Download CCSSM Algebra: Equations

CCSSM Algebra: Equations
1. Reasoning with Equations and Inequalities (A-REI)
1
Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous
step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify
a solution method.2
1
See page 65 of CCSSM
standards.
2
MP5, MP6, MP8
2. Solve equations and Inequalities in one variable
4. Solve quadratic equations in one variable.
a. Use the method of completing the square to transform
any quadratic equation in x into an equation of the form
( x − p)2 = q that has the same solutions. Derive the
quadratic formula from this form.
b. Solve quadratic equations by inspection (e.g., for x 2 =
49), taking square roots, completing the square, the quadratic
formula and factoring, as appropriate to the initial form
of the equation.3
Solving Equations
15
23
What does 32
11 x + 14 = 13 mean, where x is a real number? That
is, what is an equation in a real number x?
The statement is clearly not true for any real number x. For
example, if x = 1 then the left side of the equation is 613
154 and
23
. By the cross-multiplication
the right side of the equation is 13
algorithm (CMA), the left side is greater than the right side;
they are not equal to each other.4
3
I will assume that at this
point students are familiar
with real numbers, but not
complex numbers. Therefore, the last sentence of
the standard ( Recognize
when the quadratic formula
gives complex solutions
and write them as a ± bi for
real numbers a and b.) is
removed. I will also assume
that students have learned
about algebraic expressions
such as polynomials in x.
4 Theorem 1.1: (The version of the CMA that is
used here.)
Let A, B, C and D be nonzero whole numbers. Then
A
C
B > D ⇔ AD > BC.
CCSSM ALGEBRA: EQUATIONS
2
The statement is not true for x = 21 either. If x = 12 then the
left side of the equation is 389
154 and the left side is still greater
than the right side by CMA. By trying several more numbers
you can see, unless you are extremely lucky, that this statement
is false for all numbers that you tried.
5
We will just say an equation in x, from now on.
So, then what is an equation in a real number x?
5
An equation in x is an invitation to find a number or numbers
x that would make the equation a true statement.
The next question is, how do we handle this invitation?
6
Theorem 1.2: If A, B and
C are real numbers then the
following are true.
1. If A = B then
A + C = B + C.
2. If A = B then
AC = BC.
We know Theorem 1.2 from previous grades.6
32
23
I can think of 11
x + 15
14 as the number A and 13 as the number
B. Then in order to use the Theorem 1.2, I need to have A =
B. Unfortunately, with all my experience, I know that A is
probably not equal to B at all. In other words, I cannot use
Theorem 1.2, for this equation as it stands.
We are going to assume that the equation is true for some real
number x.
With that assumption, I can use Theorem 1.2.
Do not forget that my goal is to find a number or numbers x
that would make the equation true.
With that in mind, I can add − 15
14 to both sides of the equa15
as in the first part of the Theorem 1.2.)
tion. (Here, C = − 14
Then
32
15
x+
11
14
15
23
15
+ −
=
is also true.
+ −
14
13
14
By using properties of numbers,
23
15
32
is also true.
x=
+ −
11
13
14
By multiplying both sides by 1/(32/11),
x=
23
13
+ − 15
14
32
11
is also true.
We have repeatedly used Theorem 1.2 based on our assumption
that there is a number x so that the equation is true. What if,
SOLVING EQUATIONS
3
our assumption is false? Then everything we have done so far
is false. Therefore,
We must check and see the number x that we have found actually makes the equation true.
The left side of the equation =
Therefore, the number
23
13 +
32 
11
15
(− 14
)
32
11

23
13

+ − 15
14
 + 15 = 23
32
14
13
11
, when substituted for x makes
the equation true. Such a number is called a solution of the
23
+( − 15 )
equation. Therefore, 13 32 14 is a solution of the equation
32
11 x
+
15
14
=
11
23
13 .
An equation of the type Ax + B = Cx + D, where A, B, C, D are
fixed real numbers and A2 + C2 > 0, is called a linear equation in
x. The equation that we have solved is a linear equation where
15
32
, B = 14
, C = 0, and D = 23
A = 11
13 .
Try to find the solutions of few more linear equations of this
type. For example,
1. Show that
5 +3
2
2. Show that
43
7 −13.5
is a solution of 2x − 3 = 5.
2.3
is a solution of 2.3x + 13.5 =
43
7 .
This discussion leads to the following conjecture.
The linear equation in x, Ax + B = D, where A, B, D are fixed
−B
numbers and A 6= 0, has a solution DA
.
Do not forget to prove this conjecture.
This discussion ends with the proof of the following theorem.
Theorem 1.3: Consider a linear equation Ax + B = Cx + D,
where A, B, C, D are fixed numbers and A2 + C2 > 0.
(i) The equation has a unique solution if A 6= C and the solution
−B
is D
A−C .
(ii) The equation has no solution if A = C and B 6= D.
(iii) Every number is a solution if A = C and B = D.
Part (i) of Theorem 1.3 leads to the following corollary.
CCSSM ALGEBRA: EQUATIONS
4
Corollary 1.4: Any linear equation Ax + B = Cx + D can be
written as ax + b = 0, if A 6= C, where a and b are constants
and a is non-zero.
Therefore, by Theorems 1 and 2, any linear equation of the form
ax + b = 0 has the unique solution −ab .
For example, the unique solution of 5x − 2 = 0 is 25 .
Identities
An algebraic identity (or more simply an identity) is a statement
that two given expressions are equal for all real numbers, with
possible allowance of “a small number of exceptions".
For example, the commutative property of multiplication, ab =
ba is an identity for all real numbers a and b.
The commutative property of addition, the associative property
of addition, the associative property of multiplication, and the
distributive property are identities for all real numbers.
kn + m ℓ
The formula kℓ + m
is an identity for all real numbers
n =
ℓn
m, n, k and ℓ provided ℓ 6= 0 and n 6= 0. This is an identity
which makes allowance for the exceptions of ℓ = 0 and n =
0.
Next I will introduce some useful identities.
Theorem 2.1: For any two real numbers a and b,
1. ( a + b)2 = a2 + 2ab + b2 .
2. ( a − b)2 = a2 − 2ab + b2 .
3. ( a − b)( a + b) = a2 − b2 .
You can prove this theorem using properties of real numbers.
It is useful to look at these identities backwards. Here is the
same theorem written backwards.
Theorem 2.1: For any two real numbers a and b,
1. a2 + 2ab + b2 = ( a + b)2 .
2. a2 − 2ab + b2 = ( a − b)2 .
3. a2 − b2 = ( a − b)( a + b).
QUADRATIC EQUATIONS
5
Quadratic Equations
Quadratic equations in x are equations where the two expressions
separated by the equal sign are polynomials; one of them of
degree 2 and the other of degree at most 2 in the same number
x. Some examples are 2x2 − 4x + 1 = x2 − 4, 2x + 1 = x2 − x −
4, x2 = 4, x2 − 3x + 1 = 0.
Before, we look at finding solutions to quadratic equations, I
want to point out that the following theorem is familiar to students from previous grades.
Theorem 3.1: If A and B are real numbers and AB = 0, then
either A = 0 or B = 0.
This theorem can be proved by using the trichotomy law of real
numbers.7
We will start with simple quadratic equations of the form x 2 =
a, where a is a real number.
For example, consider the equation x2 = 4. This is an invitation to find the numbers x that would make x 2 = 4 a true
statement.
Assume x2 = 4 is true for some number x. Then by Theorem
1.2,
x2 − 4 = 0 is also true.
Then by identity 3 from Theorem 2.1,
( x − 2)( x + 2) = 0 is also true.
Then by Theorem 3.1,
x − 2 = 0 is true or x + 2 = 0 is true.
Now these are two linear equations and their solutions are 2
and −2.
Now we will check both x = 2 and x = −2 in the equation
x2 = 4 to see if our assumption is true. It turns out that both 2
and −2 are solutions of the equation x2 = 4.
Now try to find solutions to several other quadratic equations
of this type.
For example, consider x2 = 3.
7
The Trichotomy Law of
Real numbers
If A is a real number, then
A > 0, or A = 0, or A < 0.
CCSSM ALGEBRA: EQUATIONS
6
Assume x is a solution of x2 = 3. Then
x2 − 3 = 0√is true. √
=⇒ ( x −√ 3)( x + 3) =
√ 0 is true.
=⇒ x − √3 = 0 or x + 3 =√0 is true.
=⇒ x = 3 is true or x = − 3 is true.
√
After checking
with the equation, we see that both x = 3 and
√
x = − 3 are solutions of x2 − 3 = 0.
Try to find solutions to few more of this type of equations before conjecturing and proving the following theorem.
Theorem 3.2: If a is a fixed non-negative real number then the
√
√
solutions of the equation x2 = a are a and − a.
Now consider the equation ( x − 3)2 = 4.
If we look at this equation as a quadratic equation in x − 3, then
by Theorem 3.2, 2 and −2 are solutions. That is, x − 3 = 2 and
x − 3 = −2 are solutions of the quadratic equation ( x − 3)2 = 4
(in x − 3).
But then x − 3 = 2 and x − 3 = −2 are linear equations and
the solutions of those two equations are 2 + 3 and −2 + 3. By
checking these numbers with the equation ( x − 3)2 = 4, we conclude that both 2 + 3 and −2 + 3 are solutions of the quadratic
equation ( x − 3)2 = 4 in x.
Try to find solutions to several of these equations before proving the following theorem.
Theorem 3.3: If a and b are fixed real numbers√ then the solutions
of the equation ( x + a)2 = b are − a + b and − a −
√
b.
Before we look at the next type of quadratic equations, let us
look back at the identities 1 and 2 of Theorem 2.1. .
a2 + 2ab + b2 = ( a + b)2
for all real numbers a and b.
a2 − 2ab + b2 = ( a − b)2
for all real numbers a and b.
When both a and b are positive numbers, we can provide a geometrical interpretation for identity 1. Consider a square with
sides of length a + b (in some length units). We can partition
this square into 4 rectangles; a square with sides a and area a2 ,
a square with sides b and area b2 , and two rectangles with sides
a, b and area ab as shown below.
QUADRATIC EQUATIONS
b
a
7
ba
b2
a2
ab
a
b
Now consider x2 + 2hx, where both x and h are positive numbers. Then geometrically, this expression can be visualized as
three rectangles; a square with sides x and area x2 , and two
rectangles with sides x, h and area hx put together with no
gaps or overlaps as shown below.
h
x
hx
x2
hx
x
h
As you can see a piece of the square with sides x + h is missing
now. We can complete the square if we adjoin a square with
sides h and area h2 , again with no gaps or overlaps as shown
below.
CCSSM ALGEBRA: EQUATIONS
8
h
x
hx
h2
x2
hx
x
h
Algebraically, if we add h2 to x2 + 2hx, then we get a square,
namely ( x + h)2 . This is precisely the identity 1 of Theorem 2.1.
That is,
x2 + 2hx + h2 = ( x + h)2 .
Similarly, if we add h2 to x2 − 2hx, once more we get a square,
namely ( x − h)2 . This is precisely the identity 2 of Theorem 2.1.
That is,
x2 − 2hx + h2 = ( x − h)2 .
As you may have noticed, the coefficient of the x term in the
polynomial x2 + 2hx + h2 is 2h. Then h2 can be identified as
“the square of the half of the coefficient of x".
Therefore, we can always “complete the square" in any quadratic
equation if we add the square of the half of the coefficient of x
to either x2 + 2hx or x2 − 2hx according to identities 1 and 2 of
Theorem 2.1. We will use this observation in solving quadratic
equations in general.
Now Consider the quadratic equation x2 − 2x − 3 = 0.
Assume x is a solution of x2 − 2x − 3 = 0 for some number x.
Then x2 − 2x = 3 is true, by Theorem 1.2.
Now to complete the square on the left-hand expression of this
equation, we have to add ( 21 · (−2))2 [which is equal to 1] to
it. However, we want to maintain the truthfulness of the above
statement. Therefore, we will use Theorem1.2, and add 1 to
both sides of the above true statement. By doing so, we are
QUADRATIC EQUATIONS
9
killing two birds with one stone: we will complete the square
and also will maintain the truthfulness of the statement. Therefore,
x2 − 2x + 1 = 3 + 1 is still true. That is √
√
( x − 1)2 = 4 is true. By Theorem 3.3, 1 + 4 and 1 − 4 are
solutions of ( x − 1)2 = 4.
√
√
Let us check and see if 1 + 4 and 1 − 4 are indeed solutions
of x2 − 2x − 3 = 0.
(1 +
√
4) 2 − 2 ( 1 +
√
√
√
4) − 3 = 1 + 2 4 + 4 − 2 − 2 4 − 3
=0
Also,
(1 −
√
4) 2 − 2 ( 1 −
Therefore, both 1 +
3 = 0.
√
√
√
√
4) − 3 = 1 − 2 4 + 4 − 2 + 2 4 − 3
=0
4 and 1 −
√
4 are solutions of x2 − 2x −
Find solutions of several more quadratic equations of this type.
Example: Solve x2 + 4x − 7 = 0.
Assume x is a solution of x2 + 4x − 7 = 0 for some number x.
=⇒ x2 + 4x = 7 is true.
=⇒ x2 + 4x + 22 = 7 + 22 is true.
=⇒ ( x√+ 2)2 = 11 is true.
√ Therefore, by Theorem 3.3,
−2 + 11 and −2 − 11 are solutions of ( x + 2)2 = 11.
Since
(−2 +
√
11)2 + 4(−2 +
√
√
√
11) − 7 = (4 − 4 11 + 11) − 8 + 4 11 − 7
= 15 − 8 − 7
=0
and
(−2 −
√
11)2 + 4(−2 −
√
√
√
11) − 7 = (4 + 4 11 + 11) − 8 − 4 11 − 7
= 15 − 8 − 7
=0
both −2 +
0.
√
11 and −2 −
√
11 are solutions of x2 + 4x − 7 =
10
CCSSM ALGEBRA: EQUATIONS
Example: Solve x2 + 3x + 2 = 0.
Assume x is a solution of x2 + 3x + 2 = 0 for some number x.
=⇒ x2 + 3x = −2 is true.
2
2
=⇒ x2 + 3x + 32 = −2 + 32 is true.
2
=⇒ x + 32 = −2 + 49 is true.
2
=⇒ x + 32 = 41 is true. Therefore,
2
− 32 + 21 and − 23 − 21 are solutions of x + 32 = 14 .
Since
(−1)2 + 3(−1) + 2 = 1 − 3 + 2
=0
and
(−2)2 + 3(−2) + 2 = 4 − 6 + 2
=0
both −1 and −2 are solutions of x2 + 3x + 2 = 0.
Example: Solve x2 + 3x + 3 = 0.
Assume x is a solution of x2 + 3x + 3 = 0, for some number x.
=⇒ x2 + 3x = −3 is true.
2
2
=⇒ x2 + 3x + 23 = −3 + 32 is true.
2
=⇒ x + 32 = −3 + 49 is true.
2
=⇒ x + 32 = − 34 is true.
2
This is impossible as x + 23 ≥ 0. (Why?) Our assumption
that there is a number x that would make x 2 + 3x + 3 = 0
true leads us to this non-sensical statement. Therefore, our assumption must be invalid. That is, x 2 + 3x + 3 = 0 has no
solutions.
Instead of following our natural instincts of finding a general
theorem for solving quadratic equations of the type x2 + bx +
c = 0, where a and b are real numbers, we will look to find
a more general theorem by considering quadratic equations of
the form ax2 + bx + c = 0, where a, b, c are real numbers and
a 6= 0.
Example : Solve −2x2 + 3x + 3 = 0.
Assume x is a solution of −2x2 + 3x + 3 = 0 for some number
x.
=⇒ x2 − 32 x − 23 = 0 is true, by Theorem 1.2.
QUADRATIC EQUATIONS
11
At this point you should realize that this new equation is of
the type we have been working lately. Therefore, finding the
solutions must be routine.
x2 − 23 x − 32 = 0 is true =⇒ x2 − 23 x = 32 is true.
2
2
=⇒ x2 − 23 x + 34 = 23 + 34 is true.
2
9
=⇒ x − 43 = 23 + 16
is true.
2
33
3
=⇒ x − 4 = 16 is true. Therefore,
√
√
33
33
3
3
3 2
= 33
4 + 4 and 4 − 4 are solutions of x − 4
16 .
Exercise: Show that
3x + 3 = 0.
√
3+ 33
4
and
√
3− 33
4
are solutions of −2x2 +
Do few more examples of solving quadratic equations of this
type before proving the following theorems.
Theorem 3.4: The solutions of the quadratic equation ax 2 + bx +
c =√0, where a, b, √c are fixed real numbers and a 6= 0 are
2
−b+ b2 −4ac
and −b− 2ab −4ac if b2 − 4ac > 0.
2a
Theorem 3.5: Consider a quadratic equation ax2 + bx + c = 0,
where a, b and c are fixed real numbers and a 6= 0.
(i) If b2 − 4ac > 0, then ax2 + bx + c = 0 has two solutions.
(ii) If b2 − 4ac = 0, then ax2 + bx + c = 0 has one solution.
(iii) If b2 − 4ac < 0, then ax2 + bx + c = 0 has no solutions.
Theorem 3.6: Consider a quadratic equation Ax2 + Bx + C =
Dx2 + Ex + F, where A, B, C, D, E, F are fixed real numbers.8
(i) If A 6= D, then this equation can be written as ax 2 + bx + c =
0, where a and b are fixed real numbers and a is non-zero.
(ii) If A = D and B 6= E, then the quadratic equation reduces
to a linear equation.
(iii) If A = D, B = E and C 6= F, then the equation has no
solution.
(iv) If A = D, B = E and C = F, then every number is a
solution of the equation.
8
By the way, a fixed (real)
number is also known as a
constant.
12
CCSSM ALGEBRA: EQUATIONS
CCSSM Algebra: Lines
1. Graph proportional relationships, interpreting the unit rate as the slope of the graph.
Compare two different proportional relationships represented in different ways.
For example, compare a distance–time graph to a distance–time equation to determine
which of two moving objects has greater speed. (8 EE 5)
2. Use similar triangles to explain why the slope m is the same between any two distinct
points on a non–vertical line in the coordinate plane; derive the equation y = mx
for a line through the origin and the equation y = mx + b for a line intercepting the
vertical axis at b. (8 EE 6)
Proportional Relationship — Constant rate
Consider the following problem.
Example 4.1: Lavonya walked from the Grand Central Station
on 42nd Street to the Penn Station on 7th Avenue. The total
distance is 1.1 miles. She finished the walk in 25 minutes. How
far did she walk in the first 10 minutes?
How could we answer this question? As a matter of fact we
cannot answer this question without knowing more information about her walk.
Suppose we know the following information about her walk.
Lavonya walked from Grand Central Station (GCS) along 42nd
Street to an ATM machine (in between 5th and 6th avenues) in
8 minutes. She stopped at the ATM machine and spent two
minutes there getting some money. (The distance from GCS
to the ATM machine is 0.4 miles.) She reached the corner of
7th Avenue and 42nd Street 13 minutes after she left GCS. At
CCSSM ALGEBRA: LINES
14
the corner she met her friend Karega and stopped to chat with
him for 2 minutes. (The distance from GCS to 7th Avenue is
0.6 miles.) Then Lavonya walked along 7th Avenue until she
arrived at the Penn Station. Let us put this information into a
table.
Time (in Minutes)
Distance travelled (in miles)
0
8
10
13
15
25
0
0.4
0.4
0.6
0.6
1.1
Table 1
With this information, we can answer the question. She walked
exactly 0.4 miles in the first 10 minutes.
9
A definition is a common
agreement.
Is there anything you could have said with the information
given in the original wording of the Example? (That is, without
knowing the details of her walk.) We may be able to salvage
some information with the following definition.9
Definition 4.1: Suppose a person walks a distance d (miles) in
a given time interval t (minutes). Then the average speed in the
d
given time interval is in miles/minutes.
t
So we can calculate Lavonya’s average speed with the information originally given in example 1. Lavonya’s average speed in
the time interval of 25 minutes = 1.1
25 miles/min.
Definition 4.2: Suppose the average speed of a person is the
same constant C for any given time interval. Then we say that
the person is walking at a constant speed C.
Now let us reword the Example 4.1 as given in Example 4.2
below.
Example 4.2: Lavonya walked from Grand Central Station on
42nd Street to Penn Station on 7th Avenue at a constant speed.
The total distance is 1.1 miles. She finished the walk in 25 minutes. How far did she walk in the first 10 minutes?
Now this is a question we can answer. The following is the
detailed answer to the question.
PROPORTIONAL RELATIONSHIP — CONSTANT RATE
15
Answer:
Lavonya’s average speed for 25 minutes =
1.1
25
(1)
Suppose Lavonya walked y miles in 10 minutes. Then
Lavonya’s average speed for 10 minutes =
y
10
(2)
Suppose Lavonya’s constant speed is C miles/minute. Then, by
the definition of the constant speed and (1) above, we get:
1.1
=C
25
(3)
By the definition of constant speed and (2) above, we get:
y
=C
10
(4)
Now by (3) and (4) we get:
y
1.1
=
25
10
By the definition of division for real numbers:10
y=
11
.
25
10
11
miles in any 10 minutes, in particular,
25
in the first 10 minutes.
So, Lavonya walked
Let us look at this same problem more closely.
Example 4.3: Lavonya walked from Grand Central Station on
42nd Street to Penn Station on 7th Avenue at a constant speed.
The total distance is 1.1 miles. She finished the walk in 25 minutes. How far did she walk in the first x minutes?
Answer:
Lavonya’s average speed for 25 minutes =
1.1
25
(5)
Suppose Lavonya walked y miles in x minutes. Then
Lavonya’s average speed for 10 minutes =
y
x
(6)
If Lavonya’s constant speed is C miles/minute, then, by the
definition of constant speed and (5) above, we get:
1.1
=C
25
(7)
The last statement is
known as a proportional
relationship in some circles.
CCSSM ALGEBRA: LINES
16
By the definition of constant speed and (6) above, we get:
y
=C
x
(8)
Now by (7) and (8) we get:
y
1.1
=
25
x
By the definition of division for real numbers:
y=
1.1
x.
25
This last equation is an example of an equation in two variables
x and y. With this equation we can find the distance y Lavonya
walked in any given time x.
Let us look at another problem.
Example 4.4: Dave lives 15 miles from town A. One day, he
was driving at a constant speed of 50 mph from his home away
from (in the opposite direction of) the city. How far away is
Dave from the town after x hours of driving?
Answer:
Recall the definition of constant speed. That is, if the average
speed of a motion over any time interval is the same constant
then we say that the motion is in constant speed.
Since Dave is driving at a constant speed, let us assume that
this constant speed is m miles/hour. Let the distance between
Dave and the town after x hours be y miles. Then the distance
travelled by Dave in x hours is y − 15 miles. Then
The average speed of Dave in x hours =
y − 15
.
x
Since Dave is travelling at a constant speed, the average speed
for any time interval is 50 mph. That is,
y − 15
= 50
x
After using the definition of division for real numbers and collecting like terms, we have:
y = 50x + 15
CONSTANT RATE PROBLEMS
17
This is an equation in two variables x and y. With this equation
we can find the distance between Dave and the town for any
given time x.
For example, how far away was Dave from the town after one
hour?
This is an invitation to find y when x is 1. Since we know that
y = 50x + 15, y = 65 when x = 1. Therefore, Dave is 65 miles
from town after 1 hour.
Example 4.5: Find how far is Dave after 1 hour, 2 hours, 3 hours,
3.5 hours and 4.1 hours. Show your answers in a table.
Solution:
x
(hours)
linear equation in y
y
(miles)
1
2
3
3.5
4.1
y=50(1)+15
y=50(2)+15
y=50(3)+15
y=50(3.5)+15
y=50(4.1)+15
65
115
165
190
220
Table 4.2
Constant Rate Problems
In the last section we learned about constant speed. It turns
out constant speed problems are just a special case of a larger
variety of problems known as constant rate problems. You have
seen some of these problems in grade 7. The constant rate of
water pouring out of a faucet into a tub, constant rate of painting a house, constant rate of mowing a lawn, etc. First we will
define average rate.
Definition 5.1: Suppose V gallons of water flow from a faucet
in a given time interval t (minutes). Then the average rate of waV
ter flow in the given time interval is
in gallons/minute.
t
Definition 5.2: Suppose the average rate of water flow is the
same constant C for any given time interval. Then we say that
the water is flowing at a constant rate C.
18
CCSSM ALGEBRA: LINES
Definition 5.3: Suppose A square feet of lawn is mowed in
a given time interval t (minutes). Then the average rate of lawn
A
mowing in the given time interval is in square feet/minute.
t
Definition 5.4: Suppose the average rate of lawn mowing is the
same constant C for any given time interval. Then we say that
the lawn is mowed at a constant rate C.
Definition 5.5: Suppose A square feet of wall are painted in
a given time interval t (minutes). Then the average rate of wallA
painting in the given time interval is in square feet/minute.
t
Definition 5.6: Suppose the average rate of wall painting is the
same constant C for any given time interval. Then we say that
the wall is painted at a constant rate C.
As you can see all these definitions are similar. We will group
them all together and call them constant rate problems. A transcription of any of these problems lead to an equation of two
variables. Let us look at few of problems.
Example 5.1 Peter mows a lawn at a constant rate. Suppose he
mowed a 35 square foot section of lawn in 2.5 minutes. What
area of lawn does he mow in 10 minutes? in t minutes?
Answer:
Suppose Peter’s constant rate is C square feet per minute. Suppose Peter mows A square feet of lawn in 10 minutes and y
square feet of lawn in t minutes.
35
square feet/min.
2.5
A
Peter’s average rate in 10 minutes =
square feet/min.
10
Since Peter’s average rate in any time interval is the constant C,
we have
35
=C
(1)
2.5
and
A
=C
(2)
10
By (1) and (2) we get:
A
35
= .
2.5
10
We can find A from the above equation. Finding A is left as an
exercise.
Peter’s average rate in 2.5 minutes =
CONSTANT RATE PROBLEMS
Peter’s average rate in t minutes =
fore,
19
y
square feet/min. Theret
y
35
=
t
2.5
By using the definition of division for real numbers we get the
following equation of two variables for this problem.
35
t.
y=
2.5
Example 5.2: Pyumi paints a house at a constant rate. Suppose she paints a 35 square feet section of house in 2.5 minutes.
What area of the house does she paint in 10 minutes? in t minutes?
Answer:
Suppose Pyumi’s constant rate is C square feet per minute. Suppose Pyumi paints A square feet of the house in 10 minutes and
y square feet of the house in t minutes.
35
square feet/min.
2.5
A
Pyumi’s average rate in 10 minutes =
square feet/min.
10
Since Pyumi’s average rate in any time interval is the constant
C, we have
35
=C
(1)
2.5
and
A
=C
(2)
10
By (1) and (2) we get:
35
A
= .
2.5
10
We can find A from the above equation.
y
Pyumi’s average rate in t minutes = square feet/min. Theret
fore,
y
35
=
t
2.5
By using definition of division for real numbers we get the following equation of two variables for this problem.
35
t.
y=
2.5
Pyumi’s average rate in 2.5 minutes =
CCSSM ALGEBRA: LINES
20
Example 5.3: Water flows from a faucet at a constant rate. Suppose the faucet pours 35 gallons of water into a tub in 2.5 minutes. What is the amount of water pours out of the faucet in 10
minutes? in t minutes?
Answer:
Suppose constant rate of water flow is C gallons per minute.
Suppose V gallons of water flow in 10 minutes and y gallons of
water flow in t minutes.
The average rate of water flow in 2.5 minutes =
35
gallons/min.
2.5
The average rate of water flow in 10 minutes =
A
gallons/min.
10
Since average rate of water flow in any time interval is the constant C, we have
35
=C
(1)
2.5
and
V
=C
10
(2)
By equations (1) and (2) we get:
V
35
= .
2.5
10
We can find V from the above equation.
The average rate of water flow in t minutes =
Therefore,
y
gallons/min.
t
35
y
=
t
2.5
By using definition of division for real numbers we get the following equation of two variables for this problem.
y=
35
2.5
t.
Notice the seemingly identical thought pattern in Examples
5.1-5.3. We will investigate linear equations in two variables
next.
LINEAR EQUATIONS IN TWO VARIABLES
21
Linear Equations in two variables
An equation of the form ax + by = c is called a linear equation
in two variables, where a, b and c are constants and at least one
of a and b is not zero.
Once again keep in mind that x and y are numbers.
Example 6.1: −50x + y = 15 is a linear equation in x and
y.
As you can easily see, any pair of numbers x and y does not
make −50x + y = 15 a true statement. For example, if y = 5
and x = 2 then 5 − 50(2) 6= 15.
A solution of this equation is an ordered pair of numbers (x, y)
so that x and y make −50x + y = 15 a true statement.
(1, 65) is a solution of −50x + y = 15 since −50(1) + 65 =
15. Similarly, (2, 115), (3, 165), (3.5, 180) and (4.1, 220) are all
solutions of −50x + y = 15.
We can obtain many more solutions as follows. Fixed the first
number. Say x = 5. Then the resulting equation −50(5) + y =
15 is a linear equation in y. Since we know how to solve linear
equations in one variable, by solving it, we get y = 265 as the
solution of the equation y − 50(5) = 15. Therefore, (5, 265) is a
solution of −50x + y = 15.
We can find solutions to −50x + y = 15 by fixing the second
number instead of the first one. For example, let y = 75. Then
the resulting equation −50x + 75 = 15 is a linear equation in
x and the solution is x = 56 . Therefore, ( 65 , 75) is a solution of
−50x + y = 15.
Example 6.2: Find five solutions of the linear equation in two
variables 21 x + 53 y = 2.
Answer:
Let us pick few values for x and make a table.
CCSSM ALGEBRA: LINES
22
x
linear equation in y
y
1
1
3
2 (1) + 5 y
=2
5
2
2
3
1
2 (2) + 5 y
=2
5
3
3
1
3
2 (3) + 5 y
=2
5
6
4
3
1
2 (4) + 5 y
=2
0
5
1
3
2 (5) + 5 y
=2
− 56
Table 6.1
Therefore, (1, 52 ), (2, 35 ), (3, 65 ), (4, 0), and (5, − 56 ) are solutions
of 21 x + 53 y = 2.
Example 6.3: Plot the five solutions of the linear equation in
two variables 12 x + 35 y = 2 that you have found in example 12.2
on a coordinate plane.
Solution: We can visualize these solutions using a coordinate
plane. Let A be the point with coordinates (1, 25 ), B be the point
with coordinates (2, 53 ), C be the point with coordinates (3, 65 ),
D be the point with coordinates (4, 0), and E be the point with
coordinates (5, − 56 ), Then the five solutions we obtained for
the equation 12 x + 35 y = 2, that is, the five ordered pairs (1, 25 ),
(2, 53 ), (3, 65 ), (4, 0) and (5, − 56 ), can be visualized as points on
the coordinate plane.
y
3
A
2
B
C
1
x
D
−1
0
1
2
3
4
5
E
−1
−2
Figure 6.1
GRAPH OF A LINEAR EQUATION IN TWO VARIABLES
23
Graph of a Linear Equation in Two Variables
We have seen in example 6.3 that the solutions of a linear equation in two variables can be plotted on a coordinate plane as
points.
The collection of all points ( x, y) in the xy-plane so that each
( x, y) is a solution of ax + by = c is called the graph of ax + by =
c.
Obviously, we cannot plot all solutions of a linear equation of
two variables on a coordinate plane. That means we cannot
draw the graph of a linear equation.
What can we do? We can plot a few points of the graph of an
equation of two variables and then make predictions of what
the graph should look like. With that in mind let us plot some
points of a graph of a linear equation.
Example 7.1: Find a few points on the graph of 2x − 3y = 2
and plot them on a coordinate plane.
Answer:
Let x = 1. Then 2(1) − 3y = 2. Solve this equation for y. Then
y = 0. Therefore, (1, 0) is a point on the graph of 2(1) − 3y = 2.
We will continue this process for few more values of x. When
x = 2, 2(2) − 3y = 2 and y = 32 and (2, 32 ) is a point on the
graph of 2(1) − 3y = 2. Let us make a table to gather these
points on the graph.
x
y
1
0
2
2
3
3
4
3
4
2
5
8
3
Now let us plot these points of the graph of 2x − 3y = 2 on a
coordinate system.
CCSSM ALGEBRA: LINES
24
y
3
2
1
x
−1
0
1
2
3
4
5
−1
Figure 7.1
Can you predict the shape of the graph of this linear equation
based on the meager information we have so far? It looks like
the points lie on a line, but, for all we know, the graph of the
equation can be some curve passing through the five points we
have, such as the following curve.
y
3
2
1
x
−1
0
1
2
3
4
5
−1
Figure 7.2
With the hope that more points on the graph would shed more
light let us choose few more points on the graph. The following
are few additional points on the graph.
GRAPH OF A LINEAR EQUATION: HORIZONTAL AND VERTICAL LINES25
x
y
3
2
1
3
5
2
1
7
2
4
3
9
2
7
3
We will add the plots of these additional points to the same
coordinate system.
y
3
2
1
x
−1
0
1
2
3
4
5
−1
Figure 7.3
Can you make a prediction about the graph of 2x − 3y = 2?
Graph of a Linear Equation: Horizontal and Vertical Lines
It looks like the graph of a linear equation of two variables (at
least the ones that we looked at so far) are lines. Is this prediction actually true? Let us investigate to see if we can prove this
claim. That is, we want to prove, not just predict, that the graph
of a linear equation in two variables is a line.
Let us start our investigation with two special cases of linear
equations of two variables.
CCSSM ALGEBRA: LINES
26
Consider a linear equation of two variables ax + by = c, where
a, b and c are constants and both a and b cannot be 0.
Case 1: a = 1 and b = 0.
In this case, the linear equation becomes 1 · x + 0 · y = c.
What can we say about the graph of 1 · x + 0 · y = c?
Example 8.1: Let us pick a particular c, Consider the linear
equation 1 · x + 0 · y = 5.
What is the collection of all solutions of 1 · x + 0 · y = 5? We
are looking for ordered pairs ( x, y) so that 1 · x + 0 · y = 5 is
true.
Can x be any number other than 5 in a solution ( x, y) of 1 · x +
0 · y = 5? For example, can (4.5, y) be a solution of 1 · x + 0 · y =
5? Let us check.
The left side of the equation = 1 · (4.5) + 0 · y
= 4.5.
The right side of the equation = 5.
Since, 4.5 6= 5, (4.5, y) cannot be a solution of 1 · x + 0 · y = 5,
for any number y. In general, ( x, y) cannot be a solution of
1 · x + 0 · y = 5, if x 6= 5, for any number y.
But if x = 5 then (5, y) is a solution of 1 · x + 0 · y = 5, because
1 · 5 + 0 · y = 5, for any number y.
Therefore, the solutions of 1 · x + 0 · y = 5 are exactly all pairs
of the form (5, y), where y is any number.
The points with coordinates (5, y) always lie on the vertical line
passing through the point (5, 0). Since y can be any number,
the collection of all points (5, y) is the complete vertical line
passing through (5, 0).
GRAPH OF A LINEAR EQUATION: HORIZONTAL AND VERTICAL LINES27
y
(5, 5)
5
4
(5, y0 )
y0
3
2
1
x
−2
−1
0
1
2
3
4
5
6
−1
−2
(5, −2)
−3
Figure 8.1: The vertical line passing through (5, 0)
The above argument does not depend on the particular value of
c. If we replace, 5 with 3 then you will come to the conclusion
that the collection of all solutions of 1 · x + 0 · y = 3 is the
vertical line passing through the point (3, 0).
Notice that 1 · x + 0 · y = c is equivalent to x = c. Therefore, we
come to the following general conclusion. (In other words we
have proved the following theorem.)
Theorem 8.1: The graph of x = c is the vertical line passing
through (c, 0), where c is a constant.
Now let us look at vertical lines.
We claim that any vertical line has to pass through the x-axis at
some point (c, 0) for some constant c.
The following is the reason (proof) for the above statement. If
a vertical line is not passing through the x-axis, then by the
definition of parallel lines, this vertical line is parallel to the
x-axis. But, being vertical, it is parallel to the y-axis as well.
Then this implies that x-axis is parallel to the y-axis and we all
know that that is not true. Therefore, a vertical line has to pass
28
CCSSM ALGEBRA: LINES
through the x-axis.
Consider a vertical line passing through the point (c, 0), for
some constant c. We claim that this is the only vertical line
passing through (c, 0). The following is the reason (proof) for
the above claim.
The vertical line passing through (c, 0) is parallel to the y-axis.
There is exactly one line passing through (c, 0) and parrallel to
y-axis by the Parallel Postulate.
By putting the two claims together, we have:
There is exactly one vertical line passing through (c, 0).
But we have shown that this vertical line is the graph of the
linear equation x = c. With this, we have a second conclusion.
Theorem 8.2: Every vertical line is the graph of x = c, where
(c, 0) is the point of intersection of the line and the x-axis.
Case 2: a = 0 and b = 1.
In this case, the linear equation becomes 0 · x + 1 · y = c or
y = c.
What can you say about the graph of y = c?
Theorem 8.3: The graph of y = c is the horizontal line passing
through the point (0, c), where c is a constant.
Theorem 8.4: Every horizontal line is the graph of y = c, where
(c, 0) is the point of intersection of the line with the y-axis.
The proofs of above theorems are left as exercises.
Case 3: a 6= 0 and b = 0.
In this case, the linear equation becomes a · x + 0 · y = c and
after solving for x, x = ac . Since ac is a constant, we come to the
same conclusion as in case 1.
Case 4: a = 0 and b 6= 0.
In this case, the linear equation becomes 0 · x + b · y = c and
after solving for y, x = bc . Since bc is a constant, we come to the
same conclusion as in case 2.
THE SLOPE OF A NON-VERTICAL LINE
29
The Slope of a Non-Vertical Line
In the last lesson, we have seen that:
The graph of the linear equation x = c is the vertical line
passing through the point (c, 0) and every vertical line is the
graph of the equation x = c, where (c, 0) is the point of intersection of the line and the x-axis.
We have also seen that for a special non-vertical line, namely an
horizontal line, the conclusion is the "same". That is,
The graph of the linear equation y = c is the horizontal line
passing through the point (0, c), and every horizontal line is
the graph of the equation y = c, where (0, c) is the point of
intersection of the line and the x-axis.
We would like to prove a similar statement for other non-vertical
lines based on our experiments, if possible. However, this task
requires that we develop some tools first. The following is a list
of tools that we will develop to complete the task.
1. Define a number for each non-vertical line that can be
used as a measure of "steepness" or "slant" of a non-vertical
line. Once defined, this number will be called the slope of
the line.
2. Show that any two points on a non-vertical line can be
used to find the slope of that line.
3. Show that the line joining two points on the graph of a
linear equation of the form y = mx + k has the slope m.
4. Show that there is only one line passing through a given
point with a given slope.
In this section and next few sections we will develop the said
tools. With this in mind, let us look closely at non-vertical
lines.
Having excluded horizontal lines, there are two other types of
non-vertical lines. The following figures show the two types
of lines in a coordinate system. (The coordinate system is not
necessary; it is included only for clarity.) The two lines are
named ℓ1 and ℓ2 . For lack of better terms we will call the line ℓ1
a left-to-right inclining line and the line ℓ2 a left-to-right declining
line.
CCSSM ALGEBRA: LINES
30
y
ℓ1
x
y
x
ℓ2
Figure 9.1
We would like to associate a number with any non-vertical line
which indicates the amount of its "slant" or "steepness" of a
line. Our definition of such a number should assign 0 to a
horizontal line as it has no slant or steepness. Our definition
of such a number should also indicate that a vertical line has
the ultimate “steepness" even though we are not defining this
number for vertical lines.
11
We can formally achieve
this by doing the following. Let O be the origin
of the coordinate system.
Consider the translation
T of the plane along the
−→
vector OQ. Then the translated y-axis is ℓ2 . That is,
T ( y-axis) = ℓ2 . As a result, ℓ2 is a number line
with 0 at Q, positive numbers above Q and negative
numbers below Q.
Suppose a non-vertical line ℓ is given in the coordinate plane.
Let P be any point on ℓ. Draw a line ℓ1 parallel to the x-axis
through P. Locate the point Q one unit to the right of P on ℓ1 .
Draw a line ℓ2 parallel to the y-axis through Q. We will make
ℓ2 a number line with 0 at Q, positive numbers above Q and
negative numbers below Q.11
THE SLOPE OF A NON-VERTICAL LINE
31
ℓ2
y
ℓ
m
R
Q
P
1
ℓ1
x
Figure 9.2
The line ℓ, being non-vertical, will intersect ℓ2 at some point R
on ℓ2 . Let m be the number at R on ℓ2 . We will call this number
m the slope of ℓ, and this is the number we want to associate
with any non-vertical line.
In the above figure ℓ is a "left-to-right inclining" line and by
definition, the slope of such a line is a positive number.
Example 9.1: Consider the following left-to-right inclining line.
After going through the steps of the definition of the slope, we
can say that the slope of this line is 2.
3
2
R
1
P
Q
1
0
−1
−2
Figure 9.3
CCSSM ALGEBRA: LINES
32
Now consider a “left-to-right declining" line ℓ. (See the given
figure below.) By the definition of the slope of a line, the slope
of such a line is negative as seen in the following figure.
ℓ2
y
P
1
Q
ℓ1
x
m
R
ℓ
Figure 9.4
Example 9.2: Consider the following left-to-right inclining line.
After going through the steps of the definition of the slope, we
can say that the slope of this line is −3.
2
1
1
Q
P
0
−1
−2
−3
R
Figure 9.5
If ℓ is a horizontal line, then ℓ1 is same as ℓ in the definition.
Therefore, by definition, the slope of a horizontal line is 0. (See
the following figure.)
THE SLOPE OF A NON-VERTICAL LINE
33
ℓ2
y
m
P
Q=R
ℓ1 = ℓ
1
x
Figure 9.6
Even though our definition of the slope is for non-vertical lines
only, if we choose ℓ to be a vertical line and follow the definition
of the slope anyway, then ℓ and ℓ2 are parallel line and therefore, do not intersect. That is, there is no number on ℓ2 that we
can assign as a slope of ℓ. Therefore, we say that the slope of a
vertical line does not exist. (See the figure below.)
y
ℓ
ℓ2
P
Q
1
ℓ1
x
Figure 9.7
How does the slope relate to the “steepness" of a line? Consider
the two intersecting left-to-right inclining lines L1 , L2 in the
following figure. and let P be the point of intersection. We
will use P and the definition of slope to get the slopes m1 and
m2 of the two lines respectively. By construction, the line L1 is
“steeper" than L2 and it turns out that m1 > m2 .
CCSSM ALGEBRA: LINES
34
ℓ2
y
L1
m1
L2
m2
P
ℓ1
Q
1
x
Figure 9.8
Now consider the two intersecting left-to-right declining lines
L3 , L4 in the following figure. and let P be the point of intersection. Again, we will use P and the definition of slope
to get the slopes m3 and m4 of the lines, respectively. By construction, the line L3 is “steeper" than L4 and it turns out that
| m 3 | > | m 4 |.
ℓ2
y
P
1
Q
ℓ1
x
m4
m3
L4
L3
Figure 9.9
As a result of these observations, we can conclude the following.
Given two lines with slopes m and n, if |m| > |n| then the line
with slope m is steeper than the the line with slope n.
THE COMPUTATION OF THE SLOPE OF A NON-VERTICAL LINE 35
The Computation of the Slope of a Non-Vertical
Line
Now we will make a bold claim. If the coordinates of any
two distinct points P = ( p1 , p2 ) and R = (r1 , r2 ) on a nonvertical line are given, then we can compute the slope of that
line. As a matter of fact, we claim that the slope of the line is
p 2 − r2
.
p 1 − r1
Notice that ( p2 − r2 ) = −(r2 − p2 ) and ( p1 − r1 ) = −(r1 − p1 ).
Then
p 2 − r2
−(r2 − q2 )
(r − p2 )
=
= 2
.
p 1 − r1
−(r1 − p1 )
( r1 − p 1 )
Therefore, the order of using P and R to calculate the slope
would not matter, if the claim is correct.
Let us handle the easy case first. Suppose the given non-vertical
line is horizontal.
y
P
R
x
Figure 10.1
Then p2 = r2 and
p 2 − r2
0
=
.
p 1 − r1
p 1 − r1
Since P and R are distinct points, p1 6= r1 . Therefore,
p 2 − r2
= 0.
p 1 − r1
CCSSM ALGEBRA: LINES
36
Now consider a left-to-right inclining line. Suppose the slope
of the line is m. Let us call this given line L for convenience.
Let us assume that P is to the left of R and use P to get the
slope of the line by construction according to the definition of
the slope. Let Q′ be the point one unit to the right of P on the
horizontal line ℓ through P and let R′ be the point of intersection of L and the vertical line through Q′ . Let Q be the point
of intersection of ℓ and the vertical line through R. (See the
following figure.)
L
y
R
R′
Q′
P
Q
1
ℓ
x
Figure 10.2
Now consider the triangles △ PQ′ R′ and △ PQR. The angles
6 Q ′ and 6 Q are right angles by construction. The angle 6 P is
common to both triangles. Therefore, by the AA criterion, two
triangles are similar.
Let the length of the segment PQ be | PQ|, the length of the
segment RQ be | RQ|, the length of the segment PQ′ be | PQ′ |
and the length of the segment R′ Q′ be | R′ Q′ |.
Then, since △ PQ′ R′ ∼ △ PQR,
| PQ′ |
| R′ Q ′ |
=
.
| RQ|
| PQ|
(1)
By the definition of slope, | R′ Q′ | = m, | PQ′ | = 1. Since the
coordinates of P and Q are ( p1 , p2 ) and (r1 , r2 ), | RQ| = r2 − p2
THE COMPUTATION OF THE SLOPE OF A NON-VERTICAL LINE 37
and | PQ| = r1 − p1 . Therefore, by (1):
1
m
=
.
r2 − p 2
r1 − p 1
By the definition of the division for real numbers:
1
( r2 − p 2 ) .
m=
r1 − p 1
or, equivalently,
m=
r2 − p 2
.
r1 − p 1
Finally, consider a left-to-right declining line. Once again suppose the slope of the line is m and the point P is to the left
of R on the line. The following construction follows the same
sequence of thoughts.
y
P
1
Q′
Q
R′
R
x
L
Figure 10.3
Now consider the triangles △ PQ′ R′ and △ PQR again. The
angles 6 Q′ and 6 Q are right angles by construction. The angle
6 P is common to both triangles. Therefore, by the AA criterion,
the triangles are similar and
| R′ Q ′ |
| PQ′ |
=
.
| RQ|
| PQ|
(2)
By definition of slope, | R′ Q′ | = −m, | PQ′ | = 1. Since the
coordinates of P and Q are ( p1, p2 ) and (r1 , r2 ), | RQ| = p2 − r2
CCSSM ALGEBRA: LINES
38
and | PQ| = r1 − p1. Then by (2):
1
−m
=
.
p 2 − r2
r1 − p 1
Then
−m
1
=
−(r2 − p2 )
r1 − p 1
or, equivalently,
1
m
=
.
r2 − p 2
r1 − p 1
By the definition of division for real numbers:
m=
1
r1 − p 1
or
m=
( r2 − p 2 ) .
r2 − p 2
.
r1 − p 1
Therefore, we have proved the following theorem.
Theorem 10.1: Given any two distinct points P = ( p1, p2 ) and
r − p2
.
R = (r1 , r2 ) on a non-vertical line L, the slope of L is 2
r1 − p 1
The Line Joining Two Distinct Points of the Graph
of y = mx + k has Slope m
Recall our goal. We want to prove that the graph of a linear
equation of two variables ax + by = c, where a, b, and c are
constants and a 6= 0 and b 6= 0 is a non-vertical line.
We can write ax + by = c in an equivalent form as y = − ba x + bc .
The following is the reason for this.
Assume ( x, y) is a solution of ax + by = c. That is, we assume
that ax + by = c is true. Then
ax + by = c is true =⇒ ax + by − ax = c − ax is true.
=⇒ by = − ax + c is true.
c
a
=⇒ y = − x + is true, since b 6= 0.
b
b
THE LINE JOINING TWO DISTINCT POINTS OF THE GRAPH OF Y = MX + K HAS SLOPE M39
Now assume that ( x, y) is a solution of y = − ba x + bc . That is,
we assume that y = − ba x + bc is true. Then
c
a
y = − x + is true =⇒ by = − ax + c is true, since b 6= 0
b
b
=⇒ by + ax = − ax + c + ax is true.
=⇒ ax + by = c is true.
Therefore, any solution of ax + by = c is a solution of y =
− ba x + bc and any solution of y = − ba x + bc is a solution of
ax + by = c.
Let m = − ba and k = bc . m 6= 0 since a 6= 0. Then the above
linear equation becomes:
y = mx + k, m 6= 0.
We can restate our goal as
We want to prove that the graph of a linear equation of two
variables y = mx + k, where m 6= 0 and k are constants, is a
non-vertical line.
In the last two sections we took one step towards this goal by
defining the slope of a non-vertical line and learning how to
calculate the slope given any two points on the line. Now we
will take another step towards our goal by showing the following.
Theorem 11.1: The line joining any two distinct points of a
graph of y = mx + k, m 6= 0 has the slope m.
Let us look at an example first.
Example 11.1: Show that the slope of a line joining any two
distinct points of the graph of y = 2x + 3 has the slope 2.
Let P = ( p1 , p2 ) and Q = (q1 , q2 ) be any two distinct points
of the graph of y = 2x + 3. Since P is on the graph of y =
2x + 3,
p2 = 2p1 + 3.
Similarly, since Q is on the graph of y = 2x + 3,
q2 = 2q1 + 3.
By the Theorem 10.1, the slope of the line passing through P
CCSSM ALGEBRA: LINES
40
and Q is:
(2p1 + 3) − (2q1 + 3)
p 2 − q2
=
p 1 − q1
p 1 − q1
2p1 + 3 − 2q1 − 3
=
p−q
2p − 2q1
= 1
p 1 − q1
2 ( p 1 − q1 )
=
p 1 − q1
= 2, by the cancellation law for real numbers, since p1 6= q1 .
Therefore, the slope of the line passing through P and Q is
2.
Now we will handle the general proof of the Theorem 11.1.
Remember that we do not know the "shape" of the graph of
y = mx + k at this point. We guess that it is a line but we
have no proof. (Getting the proof is our goal.) Therefore, let
us not assume the graph of y = mx + k is a line until we have
a proof. In the following figure we will draw the graph of
y = mx + k as "some" curve; not a line. Let P and Q be two
points on the graph of y = mx + k so that P = ( p1, p2 ) and
Q = ( q1 , q2 ) .
y
ℓ
Q
P
graph of
y = mx + k
x
Figure 11.1
P is a point on the graph of y = mx + k. Therefore, p2 =
mp1 + k and the coordinates of P are ( p1, mp1 + k). The point
Q is also on the graph of y = mx + k. Therefore, the coordinates
of Q are (q1 , mq1 + k).
Let ℓ be the line passing through P and Q. Since P and Q are
THERE IS ONLY ONE LINE PASSING THROUGH A GIVEN POINT WITH A GIVEN SLOPE41
two distinct points on ℓ, we can use the coordinates of P and Q
to find the slope of ℓ. (Theorem 10.1.)
(mp1 + k) − (mq1 + k)
p 1 − q1
mp1 + k − mq1 − k
=
p 1 − q1
mp1 − mq1
=
p 1 − q1
m ( p 1 − q1 )
=
p 1 − q1
slope of ℓ =
= m. by using cancellation law (for real numbers), since p1 − q1 6= 0.
There is Only One Line Passing Through a Given
Point with a Given Slope
Let us recall our goal.
We want to prove that the graph of a linear equation of two
variables y = mx + k, where m 6= 0 and k are constants, is a
non-vertical line.
In the last section, we developed the following tool.
Given any two points P, Q on the graph of a linear equation
y = mx + k, where m and k are constants, the line through P
and Q has the slope m.
In this section we will develop the last tool that we need to
achieve our goal.
We want to show that there is only one line passing through a
given point with a given slope.
Theorem 12.1: If two straight lines have the same slope and
pass through the same point, then they are the same line.
Proof: Let ℓ and ℓ′ be two lines with the same slope m passing
through a point P. Since m is a number, either m > 0, m = 0 or
m < 0. We will prove the theorem for m > 0 and leave the the
other two cases as exercises.
Since m > 0, both ℓ and ℓ′ are left-to-right inclining lines.
CCSSM ALGEBRA: LINES
42
Our task is to show that ℓ and ℓ′ are the same line. As usual we
will not assume the theorem is true until proven. Therefore, we
will treat ℓ and ℓ′ as two distinct lines as given below.
Let P be the point common to both lines. Pick the point Q one
unit from P on the line passing through P and parallel to the
x-axis. Draw the line L passing through Q and parallel to the
y-axis. Then (by the definition of the slope of a line) L is a
number line with 0 at Q. Let R be the point of intersection of L
and ℓ and let R′ be the point of intersection of L and ℓ′ . Both R
and R′ lie on the positive side of the number line L.
y
R′
ℓ′
ℓ
R
P
Q
1
x
Figure 12.1
L
The slope of ℓ is m. Therefore, the number at R on the number
line L is m. The slope of ℓ′ is also m. That is, the number at R′
on the number line is m as well. But we know that if two points
on a number line have the same number then those two points
are the same. Therefore, R = R′ . This implies that both ℓ and
ℓ′ pass through the same two distinct points P and R. But there
is exactly one line passing through two given distinct points.
Therefore, ℓ and ℓ′ are the same line.
Exercises 12.1:
1. Prove the Theorem 12.1 for horizontal lines.
2. Prove the Theorem 12.1 for left-to-right declining lines.
3. In the proof of Theorem 12.1 we used the following claim.
GRAPH OF A LINEAR EQUATION IN TWO VARIABLES IS A LINE 43
R and R′ lie on the positive side of the number line L.
How do we know this? Explain.
Graph of a Linear Equation in Two Variables is
a Line
Recall our goal.
We want to prove that the graph of a linear equation of two
variables y = mx + k, where m 6= 0 and k are constants, is a
non-vertical line.
Now we are ready to prove this claim.
We will prove the claim in the following novel way. Let P, Q
be two distinct points on the graph of y = mx + k. Let ℓ be the
line passing through P and Q.
We will show that
(1) any point on the graph of y = mx + k is a point on ℓ
and
(2) any point on ℓ is a point on the graph of y = mx + k.
Why do we need to show both (1) and (2)? Let the graph of
y = mx + k be denoted by G. If G is the collection of red
segments as given in the following figure, then (1) is true but
(2) is not true. That is why we need to show both (1) and
(2).
y
ℓ
3
G
Q
2
P
1
−3
−2
−1
0
−1
1
2
3
4
5
6
Figure 13.1
Therefore, if we can prove both (1) and (2) then the graph of
y = mx + k is the line ℓ and we are done.
7
8
x
CCSSM ALGEBRA: LINES
44
Let us look at an example first to see how to prove these two
claims before proving them in general.
Example 13.1: Show that the graph of y = 2x + 3 is a nonvertical line.
Solution:
The points P(0, 3) and Q(1, 5) are on the graph of y = 2x + 3.
(Verify) Consider the line ℓ passing through P and Q. The slope
of ℓ is 2, by Theorem 17.1. Therefore, ℓ is a non-vertical left-toright inclining line.
We claim that the graph of y = 2x + 3 is ℓ.
We will prove this claim by using the following strategy.
(1) Show that any point on the graph of y = 2x + 3 is on ℓ.
(2) Show that any point on ℓ is on the graph of y = 2x + 3.
If both (1) and (2) are true then the graph of y = 2x + 3 is ℓ, as
all points on the graph of y = 2x + 3 is on ℓ and all points of ℓ
are on the graph of y = 2x + 3.
We will prove (1) first.
Let R be an arbitrary point on the graph of y = 2x + 3. We
want to show that R is on ℓ. If R is either P or Q then there
is nothing to prove. (We already know that P and Q are on
ℓ.) Let ℓ′ be the line passing through P and R. Then the slope
of ℓ′ is 2, by Theorem 16.1. Then ℓ′ and ℓ are the same line
by Theorem 18.1 as both ℓ and ℓ′ pass through P and have the
same slope 2. Therefore, R is on ℓ. We are done with the proof
of (1). The time we spent developing necessary tools is well
worth it!
Now we will prove (2).
Let S be an arbitrary point on ℓ. If S is either P or Q then there
is nothing to prove.(We know that the points P and Q are on
the graph of y = 2x + 3.) Let the coordinates of S be ( x0 , y0 ).
Let us use the coordinates of P and S to calculate the slope of
ℓ.
y −3
.
slope of ℓ = 0
x0 − 0
That is
2=
y0 − 3
.
x0
GRAPH OF A LINEAR EQUATION IN TWO VARIABLES IS A LINE 45
By the definition of division for real numbers, we get
2x0 = y0 − 3.
By adding 3 to both sides of the above equation, we get
y0 = 2x0 + 3.
Therefore, the point R( x0 , y0 ) is on the graph of y = 2x + 3. We
are done with the proof of (2).
In general we have the following theorem.
Theorem 13.1: The graph of a linear equation y = mx + k is
a non-vertical line with slope m and passing through (0, k),
where m 6= 0 and k are constants.
The proof of the Theorem 13.1 is similar to the proof of Example
13.1.
Proof of Theorem 13.1:
Let P, Q be two distinct points on the graph of y = mx + k. Let
ℓ be the line passing through P and Q.
As mentioned earlier, we will show that
(1) any point on the graph of y = mx + k is on ℓ.
(2) any point on ℓ is on the graph of y = mx + k.
Let us prove (1). Let R be any point on the graph of y = mx + k.
We want to show that R is on ℓ.
If R is either P or Q then there is nothing to prove as both P
and Q are on ℓ. Therefore, assume that R is a point distinct
from both P and Q.
The current situation in our thinking process is given in the
following figure. Let us remind ourselves that we still do not
know that the graph of the equation y = mx + k is a line. (That
is what we are trying to prove.) Therefore, we will draw some
“curve" to represent the graph of y = mx + k.
CCSSM ALGEBRA: LINES
46
y
ℓ
Q
P
graph of
y = mx + k
x
R
Figure 13.2
ℓ′
The points P and Q are on the graph of y = mx + k. Therefore,
the line ℓ passing through P and Q has the slope m, by Theorem
11.1.
Also, the points P and R are on the graph of y = mx + k. Therefore, the line ℓ′ passing through P and Q has the slope m, again
by Theorem 11.1.
That means, the two lines ℓ and ℓ′ pass through the same point
P and has the same slope m. Therefore, ℓ and ℓ′ are the same
line, by Theorem 12.1. Since R is on ℓ′ , R is on ℓ. We are done
with the proof of (1).
Now we will prove (2).
Let S be an arbitrary on ℓ. If S is either P or Q then there is
nothing to prove as both P and Q are on the graph of y = mx +
k. Therefore, assume that S is distinct from P and Q.
y
ℓ
S
Q
P
graph of
y = mx + k
Figure 13.3
x
Let the coordinates of S be (s1 , s2 ) and the coordinates of P be
GRAPH OF A LINEAR EQUATION IN TWO VARIABLES IS A LINE 47
( p1 , p2 ). Since P is on the graph of y = mx + k,
p2 = mp1 + k.
Therefore, the coordinates of P is ( p1, mp1 + k). Since both P
and S lie on ℓ, we can use the coordinates of P and S to find
the slope of ℓ.
slope of ℓ =
That is
m=
s2 − (mp1 + k)
.
s1 − p 1
s2 − (mp1 + k)
.
s1 − p 1
By the definition of division for real numbers, we get
m(s1 − p1 ) = s2 − mp1 − k.
That is:
ms1 − mp1 = s2 − mp1 − k.
By adding mp1 + k to both sides of the above equation, we
get
ms1 − mp1 + mp1 + k = s2 − mp1 − k + mp1 + k.
By collecting like terms:
ms1 + k = s2 .
That is, the point S is on the graph of y = mx + k. We are done
with the proof of (2).
By combining results of Theorems 7.1, 7.3 and 13.1 we get the
following general theorem.
Theorem 13.2: The graph of a linear equation ax + by = c is a
line, where a, b, c are constants and at least one of a and b is
not zero.
48
CCSSM ALGEBRA: LINES
Bibliography
[1] Common Core State Standards for Mathematics 2009:
http://www.corestandards.org/wp-content/uploads/Math_Standards.pdf
[2] Hung-Hsi Wu Introduction to School Algebra 2010:
http://math.berkeley.edu/~wu/Algebrasummary.pdf
[3] Hung-Hsi Wu What is different about Common Core Mathematics Standards: 2011:
http://math.berkeley.edu/~wu/CommonCoreVI.pdf