is always “value of the x co-ordinate +2” Hence the equation of the line is y = x+ 2 6. That is, y = (1)x + (2), with m =1, c

= 2 In general, y = mx + c is the equation of a line

0

1

2

3

4

5

6

-1 0 1 2 3 4

(0, 2)

(1, 3)

y

(2, 4)

Figure 2.6

x



Deduce the equation of the line, given slope, m = 1; intercept, c = 2 Use Formula y = mx + c Since m = 1, c = 2 then y = mx + c y =(1)x + 2 y = x + 2 See Figure 2.6

0

1

2

3

4

5

6

-1 0 1 2 3 4

(0, 2)

(1, 3 )

y

(2, 4)

Figure 2.6

x



The equation of a line

Putting it another way: the equation of a line may

be described as “ the formula that allows you

to calculate the y co-ordinate

when given the value of the x co-ordinate

for any point on the line”

Example 1. y = x “The y co-ordinate is equal to the x co-ordinate”. Slope, m = 1, intercept , c = 0. Example 2. y = x + 2 “The y co-ordinate is equal to the x co-ordinate plus 2”. Slope, m = 1 , intercept, c = 2

The equation of a line may be written in terms of the two characteristics:

m (slope) and c (intercept) . y = mx + c



Calculating the Horizontal Intercepts

Calculate the horizontal intercept for the line: y = mx + c The horizontal intercept is the

point where the line crosses the x -axis Use the fact that the y co-ordinate

is zero at every point on the x-axis. Substitute y = 0 into the equation

of the line 0 = mx + c and solve for x: 0 = mx + c: therefore, x = -c/m This is the formula for the

horizontal intercept

Line: y = mx + c (m > 0: c > 0)

y = mx + c Intercept = c

Slope = m

Horizontal intercept = - c/m

0, 0



Determine the slope and intercepts for a line when the equation is given in the form: ax + by +d = 0

Rearrange the equation into the form y = mx + c. Read off slope, m = Read off intercept, c =

Horizontal intercept =

Example:4x + 2y - 8 = 0

Slope, m = -2: intercept, c = 4 Horizontal intercept =

0=++ dbyax

−ab−

db

ad

−

0824 =−+ yx

224

2)4(

==−

−

axdby −−=

xba

bdy −−=

xy 482 −=

xy 24−=



Plot the line 4x+2y - 8 = 0 by calculating the horizontal and vertical intercepts: 4x+2y - 8 = 0 Rearrange the equation into the

form y = mx + c y = -2x + 4

Vertical intercept at y = 4

Horizontal intercept at x = 2

Plot these points: see Figure 2.12 Draw the line thro’ the points

Figure 2.12

-3

-2

-1

0

1

2

3

4

5

6

7

-1 0 1 2 3 4

y = -2x +4 y = 4

x = 2



Equations of Horizontal and vertical lines:

The equation of a horizontal is given by the point of intersection with the y-axis The equation of a vertical line is given by the point of intersection

with the x -axis

-4

-3

-2

-1

0

1

2

3

4

5

6

-4 -3 -2 -1 0 1 2 3 4

y = 4

x = 2 x = - 1.5

y = - 2

x

y

Figure 2.10


chapter 2: the straight line and applications. · chapter 2: the straight line and applications. ©...

Documents