ORCCA Graphing Lines Chapter Review

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)Use points from the table

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

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.

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

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-2-3\\ y\amp=\frac{1}{2}x-5 \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.

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).
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 $y\gt2x+1$

SubsectionExercises

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

$y=-\frac{5}{6} x-1$

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

$ (18,-11)$
$ (-30,24)$
$ (-6,5)$
$ (0,-1)$

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 $(-4,{9})$ and $(8,{0})\text{.}$ Find this line’s slope. If the slope does not exist, you may enter DNE or NONE.

This line’s slope is .

8

A line passes through the points $(1,-5)$ and $(-4,-5)\text{.}$ Find this line’s slope. If the slope does not exist, you may enter DNE or NONE.

This line’s slope is .

9

A line passes through the points $(-2,-2)$ and $(-2,4)\text{.}$ Find this line’s slope. If the slope does not exist, 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 $\displaystyle{ 5x-6y=12 }\text{.}$

This line’s slope is .

This line’s $y$-intercept is .

12

A line passes through the points $(0,{-2})$ and $(27,{19})\text{.}$ Find this line’s equation in point-slope form.

Using the point $(0,{-2})\text{,}$ this line’s point-slope form equation is .

Using the point $(27,{19})\text{,}$ 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 $3.6$ grams each minute. Six minutes since the experiment started, the remaining gas had a mass of $154.8$ 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.

This line’s slope-intercept equation is .
$34$ minutes after the experiment started, there would be grams of gas left.
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

$\displaystyle{8 x + 7 y = -168}$

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

16

Rewrite $y=-\frac{3}{4}x - 2$ 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 $(0,-6)$ and $(0,6)\text{.}$

Line $n$ passes points $(-1,-3)$ and $(-1,-1)\text{.}$

Determine how the two lines are related.

These two lines are

parallel
perpendicular
neither parallel nor perpendicular

20

Line $k$’s equation is $y={-{\frac{6}{7}}x+2}\text{.}$

Line $\ell$ is perpendicular to line $k$ and passes through the point $(12,{15})\text{.}$

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: .

\(x\)	\(y=-2x+5\)	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
\(-2\)	\(-2(-2)+5=\substitute{9}\)	\((-2,9)\)
\(-1\)	\(-2(-1)+5=\substitute{7}\)	\((-1,7)\)
\(0\)	\(-2(0)+5=\substitute{5}\)	\((0,5)\)
\(1\)	\(-2(1)+5=\substitute{3}\)	\((1,3)\)
\(2\)	\(-2(2)+5=\substitute{1}\)	\((2,1)\)

\(x\)	\(y\)
\(0\)	\(-4\)
\(1\)	\(-6\)
\(2\)	\(-8\)
\(3\)	\(-10\)

	\(x\)	\(y\)
	\(0\)	\(-4\)
\({}+1\rightarrow\)	\(1\)	\(-6\)	\(\leftarrow{}-2\)
\({}+1\rightarrow\)	\(2\)	\(-8\)	\(\leftarrow{}-2\)
\({}+1\rightarrow\)	\(3\)	\(-10\)	\(\leftarrow{}-2\)

\(x\)	\(y\)
\(0\)	\({-4}\)
\(1\)	\({-2}\)
\(2\)	\({0}\)
\(3\)	\({2}\)

	\(x\)-value	\(y\)-value	Location
\(y\)-intercept
\(x\)-intercept