1. Mathematics for Business and Economics - I
Chapter 3 - Lines, Parabolas, and Systems

2. Section Lines
Many relationships between quantities can be represent conveniently by straight lines.
The slope of a nonvertical line that passes through the points P(x1, y1) and Q(x2, y2) is denoted by m and is defined by y
Q(x2, y2) y2
Vertical Change
Vertical Change
P(x1, y1) y1 x1 x2 x
Horizontal Change
Horizontal Change

3. Vertical and Horizontal Lines
Either the “rise” or “run” could be zero y
m is undefined
m (slope) is positive
m = 0
m (slope) is negative x

4. EXAMPLE 1
Finding and Interpreting the Slope of a Line
Sketch the graph of the line that passes through the points P(1, –1) and Q(3, 3). Find and interpret the slope of the line.
Solution
Any two points determine a line; the graph of the line passing through the points P(1, –1) and Q(3, 3) is sketched here.
Vertical Change =4
Horizontal Change = 2

5. P(1, –1) and Q(3, 3) Solution continued Interpretation
EXAMPLE 1
Finding and Interpreting the Slope of a Line
Solution continued
P(1, –1) and Q(3, 3)
Interpretation
The slope of this line is 2; this means that the value of y increases by exactly 2 units for every increase of 1 unit in the value of x. The graph is a straight line rising by 2 units for every one unit we go to the right.

6. EXAMPLE 2
Finding and Interpreting the Slope of a Line

7. Equations of Lines
Point-Slope Form: To find the equation of a line, when you only have two points. The point-slope form of the equation of a line is
where m is the slope and (x1, y1) is a given point.

8. EXAMPLE 1
Finding an Equation of a Line with Given Point and Slope
Find the point–slope form of the equation of the line passing through the point (1, –2) and with slope m = 3. Then solve for y.
Solution
We have x1 = 1, y1 = –2, and m = 3.

9. EXAMPLE 2
Finding an Equation of a Line with Given Point and Slope
Find the equation of a line that passes through the point (1,-3) with slope of 2

10. EXAMPLE 3
Finding an Equation of a Line Passing Through Two Given Points
Find the point–slope form of the equation of the line l passing through the points (–2, 1) and (3, 7). Then solve for y.
Solution
First, find the slope.
We have x1 = 3, y1 = 7.

11. Find the equation of the line through the points (-5,7) and (4,16).
EXAMPLE 4
Finding an Equation of a Line with two Given Points
Find the equation of the line through the points (-5,7) and (4,16).
Solution:
Now use the point-slope form with m = 1 and (x1, x2) = (4,16). (We could just as well have used (-5,7)).

12. The line passes through (0, b).
EXAMPLE 5
Finding an Equation of a Line with a Given Slope and y-intercept
Find the point–slope form of the equation of the line with slope m and y-intercept b. Then solve for y.
Solution
The line passes through (0, b).

13. SLOPE–INTERCEPT FORM OF THE EQUATION OF A LINE
The slope-intercept form of the equation of the line with slope m and y-intercept b is
EXAMPLE 1
Slope – Intercept Form

14. EXAMPLE 2
Find the Slope and y-intercept of a line

15. EXAMPLE
Equations of Horizontal and Vertical Lines

16. PARALLEL AND PERPENDICULAR LINES
Let l1 and l2 be two distinct lines with slopes m1 and m2,respectively. Then
l1 is parallel to l2 if and only if m1 = m2.
l1 is perpendicular l2 to if and only if m1•m2 = –1.
Any two vertical lines are parallel, and any horizontal line is perpendicular to any vertical line.

17. Section 3-2 Applications and Linear Functions

18. EXAMPLE 1
Finding a Demand Equation

19. Note : The graph of a linear function can be drawn by using the same procedures as drawing a line.

20. EXAMPLE 2
Diet for Hens

22. EXAMPLE 3
Example
A business copier repair company charges a fixed amount plus an hourly rate for service. If a customer is billed $159 for a one-hour service call and $287 for a three hour service call, find the linear function that describes the price of a service call when x is the number of hours of service
Solution
We need to find the equation for a line of the form; y=mx+b
We know two points on the line: (1,159) and (3,287)
So, m = (287 – 159)/(3 – 1) = 64
Use point-slope approach to find the line
(y – y1) = 64(x – x1)
y = 287 – 64(3) + 65x
y = x

23. The Points are: = (40,12) , and =(25,18)
EXAMPLE 4
Finding a Demand Equation
Example : Suppose consumer will demand 40 units of a Product
when the price is $12 per unit and 25 units when The price is
$18 each. Find the demand equation Assuming that is linear.
Find the price per unit when 30 units are demanded
Solution: 
The Points are: = (40,12) , and =(25,18)
Slope Formula: x represent q and y represent p
Hence an equation of the line is

24. When q=30  Graph of the demand equation p intercept q=0 then p=28,
Q intercept p=0 then q=70 P 28 70 Q

25. EXAMPLE 5
Finding a Demand Equation

27. Example Suppose a manufacturer of shoes will place on the
Finding a Supply Equation
Example Suppose a manufacturer of shoes will place on the
market 50 (thousand pairs) when the price is 35 (dollars per
pair) and 35 when the price is 30. Find the supply equation,
assuming that price p and quantity q are linearly related
Solution:The Points are: =(50,35), and =(35,30)
Slope Formula: x represent q and y represent p
Hence the equation of the line is

28. EXAMPLE 7
Finding a Supply Equation

30. Example: Suppose the cost to produce 10 units of a product is
Finding a Cost Equation
Example: Suppose the cost to produce 10 units of a product is
$40 and the cost of 20 units is $ 70. If cost c is linearly related to
output q, find a linear equation relating c and q. Find the cost to
produce 35 units.
Solution:The line passing through (10,40) and (20,70) has slope
So an equation for the line is:
If q=35 then c=3(35)+10=115

31. The y intercept is already known
EXAMPLE 10
Finding an Equation
Office equipment was purchased for $20,000 and will have a scrap value of $2,000 after 10 years. If its value is depreciated linearly, find the linear equation that relates value (V) in dollars to time (t) in years:
Solution: When t = 0, V = 20,000 and when t = 10, V = 2,000. Thus, we have two ordered pairs (0, 20,000) and (10, 2000). We find the slope of the line using the slope formula.
The y intercept is already known
The slope is ( ,000)/(10 – 0) = -1,800.
when t = 0, V = 20,000, so the y intercept is b=20,000
Therefore, by slope-intercept form(y=mx+b)
equation is V(t) = - 1,800t + 20,000.