
## Section4.11Graphing Lines Chapter Review

### Subsection4.11.1Review of Cartesian Coordinates

##### Cartesian Coordinate System

The Cartesian coordinate system identifies the location of every point in a plane with an ordered pair.

###### Example4.11.1

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

### Subsection4.11.2Review of Graphing Equations

##### Graphing Equations

To graph an equation with two variables $x$ and $y\text{,}$ we can choose some reasonable $x$-values, then calculate the corresponding $y$-values, and then plot the $(x,y)$-pairs as points. For many (not-so-complicated) algebraic equations, connecting those points with a smooth curve will produce an excellent graph.

###### Example4.11.2

Graph the equation $y=-2x+5\text{.}$

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.

### Subsection4.11.3Review of Two-Variable Data and Rate of Change

##### Exploring Two-Variable Data and Rate of Change

For a linear relationship, by its data in a table, we can see the rate of change (slope) and the line's $y$-intercept, thus writing the equation.

###### Example4.11.5

Write an equation in the form $y=\ldots$ suggested by the pattern in the table.

 $x$ $y$ $0$ $-4$ $1$ $-6$ $2$ $-8$ $3$ $-10$

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 ouput 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 $y=-2x-4\text{.}$

### Subsection4.11.4Review of Slope

##### Slope

When $x$ and $y$ are two variables where the rate of change between any two points is always the same, we call this common rate of change the slope. Since having a constant rate of change means the graph will be a straight line, it's also called the slope of the line.

We can find a line's slope by drawing a slope-triangle on the line's graph, and then using the formula

$$m=\frac{\text{change in y}}{\text{change in x}}=\frac{\Delta y}{\Delta x}\tag{4.11.1}$$
###### Example4.11.6

Find the slope of the line in the following graph.

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 $4\text{.}$

###### Example

If we know two points on a line, we can find its slope without graphing and, instead, using the slope formula $\text{slope}=\frac{y_2-y_1}{x_2-x_1}$

A line passes the points $(-5,25)$ and $(4,-2)\text{.}$ Find this line's slope.

\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 $-3\text{.}$

### Subsection4.11.5Review of Slope-Intercept Form

##### Graphing a Line in Slope-Intercept Form

A line's equation in slope-intercept form looks like $y=mx+b\text{,}$ where $m$ is the line's slope, and $b$ is the line's $y$-intercept.

We can use a line's $y$-intercept and slope triangles to graph it.

###### Example4.11.7

Graph the line $y=-\frac{2}{3}x+10\text{.}$

##### 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 by a slope triangle. With a line's slope and $y$-intercept, we can write its equation in the form of $y=mx+b\text{.}$

###### Example4.11.9

Find the equation of the line in the graph.

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

### Subsection4.11.6Review of Point-Slope Form

##### Point-Slope Form

A line's point-slope form equation is in the form of $y=m\left(x-x_0\right)+y_0\text{,}$ where $m$ is the line's slope, and $(x_0,y_0)$ is a point on the line.

###### Example4.11.11

A line passes through $(-6,0)$ and $(9,-10)\text{.}$ Find this line's equation in both point-slope and slope-intercept form.

Solution

We will use the slope formula (4.4.3) 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*}

Next, we will change the point-slope equation into slope-intercept form:

\begin{align*} y\amp=-\frac{2}{3}(x-9)-10\\ y\amp=-\frac{2}{3}x+6-10\\ y\amp=-\frac{2}{3}x-4 \end{align*}

### Subsection4.11.7Review of Standard Form

##### Standard Form

A line's equation in standard form looks like $Ax+By=C\text{.}$ We need to convert a line's equation from standard form to slope-intercept form, and vice versa.

###### Examples

Converting from Standard Form to Slope-Intercept Form

Convert $2x+3y=6$ into slope-intercept form.

\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 $y=-\frac{2}{3}x+2\text{.}$

Converting from Slope-Intercept Form to Standard Form

Convert $y=-\frac{2}{3}x+2$ into standard form.

The line's equation in standard form is $2x+3y=6\text{.}$

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.

###### Example4.11.12

Graph $2x-3y=-6$ using its intercepts. And then use the intercepts to calculate the line's slope.

Solution

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:

### Subsection4.11.8Review of Horizontal, Vertical, Parallel, and Perpendicular Lines

##### Horizontal, Vertical, Parallel, and Perpendicular Lines

A horizontal line's equation looks like $y=k\text{,}$ while a vertical line's equation looks like $x=h\text{.}$ The following figure has graphs of a horizontal line and a vertical line.

##### Parallel and Perpendicular Lines
###### Examples

Two lines are parallel if and only if they have the same slope.

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 line $n$'s equation.

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*}

We can also easily get the line's equation in slope-intercept form:

\begin{align*} y\amp=-2(x-4)-3\\ y\amp=-2x+8-3\\ y\amp=-2x+5 \end{align*}

Two lines are perpendicular if and only if the product of their slopes is $-1\text{.}$

Line $m$'s equation is $y=-2x+20\text{.}$ Line $n$ is perpendicular to $m\text{,}$ and line $n$ also passes the point $(4,-3)\text{.}$ Find line $n$'s equation.

Since line $m$ and $n$ are perpendicular, the product of their slopes is $-1\text{.}$ Because line $m$'s slope is given as $-2\text{,}$ we can find line $n$'s slope is $\frac{1}{2}\text{.}$

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=\frac{1}{2}(x-4)+(-3)\\ y\amp=\frac{1}{2}(x-4)-3 \end{align*}

We can also easily get the line's equation in slope-intercept form:

\begin{align*} y\amp=\frac{1}{2}(x-4)-3\\ y\amp=\frac{1}{2}x-2-4\\ y\amp=\frac{1}{2}x-6 \end{align*}

### Subsection4.11.9Review of Linear Inequalities in Two Variables

##### Linear Inequalities in Two Variables

When we graph lines like $y=2x+1\text{,}$ we are graphing points which satisfy the relationship $y=2x+1\text{,}$ like $(0,1), (1,3), (2,5)\text{,}$ etc. Similarly, we can graph linear inequalities like $y\gt2x+1$ by plotting all points which satisfies the inequality, like $(0,2), (0,3), (1,4), (1,5)\text{,}$ etc. All these points form a region, instead of a line.

###### Example4.11.15

Graph $y\gt2x+1\text{.}$

There are two steps to graph an inequality.

1. Graph the line $y=2x+1\text{.}$ Because the inequality symbol is $\gt\text{,}$ (instead of $\ge$) the line should be dashed (instead of solid).
2. Next, we need to decide whether to shade the region above $y=2x+1$ or below it. We will choose a point to test whether $y\gt2x+1$ is true. As long as the line doesn't cross $(0,0)\text{,}$ we will use $(0,0)$ to test, because the number $0$ is the easiest number for calculation. \begin{gather*} y\gt2x+1\\ 0\stackrel{?}{\gt}2(0)+1\\ 0\stackrel{?}{\gt}1 \end{gather*} Because $0\gt1$ is not true, the point $(0,0)$ is not a solution and should not be shaded. As a result, we shade the region without $(0,0)\text{.}$

### Subsection4.11.10Exercises

###### 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

Consider the equation

$y=-\frac{5}{6} x-4$

Which of the following ordered pairs are solutions to the given equation? There may be more than one correct answer.

$(24,-21)$

$(-24,16)$

$(-30,24)$

$(0,-4)$

###### 3
 $x$ $y$ $0$ ${-4}$ $1$ ${-2}$ $2$ ${0}$ $3$ ${2}$

Write an equation in the form $y=\ldots$ suggested by the pattern in the table.

###### 4
 $x$ $y$ $0$ ${3}$ $1$ ${-2}$ $2$ ${-7}$ $3$ ${-12}$

Write an equation in the form $y=\ldots$ suggested by the pattern in the table.

###### 5

A line’s graph is shown below.

The slope of this line is . (If the slope does not exist, enter DNE or NONE.)

###### 6

A line’s graph is shown below.

The slope of this line is . (If the slope does not exist, enter DNE or NONE.)

###### 7

A line passes through the points $(-12,{-1})$ and $(8,{-16})\text{.}$ Find this line’s slope. If the slope does not exists, you may enter DNE or NONE.

This line’s slope is .

###### 8

A line passes through the points $(1,-4)$ and $(-3,-4)\text{.}$ Find this line’s slope. If the slope does not exists, you may enter DNE or NONE.

This line’s slope is .

###### 9

A line passes through the points $(-2,-1)$ and $(-2,5)\text{.}$ Find this line’s slope. If the slope does not exists, you may enter DNE or NONE.

This line’s slope is .

###### 10

A line’s graph is given.

This line’s slope-intercept equation is

###### 11

Find the line’s slope and $y$-intercept.

A line has equation $\displaystyle{ 7x-6y=-24 }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

###### 12

A line passes through the points $(24,{13})$ and $(0,{-8})\text{.}$ Find this line’s equation in point-slope form.

Using the point $(24,{13})\text{,}$ this line’s point-slope form equation is .

Using the point $(0,{-8})\text{,}$ this line’s point-slope form equation is .

###### 13

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. Its mass is leaking by $4.7$ grams per minute. Eight minutes since the experiment started, the remaining gas had a mass of $192.7$ 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. $36$ 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.

###### 14

Find the $y$-intercept and $x$-intercept of the line given by the equation

$\displaystyle{8 x + 3 y = -72}$

If a particular intercept does not exist, enter none into all the answer blanks for that row.

 $x$-value $y$-value Location $y$-intercept $x$-intercept
###### 15

Rewrite ${y}={2x+4}$ in standard form.

###### 16

Rewrite $y=-\frac{3}{4}x+6$ in standard form.

###### 17

Fill out this table for the equation $x=-8\text{.}$ The first row is an example.

 $x$ $y$ Points $-8$ $-3$ $\left(-8,-3\right)$ $-2$ $-1$ $0$ $1$ $2$
###### 18

A line’s graph is given.

This line’s equation is

###### 19

Line $m$ passes points $(0,-4)$ and $(0,-10)\text{.}$

Line $n$ passes points $(10,-1)$ and $(10,10)\text{.}$

Determine how the two lines are related.

These two lines are

parallel

perpendicular

neither parallel nor perpendicular

###### 20

Line $k$’s equation is $y={{\frac{6}{5}}x-1}\text{.}$

Line $\ell$ is perpendicular to line $k$ and passes through the point $(18,{-11})\text{.}$

Find an equation for line $\ell$ in both slope-intercept form and point-slope forms.

An equation for $\ell$ in slope-intercept form is: .

An equation for $\ell$ in point-slope form is: .