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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 \(y=-2x+5\text{.}\)

\(x\) \(y=-2x+5\) Point
\(-2\) \(\phantom{-2(-2)+5=\substitute{9}}\) \(\phantom{(-2,9)}\)
\(-1\) \(\phantom{-2(-1)+5=\substitute{7}}\) \(\phantom{(-1,7)}\)
\(0\) \(\phantom{-2(0)+5=\substitute{5}}\) \(\phantom{(0,5)}\)
\(1\) \(\phantom{-2(1)+5=\substitute{3}}\) \(\phantom{(1,3)}\)
\(2\) \(\phantom{-2(2)+5=\substitute{1}}\) \(\phantom{(2,1)}\)
\(x\) \(y=-2x+5\) Point
\(-2\) \(-2(-2)+5=\substitute{9}\) \((-2,9)\)
\(-1\) \(-2(-1)+5=\substitute{7}\) \((-1,7)\)
\(0\) \(-2(0)+5=\substitute{5}\) \((0,5)\)
\(1\) \(-2(1)+5=\substitute{3}\) \((1,3)\)
\(2\) \(-2(2)+5=\substitute{1}\) \((2,1)\)
(a)Set up the table
(b)Complete the table
Figure4.11.3Making a table for \(y=-2x+5\)

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.

a Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1)
a Cartesian grid with points (-2,9),(-1,7),(0,5),(1,3),(2,1)
(a)Use points from the table
(b)Connect the points in whatever pattern is apparent
Figure4.11.4Graphing the Equation \(y=-2x+5\)

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

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

Find the slope of the line in the following graph.

This is a grid with a line, passing the points (3,15) and (6,27). A slope triangle is drawn, starting at (3,15), passing (6,15), and ends at (6,27). The label from (3,15) to (6,15) is "6-3=3"; the label from (6,15) to (6,27) is "27-15=12".

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 \(4\text{.}\)

Finding a Line's Slope by the Slope Formula
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 \(y=-\frac{2}{3}x+10\text{.}\)

(a)First, plot the line's \(y\)-intercept, \((0,10)\text{.}\)
(b)The slope is \(-\frac{2}{3}=\frac{-2}{3}=\frac{2}{-3}\text{.}\) So we can try using a “run” of \(3\) and a “rise” of \(-2\) or a “run” of \(-3\) and a “rise” of \(2\text{.}\)
(c)Arrowheads and labels are encouraged.
Figure4.11.8Graphing \(y=-\frac{2}{3}x+10\)
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.

(a)Graph of a line
(b)Identify the line's \(y\)-intercept, \(10\text{.}\)
(c)Identify the line's slope by a slope triangle. Note that we can pick any two points on the line to create a slope triangle. We would get the same slope: \(-\frac{2}{3}\)
Figure4.11.10Find 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.

\begin{align*} y\amp=-\frac{2}{3}x+2\\ \multiplyleft{3}y\amp=\multiplyleft{3}(-\frac{2}{3}x+2)\\ 3y\amp=3\cdot(-\frac{2}{3}x)+3\cdot2\\ 3y\amp=-2x+6\\ 3y\addright{2x}\amp=-2x+6\addright{2x}\\ 2x+3y\amp=6 \end{align*}

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:

This is a grid with the graph of line 2x-3y=-6. The following points are plotted: (0,2), (-3,0). The segment from (0,0) to (0,2) is labeled "2-0=0", and the segment from (-3,0) to (0,0) is labeled "0-(-3)=3".
Figure4.11.13Graph of \(2x-3y=-6\)

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.

a horizontal line at y=3 and a vertical line at x=5
Figure4.11.14Graphs of \(y=3\) and \(x=5\)
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 \((0,0)\text{.}\)
This is a coordinate plane with y=2x+1 graphed.
This is a coordinate plane with y=2x+1 graphed. The region above the line is shaded.
Figure4.11.16Step 1 of graphing \(y\gt2x+1\)
Figure4.11.17Step 2 of graphing \(y\gt2x+1\)

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: .