Elementary Algebra Properties of Lines

Section 3.5 Properties of Lines

Subsection Parallel and Perpendicular Lines

Lines that lie in the same plane but never intersect are called parallel lines. It is easy to understand that parallel lines have the same slope.

Exercise 3.57.

Calculate the slopes of the two parallel lines in the figure. (We denote the slope of the first line by $m_1$ and the slope of the second line by $m_2\text{.}$)

\begin{align*} m_1 \amp= \dfrac{\Delta y}{\Delta x} =\hphantom{00000} \\ m_2 \amp= \dfrac{\Delta y}{\Delta x} = \end{align*}

$parallel lines$

Lines that intersect at right angles are called perpendicular lines. It is a little harder to see the relationship between the slopes of perpendicular lines.

Exercise 3.58.

Calculate the slopes of the two perpendicular lines in the figure.

\begin{align*} m_1 \amp= \dfrac{\Delta y}{\Delta x} =\hphantom{00000} \\ m_2 \amp= \dfrac{\Delta y}{\Delta x} = \end{align*}

$perpendicular lines$

In the Exercise above, note that the product of $m_1$ and $m_2$ is $-1\text{,}$ that is,

\begin{equation*} m_1 m_2 = \dfrac{1}{2} (-2) = -1 \end{equation*}

This relationship holds for any pair of perpendicular lines.

Parallel and Perpendicular Lines.

Two lines are parallel if their slopes are equal, that is, if

\begin{equation*} \blert{m_1 = m_2} \end{equation*}

or if both lines are vertical.
Two lines are perpendicular if the product of their slopes is $-1\text{,}$ that is, if

\begin{equation*} \blert{m_1 m_2 = -1} \end{equation*}

or if one of the lines is horizontal and one is vertical.

QuickCheck 3.59.

What do we call two lines that lie in the same plane but never intersect?

Answer.

Parallel lines

QuickCheck 3.60.

What are perpendicular lines?

Answer.

Lines that intersect at right angles

Example 3.61.

Decide whether the lines

\begin{equation*} 2x+3y=6~~~~~~~~~ \text{and} ~~~~~~~~~3x-2y=6 \end{equation*}

are parallel, perpendicular, or neither.

Solution.

We could graph the lines, but we can’t be sure from a graph if the lines are exactly parallel or exactly perpendicular. A more accurate way to answer the question is to find the slope of each line. To do this we write each equation in slope-intercept form, that is, we solve for $y\text{.}$

\begin{align*} 2x+3y \amp = 6\\ 3y \amp = -2x+6\\ y \amp = \dfrac{-2}{3}x+2 \end{align*}

\begin{align*} 3x-2y \amp = 6\\ -2y \amp = -3x+6\\ y \amp = \dfrac{3}{2}x-3 \end{align*}

The slope of the first line is $\dfrac{-2}{3}\text{,}$ and the slope of the second line is $\dfrac{3}{2}\text{.}$ The slopes are not equal, so the lines are not parallel. However, the product of the slopes is

\begin{equation*} m_1 m_2 = (\dfrac{-2}{3}) (\dfrac{3}{2}) = -1 \end{equation*}

so the lines are perpendicular.

Look Closer.

Another way to state the condition for perpendicular lines is

\begin{equation*} m_2 = \dfrac{-1}{m_1} \end{equation*}

Because of this relationship, we often say that the slope of one perpendicular line is the negative reciprocal of the other.

QuickCheck 3.62.

What is true about the slopes of parallel lines?

Answer.

They are equal.

QuickCheck 3.63.

What do we call the slopes of perpendicular lines?

Answer.

Negative reciprocals

Subsection Equations for Horizontal and Vertical Lines

In Lesson 3.3 we learned that the slope of a horizontal line is zero. What does this tell us about the equation of a horizontal line that passes through a particular point?

Example 3.64.

Find the equation of the horizontal line that passes through $(5,3)\text{.}$

Solution.

Looking at the graph of the line shown at right, we see that the $y$-coordinate of every point on the line is 3.

In particular, the $y$-intercept of the line is the point $(0,3)\text{,}$ so $b=3\text{.}$ As we noted above, the slope of the line is $m=0\text{.}$ Substituting these values into the slope-intercept form gives us the equation

\begin{equation*} y= 0 \cdot x + 3 \end{equation*}

or just $y=3\text{.}$ The fact that $x$ does not appear in the equation means that $y=3$ for every point on the line, no matter what the value of $x$ is.

Look Closer.

What about the equation of a vertical line? The slope of a vertical line is undefined; a vertical line does not have a slope. We cannot use the slope-intercept form to write the equation of a vertical line. However, we can use what we learned about horizontal lines in Example 3.64.

Look at the graph of the vertical line at right. Every point on the line has $x$-coordinate $-2\text{,}$ no matter what the $y$-coordinate is. An equation for this line is $x=-2\text{.}$ The value of $x$ does not depend upon $y\text{;}$ it is constant, so $y$ does not appear in the equation.

Note 3.65.

We make two observations about the examples above.

The $y$-intercept of the horizontal line $y=3$ is $(0,3)\text{;}$ it has no $x$-intercept.
The $x$-intercept of the vertical line $x=-2$ is $(-2,0)\text{;}$ it has no $y$-intercept.

Horizontal and Vertical Lines.

The equation of the horizontal line passing through $(0,b)$ is

\begin{equation*} \blert{y=b} \end{equation*}
The equation of the vertical line passing through $(a,0)$ is

\begin{equation*} \blert{x=a} \end{equation*}

Caution 3.66.

The equation for a line is not the same thing as the slope of the line!

The slope of every horizontal line is zero, but the equation of a horizontal line has the form $y=b\text{,}$ where $b$ is the $y$-coordinate of every point on the line.
Similarly, the equation of a vertical line has the form $x=a\text{,}$ but the slope of a vertical line is undefined.

QuickCheck 3.67.

Give an example of an equation of a vertical line.

Answer.

$x=k$

QuickCheck 3.68.

Give an example of an equation of a horizontal line.

Answer.

$y=k$

Subsection Distance Between Points

It is easy to compute the distance between two points that lie on the same horizontal or vertical line. We subtract the smaller coordinate from the larger one.

Example 3.69.

The distance between the points $A(6, \blert{5})$ and $B(6, \blert{-7})$ in the figure is

\begin{equation*} \Delta y = 5-(-7)=12 \end{equation*}

and the distance between points $C(\blert{-7}, -3)$ and $D(\blert{-2}, -3)$ is

\begin{equation*} \Delta x = -2-(-7)=5 \end{equation*}

$a vertical segment and a horizontal segment$

Note that the distance between two points is always a positive number. That is why we always subtract the smaller coordinate from the larger one to compute distance. However, when we compute a slope, we need more information.

Look Closer.

When we compute slope, the direction in which we move on the line makes a difference. We will call this the directed distance, and it can be either positive or negative.

If we move from $C$ to $D\text{,}$ we have moved in the positive $x$-direction, so the directed distance is positive:

\begin{equation*} \Delta x = -2-(-7)=5 \end{equation*}

If we move from $D$ to $C\text{,}$ we have moved in the negative $x$-direction, so the directed distance is negative:

\begin{equation*} \Delta x = -7-(-2)=-5 \end{equation*}

Directed Distance.

To find the directed distance between two points on a number line, we subtract the initial coordinate from the final coordinate.

QuickCheck 3.70.

How do we find the distance between two points that lie on the same vertical line?

Answer.

Subtract the smaller $y$-coordinate from the larger one

QuickCheck 3.71.

A directed distance can be either or .

Answer.

positive, negative

Look Ahead.

So far we have computed the slope of a line by finding $\Delta x$ and $\Delta y$ on the graph. Now we can use directed distance to develop a formula for slope.

Example 3.72.

Compute the slope of the line segment joining $P$ and $R$ in two ways:

Find $\Delta y$ and $\Delta x$ using the graph
Find $\Delta y$ and $\Delta x$ using coordinates.

$graph$

Solution.

As we move from $P$ to $Q\text{,}$ we move up 2 squares on the graph, so $\Delta y = 2\text{.}$ As we move from $Q$ to $R\text{,}$ we move 8 squares to the right, so $\Delta x = 8\text{.}$ Thus, the slope of the line is

\begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{2}{8} = \dfrac{1}{4} \end{equation*}
First we write down the coordinates of $P$ and $Q\text{:}$

\begin{equation*} P(-3,\blert{-4})~~~~~~\text{and}~~~~~~Q(-3,\blert{-2}) \end{equation*}

and compute the directed distance from $P$ to $Q\text{:}$

\begin{equation*} \Delta y = -2-(-4) = 2~~~~~~~~~~~~~~~~~~\blert{\text{final} - \text{initial}} \end{equation*}

Then we write down the coordinates of $Q$ and $R\text{:}$

\begin{equation*} Q(\blert{-3},-2)~~~~~~\text{and}~~~~~~R(\blert{5},-2) \end{equation*}

and compute the directed distance from $Q$ to $R\text{:}$

\begin{equation*} \Delta x = 5-(-3) = 8~~~~~~~~~~~~~~~~~~\blert{\text{final} - \text{initial}} \end{equation*}

We get the same value for the slope as in part (a),

\begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{2}{8} = \dfrac{1}{4} \end{equation*}

Subsection Subscript Notation

When we compute the slope of a line joining two points, we must be careful to subtract their $x$-coordinates and their $y$-coordinates in the same order, final minus initial. To distinguish between the coordinates of different points, we use a new notation called subscripts.

Subscripts.

We denote the coordinates of a particular point by $(x_1,y_1)$ and the coordinates of a second point by $(x_2,y_2)\text{.}$

Example 3.73.

Compute the slope between point $D$ and point $F$ in the figure.

$graph$

Solution.

We’ll call $D$ the first point and $F$ the second point. Then their coordinates are

\begin{align*} D:~~~x_1 \amp =-6 \amp\amp \text{and}\amp y_1 \amp =-2\\ F:~~~x_2 \amp = 9 \amp\amp \text{and}\amp y_2 \amp =4 \end{align*}

Now we can describe the formula for slope in an organized way. We compute $\Delta y$ and $\Delta x$ as directed distances. First, we observe that the coordinates of point $E$ are $(-6,4)$ or $(x_1,y_2)\text{.}$ From the figure, we can see that

\begin{equation*} \Delta y = y_2-y_1 = 4-(-2)=6 \end{equation*}

and

\begin{equation*} \Delta x = x_2-x_1 = 9-(-6)=15 \end{equation*}

Thus, the slope of the line segment joining $D$ and $F$ is

\begin{equation*} m=\dfrac{y_2-y_1}{x_2-x_1} = \dfrac{6}{15} = \dfrac{2}{5} \end{equation*}

Caution 3.74.

The subscript 1 on $x_1\text{,}$ for instance, has nothing to do with the value of the coordinate; it merely identifies this coordinate as the $x$-coordinate of the first point.

QuickCheck 3.75.

What do the subscripts in $(x_1,y_1)$ mean?

Answer.

Coordinates of the first point

QuickCheck 3.76.

Write a formula for computing $\Delta y\text{.}$

Answer.

$\Delta y= y_2-y_1$

Subsection A New Formula for Slope

The method described in Example 3.73 gives us a new formula for computing slope.

Two-Point Formula for Slope.

The slope of the line joining points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ is

\begin{equation*} \blert{m=\dfrac{y_2-y_1}{x_2-x_1}}~~~~~~\text{if}~~~~x_2 \not= x_1 \end{equation*}

We don’t need a graph in order to use this formula, just the coordinates of two points on the line.

Example 3.77.

Compute the slope of the line joining the points $(6,-2)$ and $(3,-1)\text{.}$

Solution.

It doesn’t matter which point is $P_1$ and which is $P_2\text{,}$ so we choose $P_1$ to be $(6,-2)\text{.}$ Then $(x_1,y_1)=(6,-2)$ and $(x_2,y_2)=(3,-1)\text{.}$ Thus,

\begin{align*} m \amp = \dfrac{y_2-y_1}{x_2-x_1}\\ \amp = \dfrac{-1-2}{3-(-6)} = \dfrac{-3}{9} = \dfrac{-1}{3} \end{align*}

Caution 3.78.

In Example 3.77, we can reverse the order of both subtractions to find

\begin{align*} m \amp = \dfrac{y_1-y_2}{x_1-x_2}\\ \amp = \dfrac{2-(-1)}{-6-3)} = \dfrac{3}{-9} = \dfrac{-1}{3} \end{align*}

the same answer as before. The order of the points does not matter, but we must be consistent and use the same order when computing $\Delta y$ and $\Delta x\text{.}$

QuickCheck 3.79.

What is wrong with this formula for slope: $m = \dfrac{y_2-y_1}{x_1-x_2}\text{?}$

Answer.

The $x$- and $y$-coordinates should be subtracted in the same order.

Skills Warm-Up 3.80.

For Problems 1-8, find the negative reciprocal.
1. $\displaystyle \dfrac{3}{4}$
2. $\displaystyle \dfrac{-1}{3}$
3. $\displaystyle -6$
4. $\displaystyle \dfrac{11}{8}$
5. $\displaystyle 1\dfrac{2}{3}$
6. $\displaystyle -2\dfrac{1}{2}$
7. $\displaystyle -3.2$
8. $\displaystyle 0.625$
For Problems 9-14, simplify.
1. $\displaystyle \dfrac{10-2}{2-9}$
2. $\displaystyle \dfrac{-5(-5)}{2-8}$
3. $\displaystyle \dfrac{3}{2}(4-7)+\dfrac{1}{2}$
4. $\displaystyle \dfrac{-6-(-12)}{3-(-5)}$
5. $\displaystyle -3(-4-2)-6$
6. $\displaystyle \dfrac{5}{3}(5-8)+3$

Answer.

1. $\displaystyle \dfrac{-4}{3}$
2. $\displaystyle 3$
3. $\displaystyle \dfrac{1}{6}$
4. $\displaystyle \dfrac{-8}{11}$
5. $\displaystyle \dfrac{-3}{5}$
6. $\displaystyle \dfrac{2}{5}$
7. $\displaystyle 0.3125$
8. $\displaystyle -1.6$
1. $\displaystyle \dfrac{-4}{3}$
2. $\displaystyle 0$
3. $\displaystyle -4$
4. $\displaystyle \dfrac{3}{4}$
5. $\displaystyle 12$
6. $\displaystyle -2$

Subsection Lesson

Activity 3.15. Parallel and Perpendicular Lines.

Put each equation into slope-intercept form.
1. $\displaystyle 2x-3y+3$
2. $\displaystyle 2x+3y-6 = 0$
3. $\displaystyle 3x+2y-2$
4. $\displaystyle 2x-3y-2 = 0$
Which of the four lines in question (1) are parallel? How do you know?
Which of the four lines in question (1) are perpendicular? How do you know?

Activity 3.16. Horizontal and Vertical Lines.

Sketch a graph of the vertical line passing through $(-4,-1)\text{,}$ then find its equation.

What is the slope of the line?
Sketch a graph of the horizontal line passing through $(-4,-1)\text{,}$ then find its equation.

What is the slope of the line?

Activity 3.17. Slope.

Compute the slope of the line segment joining $A$ and $C$ in two ways:

$line segment on grid$
1. Using the graph.
  
  Draw the line through $A$ and $C\text{.}$ Use point $B$ to find $\Delta y$ and $\Delta x\text{.}$
  
  $\Delta y = \fillinmath{XXXXXX},~~~~ \Delta x = \fillinmath{XXXXXX}$
  
  $m = \dfrac{\Delta y}{\Delta x} = $
2. Using coordinates.
  - Step 1 Write down the coordinates of $A~ \fillinmath{XXXXXX}$
    
    $\hphantom{000000}$ Write down the coordinates of $B~ \fillinmath{XXXXXX}$
  - Step 2 Compute $\Delta y$ and $\Delta x\text{.}$
    
    \begin{gather*} \Delta x~ =~ \text{final} - \text{initial}~ =~ \fillinmath{XXXXXX}\\ \Delta y~ = ~\text{final} - \text{initial}~ = ~\fillinmath{XXXXXX} \end{gather*}
  - Step 3 Compute the slope: $~~m = \dfrac{\Delta y}{\Delta x} = $
3. Do you get the same answers for parts (a) and (b)? You should!
Follow the steps to compute the slope of the line segment joining $H$ and $K\text{.}$

$line segment on grid$
1. Step 1 Let $H$ be the first point and $K$ the second point. Write down their coordinates.
  
  \begin{gather*} H(x_1,y_1) = \\ K(x_2,y_2) = \end{gather*}
2. Step 2 Fill in the blanks:
  
  \begin{gather*} y_2 = \fillinmath{XXXXXX},~~y_1 = \fillinmath{XXXXXX}\\ x_2 = \fillinmath{XXXXXX},~~x_1 = \fillinmath{XXXXXX} \end{gather*}
3. Step 3 Compute $\Delta y$ and $\Delta x\text{.}$
  
  \begin{gather*} \Delta y~ =~ y_x-y_1~ =~ \fillinmath{XXXXXX}\\ \Delta x~ = ~x_2-x_1~ = ~\fillinmath{XXXXXX} \end{gather*}
4. Step 4 Compute the slope: $~~m = \dfrac{y_2-y_1}{x_2-x_1} = $
5. Illustrate $\Delta y$ and $\Delta x$ on the graph. Is your value for the slope reasonable?
Use the formula $~~m = \dfrac{y_2-y_1}{x_2-x_1}$ to compute the slope of the line joining the points $(-4,-7)$ and $(2,-3)\text{.}$

Subsubsection Wrap-Up

Objectives.

In this Lesson we practiced the following skills:

Finding equations for parallel or perpendicular lines
Finding equations for horizontal or vertical lines
Using the slope formula

Questions.

How can you decide if two lines are parallel, perpendicular, or neither?
How many points do you need to find the equation of a horizontal or a vertical line?
Explain the difference between the equation of a horizontal or vertical line and the slope of the line.
Explain why the two-point formula for slope is the same as our old formula, $~~m = \dfrac{\Delta y}{\Delta x}.$

Activity 3.18. Homework Preview.

Are the lines parallel, perpendicular, or neither?

\begin{equation*} 3x-4y=6~~~~~~~~\text{and}~~~~~~~~4x-3y=2 \end{equation*}
Sketch a graph of each equation, label the coordinates of its intercept, and state the slope of the line.
1. $\displaystyle y=-6$
2. $\displaystyle x=-8$
Find the slope of the line through $(-2,3)$ and $(1,-5)\text{.}$ Graph the line.

$m = $
Find the slope of the line through $(5,-4)$ and $(-2,-4)\text{.}$ Graph the line.

$m = $

Answers to Homework Preview

Neither
1. $\displaystyle (0,-6),~m = 0$
2. $(-2,0),~m~ $ is undefined
$\displaystyle m=\dfrac{-8}{3}$
$~m~ $ is undefined

Exercises Homework 3.5

Exercise Group.

For Problems 1–3, decide whether the lines are parallel perpendicular, or neither.

1.

\begin{gather*} 3x-4y=2\\ 8y-6x=6 \end{gather*}

2.

\begin{gather*} 3x-4y=2\\ 8y-6x=6 \end{gather*}

3.

\begin{gather*} 2x=4-5y\\ 2y=4x-5 \end{gather*}

4.

The slopes of several lines are given below. Which of the lines are parallel to the graph of $y=0.75x+2\text{,}$ and which are perpendicular to it?

$\displaystyle \dfrac{3}{4}$
$\displaystyle \dfrac{8}{6}$
$\displaystyle \dfrac{-20}{15}$
$\displaystyle \dfrac{-39}{52}$
$\displaystyle \dfrac{4}{3}$
$\displaystyle \dfrac{-16}{12}$
$\displaystyle \dfrac{36}{48}$
$\displaystyle \dfrac{9}{12}$

Exercise Group.

For Problems 5–7,

Sketch a rough graph of the equation, and label its intercept.
State the slope of the line.

5.

$y=-3$

6.

$2x=8$

7.

$x=0$

Exercise Group.

For Problems 8–12, give the equation of the line described.

8.

Horizontal, passes through $(6,-5)$

9.

$m$ is undefined, passes through $(2,1)$

10.

Parallel to $x=4\text{,}$ passes through $(-8,3)$

11.

The $x$-axis

12.

Perpendicular to the $y$-axis, passes through $(4,-12)$

Exercise Group.

For Problems 13–16, state the $y$-intercept and the slope of the line.

13.

$y=-6$

14.

$y=-6x$

15.

$x=-6y$

16.

$x=6-y$

Exercise Group.

For Problems 17–20, compute the slope of the line joining the given points.

17.

$(5,2),~(8,7)$

18.

$(3,-2),~(0,1)$

19.

$(6,-2),~(-3,-3)$

20.

$(3,-5),~(8,-5)$

Exercise Group.

For Problems 21–23,

Find the $x$- and $y$-intercepts of each line.
Use the intercepts to compute the slope of the line.

21.

$\dfrac{x}{5}+\dfrac{y}{7}=1$

22.

$\dfrac{-x}{2.4}+\dfrac{y}{1.6}=1$

23.

$3x+\dfrac{2}{7}y=1$

Exercise Group.

For Problems 24–25, you are given the slope of a line and one point on it. Use the grids to help you find the missing coordinates of the other points on the line.

24.

$m=2,~~(1,-1)$

$\displaystyle (0,?)$
$\displaystyle (?,-7)$

25.

$m=\dfrac{-1}{2},~~(2,-6)$

$\displaystyle (6,?)$
$\displaystyle (?,-3)$

Exercise Group.

For Problems 26–29, calculate the slope of the line described, including units, and explain its meaning for the problem.

26.

Rani and Larry start a college fund for their son, Colby. When Colby is five years old, the fund has $11,000. When he is 10 years old, the fund has grown to $17,000. The graph of the amount, $A\text{,}$ in the fund when Colby is $n$ years old is a straight line.

27.

It cost Melanie $6.88 to copy and bind her 26-page paper at CopyQuik. It cost Nelson $8.80 to copy and bind his 50-page portfolio. The graph of the cost, $C\text{,}$ of producing a document of length $p$ pages at CopyQuik is a straight line.

28.

Dean is cycling down Mt. Whitney. After five minutes, his elevation is 9200 feet, and after 21 minutes, his elevation is 1200 feet. The graph of Dean’s elevation, $h\text{,}$ after cycling for $t$ minutes is approximately a straight line.

29.

At 2 pm the temperature was $16 \degree\text{,}$ and by midnight it had dropped to $14 \degree\text{.}$ The graph of temperature, $T\text{,}$ at $h$ hours after noon is approximately a straight line.

30.

The figure shows the wholesale cost, $C\text{,}$ in dollars, of $p$ pounds of dry-roasted peanuts.

$graph of line$

What does the slope of this graph tell you about dry-roasted peanuts?
Compute the slope, including units.
If the Lone Star Barbecue increases its weekly order of peanuts from 100 to 120 pounds, how much will the cost increase?

31.

Roy is traveling home by train. The figure shows his distance $d$ from home in miles $h$ hours after the train started.

$graph of line$

Compute the slope of the graph, including units.
What does the slope tell you about Roy’s journey?
Find the vertical intercept of the graph. What is its meaning for the problem?
Compute the horizontal intercept of the graph. What is its meaning for the problem?

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