## Section3.5Properties of Lines

### SubsectionParallel 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.

###### Exercise3.23.

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*}

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.

###### Exercise3.24.

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*}

In the Checkpoint 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.
1. 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.

2. 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.

###### 1.

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

###### 2.

What are perpendicular lines?

###### Example3.25.

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

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 $m_2 = \dfrac{-1}{m_1}$ Because of this relationship, we often say that the slope of one perpendicular line is the negative reciprocal of the other.

###### 3.

What is true about the slopes of parallel lines?

###### 4.

What do we call the slopes of perpendicular lines?

### SubsectionEquations 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?

###### Example3.26.

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

Solution

Looking at the graph of the line shown below, 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 3.26.

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.

###### Note3.27.

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

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

\begin{equation*} \blert{x=a} \end{equation*}
###### Caution3.28.

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.

###### 5.

Give an example of an equation of a vertical line.

###### 6.

Give an example of an equation of a horizontal line.

### SubsectionDistance 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.

###### Example3.29.

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*}

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*}

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

###### 7.

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

###### 8.

A directed distance can be either or .

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.

###### Example3.30.

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

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

2. Find $\Delta y$ and $\Delta x$ using coordinates.

Solution
1. 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*}
2. 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*}

### SubsectionSubscript 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.

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

###### Example3.31.

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

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*}
###### Caution3.32.

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.

###### 9.

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

###### 10.

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

### SubsectionA New Formula for Slope

The method described in Example 3.31 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.

###### Example3.33.

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{equation*} \begin{aligned} m \amp = \dfrac{y_2-y_1}{x_2-x_1} \\ \amp = \dfrac{-1-2}{3-(-6)} = \dfrac{-3}{9} = \dfrac{-1}{3} \end{aligned} \end{equation*}
###### Caution3.34.

In Example 3.33, we can reverse the order of both subtractions to find

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

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$\Delta x\text{.}$

###### 11.

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

### SubsectionSkills Warm-Up

#### ExercisesExercises

For Problems 1-8, find the negative reciprocal.

###### 1.
$\dfrac{3}{4}$
###### 2.
$\dfrac{-1}{3}$
###### 3.
$-6$
###### 4.
$\dfrac{11}{8}$
###### 5.
$1\dfrac{2}{3}$
###### 6.
$-2\dfrac{1}{2}$
###### 7.
$-3.2$
###### 8.
$0.625$

For Problems 9-14, simplify.

###### 9.
$\dfrac{10-2}{2-9}$
###### 10.
$\dfrac{-5(-5)}{2-8}$
###### 11.
$\dfrac{3}{2}(4-7)+\dfrac{1}{2}$
###### 12.
$\dfrac{-6-(-12)}{3-(-5)}$
###### 13.
$-3(-4-2)-6$
###### 14.
$\dfrac{5}{3}(5-8)+3$

### ExercisesHomework 3.5

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?

1. $\dfrac{3}{4}$
2. $\dfrac{8}{6}$
3. $\dfrac{-20}{15}$
4. $\dfrac{-39}{52}$
5. $\dfrac{4}{3}$
6. $\dfrac{-16}{12}$
7. $\dfrac{36}{48}$
8. $\dfrac{9}{12}$

For Problems 5–7,

1. Sketch a rough graph of the equation, and label its intercept.
2. State the slope of the line.
###### 5.
$y=-3$
###### 6.
$2x=8$
###### 7.
$x=0$

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 $(4,-12)$

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

###### 13.
$y=-6$
###### 14.
$y=-6x$
###### 15.
$x=-6y$
###### 16.
$x=6-y$

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

For Problems 21–23,

1. Find the $x$- and $y$-intercepts of each line.
2. 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.
$3x+\dfrac{2}{7}y=1$

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.

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

1. $(0,?)$
2. $(?,-7)$
###### 25.

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

1. $(6,?)$
2. $(?,-3)$

For Problems 26–29, 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 linecalculate 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.

1. What does the slope of this graph tell you about dry-roasted peanuts?
2. Compute the slope, including units.
3. 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.

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