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# Graphs of Linear Equations

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Practice Graphs of Linear Equations
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Equations of Lines from Graphs

Write the equation, in standard form, of the following graph:

### Guidance

You can determine the equation of a line from a graph by counting. Find the y-intercept ( $b$ ) first and then a second point on the line. Use the y-intercept and second point to determine the slope ( $m$ ). Then, write the equation in slope intercept form: $y=mx+b$ .

The $y$ -intercept of the graph is (0, –5). The slope of the line is $\frac{3}{4}$ . The equation of the line in slope-intercept form is $\boxed{y=\frac{3}{4}x-5}$ .

If you cannot determine the y-intercept, you can algebraically determine the equation of a line by using the coordinates of two points on the graph. These two points can be used to calculate the slope of the line by counting and then the y-intercept can then be determined algebraically.

To write the equation of a line in standard form, the value of the $y$ -intercept is not needed. The slope can be determined by counting. The value of the slope and the coordinates of one other point on the line are used in the function $y-y_1=m(x-x_1)$ . This equation is then set equal to 0 to write the equation in standard form.

#### Example A

Determine the equation of the following graph. Write the equation in slope-intercept form.

Solution:

The $y$ -intercept is (0, 4) so $b=4$ . The slope has a run of five units to the right and a rise of 2 units downward. The slope of the line is $-\frac{2}{5}$ . The equation of the line in slope-intercept form is $y=mx+b$ so $y=-\frac{2}{5}x+4.$

#### Example B

Determine the equation in slope-intercept form of the line shown on the following graph:

Solution:

The $y$ -intercept is not an exact point on this graph. The value of fractions on a Cartesian grid can only be estimated. Therefore, the points (3, –1) and (9, –6) will be used to determine the slope of the line. The slope is $-\frac{5}{6}$ . The slope and one of the points will be used to algebraically calculate the $y$ -intercept of the line.

$y&=mx+b\\-1&=\left(\frac{-5}{6}\right)(3)+b\\{\color{red}-1}&=\left(\frac{{\color{red}-5}}{{\color{red}\cancel{6}_2}}\right)({\color{red}\cancel{3}})+b\\-1&=\frac{-5}{2}+b\\-1{\color{red}+\frac{5}{2}}&=\frac{-5}{2}{\color{red}+\frac{5}{2}}+b\\-1+\frac{5}{2}&=b\\{\color{red}\frac{-2}{2}}+\frac{5}{2}&=b\\\frac{3}{2}&=b$

The equation in slope-intercept form is $\boxed{y=-\frac{5}{6}x+\frac{3}{2}}$

#### Example C

Determine the equation, in standard form, for the line on the following graph:

Solution:

The $y$ -intercept is not an exact point on this graph. Therefore, the points (4, 0) and (–1, –3) will be used to determine the slope of the line. The slope is $\frac{3}{5}$ . The slope and one of the points will be used to algebraically calculate the equation of the line in standard form.

$y-y_1&=m(x-x_1) && \text{Use this formula to determine the equation in standard form.}\\y-{\color{red}0}&={\color{red}\frac{3}{5}}(x-{\color{red}4}) && \text{Fill in the value for} \ m \ \text{of} \ \frac{3}{5} \ \text{and} \ \begin{pmatrix} x_1, & y_1 \\ 4, & 0 \end{pmatrix}\\y&=\frac{3}{5}{\color{red}x}-{\color{red}\frac{12}{5}}\\{\color{red}5}(y)&={\color{red}5}\left(\frac{3}{5}x\right)-{\color{red}5}\left(\frac{12}{5}\right) && \text{Multiply every term by 5.}\\{\color{red}5}(y)&={\color{red}\cancel{5}}\left(\frac{3}{\cancel{5}}x\right)-{\color{red}\cancel{5}}\left(\frac{12}{\cancel{5}}\right) && \text{Simplify and set the equation equal to zero.}\\\\5y&=3x-12\\5y{\color{red}-3x}&=3x{\color{red}-3x}-12\\5y{\color{red}-3x}&=-12\\5y-3x{\color{red}+12}&=-12{\color{red}+12}\\5y-3x+12&=0\\{\color{red}-3x}+5y+12&=0 && \text{The coefficient of} \ x \ \text{cannot be a negative value.}\\3x-5y-12&=0$

The equation of the line in standard form is $\boxed{3x-5y-12=0}$ .

#### Concept Problem Revisited

Write the equation, in standard form, of the following graph:

The first step is to determine the slope of the line.

The slope of the line is $\frac{3}{4}$ . The coordinates of one point on the line are (2, 5).
$y-y_1&=m(x-x_1)\\y-5&=\frac{3}{4}(x-2)\\y-5&=\frac{3}{4}x-\frac{6}{4}\\4(y)-4(5)&=4\left(\frac{3}{4}\right)x-4\left(\frac{6}{4}\right)\\4(y)-4(5)&=\cancel{4}\left(\frac{3}{\cancel{4}}\right)x-\cancel{4}\left(\frac{6}{\cancel{4}}\right)\\4y-20&=3x-6\\-3x+4y-20&=3x-3x-6\\-3x+4y-20&=-6\\-3x+4y-20+6&=-6+6\\-3x+4y-14&=0\\3x-4y+14&=0$

The equation of the line in standard form is $\boxed{3x-4y+14=0}$ .

### Guided Practice

1. Write the equation, in slope-intercept form, of the following graph:

2. Write the equation, in slope-intercept form, of the following graph:

3. Rewrite the equation of the line from #2 in standard form.

1. The first step is to determine the coordinates of the $y$ -intercept. The $y$ -intercept is (0, –3) so $b=-3$ . The second step is to count to determine the value of the slope. Another point on the line is $(7,1)$ so the slope is $\frac{4}{7}$ . The equation of the line in slope-intercept form is $\boxed{y=\frac{4}{7}x-3}$

2. The $y$ -intercept is not an exact point on the graph. Therefore begin by determining the slope of the line by counting between two points on the line. The coordinates of two points on the line are (1, 0) and (6, –4). The slope is is $-\frac{4}{5}$ . The $y$ -intercept of the line must be calculated by using the slope and one of the points on the line.

$y&=mx+b\\{\color{red}0}&={\color{red}\frac{-4}{5}}({\color{red}1})+b\\0&=\frac{-4}{5}+b\\0{\color{red}+\frac{4}{5}}&=\frac{-4}{5}{\color{red}+\frac{4}{5}}+b\\\frac{4}{5}&=b$

The equation of the line in slope-intercept form is $\boxed{y=-\frac{4}{5}x+\frac{4}{5}}$

3. To rewrite the equation in standard form, first multiply the equation by 5 to get rid of the fractions. Then, set the equation equal to 0.

$y&=-\frac{4}{5}x+\frac{4}{5}\\5y&=-4x+4\\4x+5y-4&=0$

### Explore More

For each of the following graphs, write the equation in slope-intercept form:

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For each of the following graphs, write the equation in slope-intercept form:

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For each of the following graphs, write the equation standard form:

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1. Can you always find the equation of a line from its graph?
2. How do you find the equation of a vertical line? What about a horizontal line?
3. Rewrite the equation $y=\frac{1}{4}x-5$ in standard form.
4. Rewrite the equation $y=\frac{2}{3}x+1$ in standard form.
5. Rewrite the equation $y=\frac{1}{3}x-\frac{3}{7}$ in standard form.