
SECTION 13.1: SLOPE OR THE AVERAGE RATE OF
CHANGE OF Y WITH RESPECT TO X

figure 1
Take a moment to examine figure 1 to the right. There are a many symbols used in figure 1 that you must get use to using. One of those important symbols is the delta symbol, Δ, which is used when one wants to represents a difference between two numbers. Let's get started.
Let's consider two ordered pairs (6, 6) and (2, 3) and their associated points placed on the Cartesian Coordinate Plane.
 x_{1} and y_{1 } are used to name the x and y coordinates respectively of point 1. Hence, x_{1} = 2 and y_{1} = 3.
 x_{2 } and y_{2 } are used to name the x and y coordinates respectively of point 2. Hence, x_{2} = 6 and y_{2} = 6
 Δx is the symbol used to name the difference between the two x values x_{2 } and x_{1 }: Δx = x_{2 }_{} x_{1 }.
Hence Δx = x_{2 }_{} x_{1} = 6  2 = 4 or simply Δx = 4.
 Δy is the symbol used to represent the difference between the two y values y_{2 } and y_{1}: Δy = y_{2 }_{} y_{1}.
Hence Δy = y_{2 }_{} y_{1} = 6  3 = 3 or simply Δy = 3.
The Slope Formula

The Slope Formula
(in general, used to represent the steepness and tilt of a line)
m

= 
y_{2}  y_{1 } 
x_{2 }  x_{1 } 
IMPORTANT INFORMATION ON THE "TILT" OF A LINE!

The
Formula For The Average Rate of Change of y
With Respect To x

The Average Rate of Change of y with respect to x Formula
(in general, used in real world problems such speed, weight loss, etc.)
Δy 
= 
y_{2}  y_{1 } 
Δx 
x_{2 }  x_{1 } 
Teacher Comment: The "average rate of change" and the "slope" are two different names for the same formula (y2  y1) / (x2  x1).
In calculus and precalculus the phrase "average rate of change" is used in real world problems.

ex. 1) Tom left Deering high school and walked along Stevens Avenue. At 10 seconds Tom had walked 100 feet from Deering and at 25 seconds he had walked 300 feet from Dering. Determine Tom's Average Rate of Change of Distance with respect to Time from 10 seconds to 25 seconds.
ans: We are asked to find Average Rate of Change of Distance with respect to Time. In the phrase "Average Rate of Change of Distance with respect to Time" the first concept mentioned, Distance, is placed along the vertical axis(Distance axis) and the second concept mentioned, time, is placed along the horizontal axis(time axis).
The Average Rate of Change of Distance with respect to Time is 13.333 feet/second. Now what does this mean to you? Any answers?
figure 2
ex. 2) A scientist decided to determine how bacteria were thriving in a petri dish in a very cold freezer. Aftert 2 hours there were 600 live bacteria in a petrie dish; after 6 hours there were only 100 live bacteria in the petri dish. Determine the average rate of change of the bacteria with respect to time from 2 to 6 hours and explain your answer.
ans:
Since the average rate of change over the 4 hour period of time is negative, 125 bacteria/hour, it appears that the bacteria are dying at an average rate of 125 bacteria every hour.
Notice how the top of the line tilts to the left, hence the negative slope!
ex. 3) Determine the slope of the following two line segments in the image below. Also, which line is steeper
figure 3
ans:
I'll use the slope formula:
m

= 
y_{2}  y_{1 } 
x_{2 }  x_{1 } 
to find the slopes of line 1 and line 2.
Let's find the slope of line 1.
Point 1 is assigned the ordered pair (5, 3). Hence x_{1 } = 5 and y_{1 } = 3.
Point 2 is assigned the ordered pair (3, 4). Hence x_{2 } = 3 and y_{2 } = 4.
m_{ line 1}

= 
y_{2}  y_{1 } 
= 
4_{}  3_{ } 
= 
7_{} 
x_{2 }  x_{1 } 
3_{ }  5_{ } 
2_{} 
Hence, m_{ line 1} = 3.5
Let's find the slope of line 2.
Point 1 is assigned the ordered pair (1, 1). Hence x_{1 } = 1 and y_{1 } = 1.
Point 2 is assigned the ordered pair (5, 3). Hence x_{2 } = 5 and y_{2 } = 3.
m_{line 2}

= 
y_{2}  y_{1 } 
= 
3_{}  1_{} 
= 
2_{} 
x_{2 }  x_{1 } 
5_{ }  1_{} 
4_{} 
Hence, m_{ line 2} = .5
Important Comment: Notice in figure 3 above that line 1 is "steeper" than line 2. If you take the absolute value of the slope of any two lines, the slope with the largest absolute value will have the steeper line.
m_{ line 1} = 3.5 >  3.5  = 3.5
m_{line 2} = .5 >  .5  = .5
 m_{ line 1} >  m_{line 2}  therefore line 1 will be steeper than line 2 which can be verified by figure 3 above.
ex. 4) Determine the slope of the following two line segments in figure 4 below. Which line is steeper?
figure 4
ans: I'll use the slope formula:
m

= 
y_{2}  y_{1 } 
x_{2 }  x_{1 } 
to find the slopes of line 1 and line 2.
Let's find the slope of line 1.
Point 1 is assigned the ordered pair (2, 2). Hence x_{1 } = 2 and y_{1 } = 2.
Point 2 is assigned the ordered pair (3, 4). Hence x_{2 } = 3 and y_{2 } = 4.
m_{ line 1}

= 
y_{2}  y_{1 } 
= 
4_{}  2_{ } 
= 
6_{} 
x_{2 }  x_{1 } 
3_{ }  2_{ } 
1_{} 
Hence, m_{ line 1} = 6
Let's find the slope of line 2.
Point 1 is assigned the ordered pair (1, 1). Hence x_{1 } = 1 and y_{1 } = 1.
Point 2 is assigned the ordered pair (5, 3). Hence x_{2 } = 5 and y_{2 } = 3.
m_{line 2}

= 
y_{2}  y_{1 } 
= 
3_{}  1_{} 
= 
2_{} 
x_{2 }  x_{1 } 
5_{ }  1_{} 
4_{} 
Hence, m_{ line 2} = .5
Important Comment: Notice in figure 3 above that line 1 is "steeper" than line 2. If you take the absolute value of the slope of any two lines, the slope with the largest absolute value will have the steeper line.
m_{ line 1} = 6 >  6  = 6
m_{line 2} = .5 >  .5  = .5
 m_{ line 1} >  m_{line 2}  therefore line 1 will be steeper than line 2 which can be verified by figure 4 above.
Important Comment 2: Since the slope of line 1 is negative I KNOW the top of the line will tilt to the left; since line 2 has a positive slope I know the top of the line will tilt to the right.
SECTION 13.2: RISE OVER THE RUN
DEFINITIONS OF THE RISE AND THE RUN 
The "rise" is another name for Δy = y_{2 } y_{1}.
The "run" is another name for Δx = x_{2 } x_{1}.

In the picture below, figure 1, is a dotted RIGHT triangle highlighted in red. I've extracted the highlighted triangle in figure 2. Please notice that the lengths of base and height of the right triangle below and the symbols used to describe these lengths!
figure 1


figure 2

EXAMINE FIGURE 2 ABOVE ONE MORE TIME! CAN YOU SEE THAT ΔX AND ΔY HAPPEN TO BE THE BASE AND THE HEIGHT OF THE TRIANGLE IN FIGURE 2.
Please note that the rise and the run are NOT NECESSARILY the lengths of the base and height of a right triangle as seen above. Let me clarify with an figure 3 below.
If you examine figure 3 below you see that Δy is 7 and Δx is 6. You can also see that the height of the right triangle is 7 and is the same value of Δy; but the base of the triangle is 6 though Δx has a value of 6. Hence, Δy and Δx do not necessarily represent the base and height of a right triangle, but there is some connection don't you agree? How would you explain that connection?
If you go over this section once again I believe you will see that we have "three different ways of naming the same idea."
THREE DIFFERENT WAYS OF NAMING THE SAME IDEA 
slope(m) 
= 
average rate of change 
= 

= 
y_{2 }  y_{1} 
x_{2 }  x_{1} 

!!! IMPORTANT !!!

FROM THIS POINT ON THE ACRONYM A.S.R. WILL BE USED
INTERCHANGEABLY TO REFER TO
THE AVERAGE RATE OF CHANGE OR THE SLOPE
OR THE RISE OVER THE RUN


SECTION 13.3: PYTHAGOREAN THEOREM AND ITS CONNECTION
TO THE DISTANCE FORMULA

You learned about the Pythagorean Theorem in geometry. This theorem dealt with right triangles. Note the right triangle below (figure 1). The two line segments that meet to form a 90° angle are called "legs." the line segment that connects the two legs is called the "hypotenuse."
figure 1

PYTHAGOREAN FORMULA
From the Pythagorean formula, the following three
formulas can be deduced mathematically


A PROOF OF THE DISTANCE BETWEEN TWO POINTS
AC USING THE PYTHAGOREAN THEOREM

In figure 2 below you can see the right triangle ABC on a coordinate system. I will be using the Pythagorean Theorem to do this proof.
figure 2
GIVEN:
 length of leg 1 = x_{2 }  x_{1 } eq. 1
 length of leg 2 = y_{2 }  y_{1 } eq. 2
 AC, the distance between the two points, is equal to the length of the hypotenuse
 From the Pythagorean Theorem
Therefore, by substitution
Q.E.D.
