# Open Resources for Community College Algebra

## Section3.10Graphing Lines Chapter Review

### Subsection3.10.1Cartesian Coordinates

In Section 1 we covered the definition of the Cartesian Coordinate System and how to plot points using the $$x$$- and $$y$$-axes.

#### Example3.10.1.

On paper, sketch a Cartesian coordinate system with units, and then plot the following points: $$(3,2),(-5,-1),(0,-3),(4,0)\text{.}$$
Explanation.

### Subsection3.10.2Graphing Equations

In Section 2 we covered how to plot solutions to equations to produce a graph of the equation.

#### Example3.10.3.

Graph the equation $$y=-2x+5\text{.}$$
Explanation.
We use points from the table to graph the equation. First, plot each point carefully. Then, connect the points with a smooth curve. Here, the curve is a straight line. Lastly, we can communicate that the graph extends further by sketching arrows on both ends of the line.

### Subsection3.10.3Exploring Two-Variable Data and Rate of Change

In Section 3 we covered how to find patterns in tables of data and how to calculate the rate of change between points in data.

#### Example3.10.6.

Write an equation in the form $$y=\ldots$$ suggested by the pattern in the table.
Explanation.
We consider how the values change from one row to the next. From row to row, the $$x$$-value increases by $$1\text{.}$$ Also, the $$y$$-value decreases by $$2$$ from row to row.
 $$x$$ $$y$$ $$0$$ $$-4$$ $${}+1\rightarrow$$ $$1$$ $$-6$$ $$\leftarrow{}-2$$ $${}+1\rightarrow$$ $$2$$ $$-8$$ $$\leftarrow{}-2$$ $${}+1\rightarrow$$ $$3$$ $$-10$$ $$\leftarrow{}-2$$
Since row-to-row change is always $$1$$ for $$x$$ and is always $$-2$$ for $$y\text{,}$$ the rate of change from one row to another row is always the same: $$-2$$ units of $$y$$ for every $$1$$ unit of $$x\text{.}$$
We know that the output for $$x = 0$$ is $$y = -4\text{.}$$ And our observation about the constant rate of change tells us that if we increase the input by $$x$$ units from $$0\text{,}$$ the output should decrease by $$\overbrace{(-2)+(-2)+\cdots+(-2)}^{x\text{ times}}\text{,}$$ which is $$-2x\text{.}$$ So the output would be $$-4-2x\text{.}$$
So the equation is $$y=-2x-4\text{.}$$

### Subsection3.10.4Slope

In Section 4 we covered the definition of slope and how to use slope triangles to calculate slope. There is also the slope formula which helps find the slope through any two points.

#### Example3.10.8.

Find the slope of the line in the following graph.
Explanation.
We picked two points on the line, and then drew a slope triangle. Next, we will do:
\begin{equation*} \text{slope}=\frac{12}{3}=4 \end{equation*}
The line’s slope is $$4\text{.}$$

#### Example3.10.11.Finding a Line’s Slope by the Slope Formula.

Use the slope formula to find the slope of the line that passes through the points $$(-5,25)$$ and $$(4,-2)\text{.}$$
Explanation.
\begin{align*} \text{slope}\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{-2-(25)}{4-(-5)}\\ \amp=\frac{-27}{9}\\ \amp=-3 \end{align*}
The line’s slope is $$-3\text{.}$$

### Subsection3.10.5Slope-Intercept Form

In Section 5 we covered the definition of slope intercept-form and both wrote equations in slope-intercept form and graphed lines given in slope-intercept form.

#### Example3.10.12.

Graph the line $$y=-\frac{5}{2}x+4\text{.}$$
Explanation.

#### Writing a Line’s Equation in Slope-Intercept Form Based on Graph.

Given a line’s graph, we can identify its $$y$$-intercept, and then find its slope using a slope triangle. With a line’s slope and $$y$$-intercept, we can write its equation in the form of $$y=mx+b\text{.}$$

#### Example3.10.14.

Find the equation of the line in the graph.
Explanation.
With the line’s slope $$-\frac{2}{3}$$ and $$y$$-intercept $$10\text{,}$$ we can write the line’s equation in slope-intercept form: $$y=-\frac{2}{3}x+10\text{.}$$

### Subsection3.10.6Point-Slope Form

In Section 6 we covered the definition of point-slope form and both wrote equations in point-slope form and graphed lines given in point-slope form.

#### Example3.10.18.

A line passes through $$(-6,0)$$ and $$(9,-10)\text{.}$$ Find this line’s equation in point-slope form. .
Explanation.
We will use the slope formula to find the slope first. After labeling those two points as $$(\overset{x_1}{-6},\overset{y_1}{0}) \text{ and } (\overset{x_2}{9},\overset{y_2}{-10})\text{,}$$ we have:
\begin{align*} \text{slope}\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{-10-0}{9-(-6)}\\ \amp=\frac{-10}{15}\\ \amp=-\frac{2}{3} \end{align*}
Now the point-slope equation looks like $$y=-\frac{2}{3}(x-x_0)+y_0\text{.}$$ Next, we will use $$(9,-10)$$ and substitute $$x_0$$ with $$9$$ and $$y_0$$ with $$-10\text{,}$$ and we have:
\begin{align*} y\amp=-\frac{2}{3}(x-x_0)+y_0\\ y\amp=-\frac{2}{3}(x-9)+(-10)\\ y\amp=-\frac{2}{3}(x-9)-10 \end{align*}

### Subsection3.10.7Standard Form

In Section 7 we covered the definition of standard form of a linear equation. We converted equations from standard form to slope-intercept form and vice versa. We also graphed lines from standard form by finding the intercepts of the line.

#### Example3.10.19.

1. Convert $$2x+3y=6$$ into slope-intercept form.
2. Convert $$y=-\frac{4}{7}x-3$$ into standard form.
Explanation.
1. \begin{align*} 2x+3y\amp=6\\ 2x+3y\subtractright{2x}\amp=6\subtractright{2x}\\ 3y\amp=-2x+6\\ \divideunder{3y}{3}\amp=\divideunder{-2x+6}{3}\\ y\amp=\frac{-2x}{3}+\frac{6}{3}\\ y\amp=-\frac{2}{3}x+2 \end{align*}
The line’s equation in slope-intercept form is $$y=-\frac{2}{3}x+2\text{.}$$
The line’s equation in standard form is $$4x+7y=-21\text{.}$$
To graph a line in standard form, we could first change it to slope-intercept form, and then graph the line by its $$y$$-intercept and slope triangles. A second method is to graph the line by its $$x$$-intercept and $$y$$-intercept.

#### Example3.10.20.

Graph $$2x-3y=-6$$ using its intercepts. And then use the intercepts to calculate the line’s slope.
Explanation.
We calculate the line’s $$x$$-intercept by substituting $$y=0$$ into the equation
\begin{align*} 2x-3y\amp=-6\\ 2x-3(\substitute{0})\amp=-6\\ 2x\amp=-6\\ x\amp=-3 \end{align*}
So the line’s $$x$$-intercept is $$(-3,0)\text{.}$$
Similarly, we substitute $$x=0$$ into the equation to calculate the $$y$$-intercept:
\begin{align*} 2x-3y\amp=-6\\ 2(\substitute{0})-3y\amp=-6\\ -3y\amp=-6\\ y\amp=2 \end{align*}
So the line’s $$y$$-intercept is $$(0,2)\text{.}$$
With both intercepts’ coordinates, we can graph the line:
Now that we have graphed the line we can read the slope. The rise is $$2$$ units and the run is $$3$$ units so the slope is $$\frac{2}{3}\text{.}$$

### Subsection3.10.8Horizontal, Vertical, Parallel, and Perpendicular Lines

In Section 8 we studied horizontal and vertical lines. We also covered the relationships between the slopes of parallel and perpendicular lines.

#### Example3.10.22.

Line $$m$$’s equation is $$y=-2x+20\text{.}$$ Line $$n$$ is parallel to $$m\text{,}$$ and line $$n$$ also passes the point $$(4,-3)\text{.}$$ Find an equation for line $$n$$ in point-slope form.
Explanation.
Since parallel lines have the same slope, line $$n$$’s slope is also $$-2\text{.}$$ Since line $$n$$ also passes the point $$(4,-3)\text{,}$$ we can write line $$n$$’s equation in point-slope form:
\begin{align*} y\amp=m(x-x_1)+y_1\\ y\amp=-2(x-4)+(-3)\\ y\amp=-2(x-4)-3 \end{align*}
Two lines are perpendicular if and only if the product of their slopes is $$-1\text{.}$$

#### Example3.10.23.

Line $$m$$’s equation is $$y=-2x+20\text{.}$$ Line $$n$$ is perpendicular to $$m\text{,}$$ and line $$q$$ also passes the point $$(4,-3)\text{.}$$ Find an equation for line $$q$$ in slope-intercept form.
Explanation.
Since line $$m$$ and $$q$$ are perpendicular, the product of their slopes is $$-1\text{.}$$ Because line $$m$$’s slope is given as $$-2\text{,}$$ we can find line $$q$$’s slope is $$\frac{1}{2}\text{.}$$
Since line $$q$$ also passes the point $$(4,-3)\text{,}$$ we can write line $$q$$’s equation in point-slope form:
\begin{align*} y\amp=m(x-x_1)+y_1\\ y\amp=\frac{1}{2}(x-4)+(-3)\\ y\amp=\frac{1}{2}(x-4)-3 \end{align*}
We can now convert this equation to slope-intercept form:
\begin{align*} y\amp=\frac{1}{2}(x-4)-3\\ y\amp=\frac{1}{2}x-2-3\\ y\amp=\frac{1}{2}x-5 \end{align*}

### Exercises3.10.9Exercises

#### Exercise Group.

##### 1.
Sketch the points $$(8,2)\text{,}$$ $$(5,5)\text{,}$$ $$(-3,0)\text{,}$$ $$\left(0,-\frac{14}{3}\right)\text{,}$$ $$(3,-2.5)\text{,}$$ and $$(-5,7)$$ on a Cartesian plane.
##### 2.
Locate each point in the graph:
Write each point’s position as an ordered pair, like $$(1,2)\text{.}$$
 $$A=$$ $$B=$$ $$C=$$ $$D=$$
##### 3.
Consider the equation
$$y=-\frac{5}{8} x-2$$
Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.
• $$\displaystyle (0,-2)$$
• $$\displaystyle (-16,8)$$
• $$\displaystyle (-8,4)$$
• $$\displaystyle (16,-7)$$
##### 4.
Consider the equation
$$y=-\frac{3}{8} x-4$$
Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.
• $$\displaystyle (-8,3)$$
• $$\displaystyle (0,-4)$$
• $$\displaystyle (-32,8)$$
• $$\displaystyle (40,-14)$$
##### 5.
Write an equation in the form $$y=\ldots$$ suggested by the pattern in the table.
 $$x$$ $$y$$ $$0$$ $${-4}$$ $$1$$ $${-2}$$ $$2$$ $${0}$$ $$3$$ $${2}$$
##### 6.
Write an equation in the form $$y=\ldots$$ suggested by the pattern in the table.
 $$x$$ $$y$$ $$0$$ $${3}$$ $$1$$ $${-2}$$ $$2$$ $${-7}$$ $$3$$ $${-12}$$

#### Exercise Group.

##### 7.
Find the slope of the line.
##### 8.
Find the slope of the line.
##### 9.
Find the slope of the line.
##### 10.
Below is a line’s graph.
The slope of this line is .
##### 11.
Below is a line’s graph.
The slope of this line is .
##### 12.
Below is a line’s graph.
The slope of this line is .

#### Exercise Group.

##### 13.
A line passes through the points $$(-4,{18})$$ and $$(6,{-27})\text{.}$$ Find this line’s slope.
##### 14.
A line passes through the points $$(-8,{-4})$$ and $$(8,{-10})\text{.}$$ Find this line’s slope.
##### 15.
A line passes through the points $$(2,-7)$$ and $$(-4,-7)\text{.}$$ Find this line’s slope.
##### 16.
A line passes through the points $$(5,-5)$$ and $$(-2,-5)\text{.}$$ Find this line’s slope.
##### 17.
A line passes through the points $$(-2,-2)$$ and $$(-2,1)\text{.}$$ Find this line’s slope.
##### 18.
A line passes through the points $$(0,-4)$$ and $$(0,3)\text{.}$$ Find this line’s slope.
##### 19.
A line’s graph is given. What is this line’s slope-intercept equation?
##### 20.
A line’s graph is given. What is this line’s slope-intercept equation?
##### 21.
Find the line’s slope and $$y$$-intercept.
A line has equation $$\displaystyle{ 3x-7y=21 }\text{.}$$
This line’s slope is .
This line’s $$y$$-intercept is .
##### 22.
Find the line’s slope and $$y$$-intercept.
A line has equation $$\displaystyle{ 3x-8y=40 }\text{.}$$
This line’s slope is .
This line’s $$y$$-intercept is .
##### 23.
A line passes through the points $$(6,{4})$$ and $$(-6,{2})\text{.}$$ Find this line’s equation in point-slope form.
Using the point $$(6,{4})\text{,}$$ this line’s point-slope form equation is .
Using the point $$(-6,{2})\text{,}$$ this line’s point-slope form equation is .
##### 24.
A line passes through the points $$(-3,{-11})$$ and $$(3,{-7})\text{.}$$ Find this line’s equation in point-slope form.
Using the point $$(-3,{-11})\text{,}$$ this line’s point-slope form equation is .
Using the point $$(3,{-7})\text{,}$$ this line’s point-slope form equation is .

#### 25.

Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose $$8.6$$ grams. Seven minutes since the experiment started, the remaining gas had a mass of $$301$$ grams.
Let $$x$$ be the number of minutes that have passed since the experiment started, and let $$y$$ be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.
1. This line’s slope-intercept equation is .
2. $$37$$ minutes after the experiment started, there would be grams of gas left.
3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.

#### 26.

Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose $$5.5$$ grams. Ten minutes since the experiment started, the remaining gas had a mass of $$187$$ grams.
Let $$x$$ be the number of minutes that have passed since the experiment started, and let $$y$$ be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.
1. This line’s slope-intercept equation is .
2. $$37$$ minutes after the experiment started, there would be grams of gas left.
3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.

#### 27.

Find the $$y$$-intercept and $$x$$-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
\begin{equation*} 4 x + 3 y = -12 \end{equation*}
 $$x$$-value $$y$$-value Location (as an ordered pair) $$y$$-intercept $$x$$-intercept
 $$x$$-intercept and $$y$$-intercept of the line $$4 x+3 y=-12$$

#### 28.

Find the $$y$$-intercept and $$x$$-intercept of the line given by the equation. If a particular intercept does not exist, enter none into all the answer blanks for that row.
\begin{equation*} 6 x + 7 y = -126 \end{equation*}
 $$x$$-value $$y$$-value Location (as an ordered pair) $$y$$-intercept $$x$$-intercept
 $$x$$-intercept and $$y$$-intercept of the line $$6 x+7 y=-126$$

#### Exercise Group.

##### 29.
Find the line’s slope and $$y$$-intercept.
A line has equation $$\displaystyle{ -{8}x+y= -1 }\text{.}$$
This line’s slope is .
This line’s $$y$$-intercept is .
##### 30.
Find the line’s slope and $$y$$-intercept.
A line has equation $$\displaystyle{ -x-y= 7 }\text{.}$$
This line’s slope is .
This line’s $$y$$-intercept is .
##### 31.
Find the line’s slope and $$y$$-intercept.
A line has equation $$\displaystyle{ 24x+20y=3 }\text{.}$$
This line’s slope is .
This line’s $$y$$-intercept is .
##### 32.
Find the line’s slope and $$y$$-intercept.
A line has equation $$\displaystyle{ 3x+12y=1 }\text{.}$$
This line’s slope is .
This line’s $$y$$-intercept is .
##### 33.
Fill out this table for the equation $$x=-9\text{.}$$ The first row is an example.
 $$x$$ $$y$$ Points $$-9$$ $$-3$$ $$\left(-9,-3\right)$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$
 Values of $$x$$ and $$y$$ satisfying the equation $$x=-9$$
##### 34.
Fill out this table for the equation $$x=-8\text{.}$$ The first row is an example.
 $$x$$ $$y$$ Points $$-8$$ $$-3$$ $$\left(-8,-3\right)$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$
 Values of $$x$$ and $$y$$ satisfying the equation $$x=-8$$
##### 35.
A line’s graph is shown. Write an equation for the line.
##### 36.
A line’s graph is shown. Write an equation for the line.
##### 37.
Line $$m$$ passes points $$(2,4)$$ and $$(2,0)\text{.}$$
Line $$n$$ passes points $$(6,0)$$ and $$(6,2)\text{.}$$
These two lines are
• parallel
• perpendicular
• neither parallel nor perpendicular
.
##### 38.
Line $$m$$ passes points $$(4,-4)$$ and $$(4,9)\text{.}$$
Line $$n$$ passes points $$(1,0)$$ and $$(1,-8)\text{.}$$
These two lines are
• parallel
• perpendicular
• neither parallel nor perpendicular
.
##### 39.
Line $$k$$’s equation is $$y={-{\frac{8}{9}}x-3}\text{.}$$
Line $$\ell$$ is perpendicular to line $$k$$ and passes through the point $$(-3,{-{\frac{19}{8}}})\text{.}$$
Find an equation for line $$\ell$$ in both point-slope form and slope-intercept form.
An equation for $$\ell$$ in point-slope form is: .
An equation for $$\ell$$ in slope-intercept form is: .
##### 40.
Line $$k$$’s equation is $$y={{\frac{9}{7}}x+5}\text{.}$$
Line $$\ell$$ is perpendicular to line $$k$$ and passes through the point $$(-2,{-{\frac{22}{9}}})\text{.}$$
Find an equation for line $$\ell$$ in both point-slope form and slope-intercept form.
An equation for $$\ell$$ in point-slope form is: .
An equation for $$\ell$$ in slope-intercept form is: .
##### 41.
Graph the linear inequality $$y\gt \frac{4}{3}x+1\text{.}$$
##### 42.
Graph the linear inequality $$y\leq -\frac{1}{2}x-3\text{.}$$
##### 43.
Graph the linear inequality $$y\geq 3\text{.}$$
##### 44.
Graph the linear inequality $$3x+2y\lt -6\text{.}$$