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# Graphs of Linear Equations

## Graph lines presented in ax+by = c form

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Equations of Lines from Graphs

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

### Equations of Lines from Graphs

You can determine the equation of a line from a graph by counting and performing small calculations. One strategy is to find the y-intercept (b\begin{align*}b\end{align*}) first and then a second point on the line. Then, you can use the y-intercept and second point to determine the slope (m\begin{align*}m\end{align*}) and finally write the equation in slope-intercept form: y=mx+b\begin{align*}y=mx+b\end{align*}.

For example, the y\begin{align*}y\end{align*}-intercept of the graph above is (0, –5). The slope of the line is 34\begin{align*}\frac{3}{4}\end{align*}. The equation of the line in slope-intercept form is \begin{align*}\boxed{y=\frac{3}{4}x-5}\end{align*}

If you cannot determine the y-intercept because it does not lie on a precise point, you can algebraically determine the equation of the line by using the coordinates of two point on the graph. You can use these two points to calculate the slope of the line either by counting or algebraically. Then, you can plug one of the points into the slope-intercept equation and solve for the y-intercept \begin{align*}(b)\end{align*} algebraically.

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

#### Let's find the equation of the line for each of the following graphs:

1.

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

1.

The \begin{align*}y\end{align*}-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 \begin{align*}-\frac{5}{6}\end{align*}. The slope and one of the points will be used to algebraically calculate the \begin{align*}y\end{align*}-intercept of the line.

\begin{align*}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\end{align*}

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

1.

The \begin{align*}y\end{align*}-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 \begin{align*}\frac{3}{5}\end{align*}. The slope and one of the points will be used to algebraically calculate the equation of the line in standard form.

\begin{align*}y-y_1&=m(x-x_1) && \text{Start with point-slope 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\end{align*}

The equation of the line in standard form is \begin{align*}\boxed{3x-5y-12=0}\end{align*}

### Examples

#### Example 1

Earlier, you were asked to 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 \begin{align*}\frac{3}{4}\end{align*}. The coordinates of one point on the line are (2, 5).
\begin{align*}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\end{align*}

The equation of the line in standard form is \begin{align*}\boxed{3x-4y+14=0}\end{align*}

#### Example 2

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

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

#### Example 3

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

The \begin{align*}y\end{align*}-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 \begin{align*}-\frac{4}{5}\end{align*}. The \begin{align*}y\end{align*}-intercept of the line must be calculated by using the slope and one of the points on the line.

\begin{align*}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\end{align*}

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

#### Example 4

Rewrite the equation of the line from Example 3 in standard form.

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.

\begin{align*}y&=-\frac{4}{5}x+\frac{4}{5}\\ 5y&=-4x+4\\ 4x+5y-4&=0\end{align*}

### Review

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

1. .

1. .

1. .

1. .

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

1. .

1. .

1. .

1. .

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

1. .

1. .

1. .

1. .

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 \begin{align*}y=\frac{1}{4}x-5\end{align*} in standard form.
4. Rewrite the equation \begin{align*}y=\frac{2}{3}x+1\end{align*} in standard form.
5. Rewrite the equation \begin{align*}y=\frac{1}{3}x-\frac{3}{7}\end{align*} in standard form.

To see the Review answers, open this PDF file and look for section 4.5.

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### Vocabulary Language: English

Cartesian Plane

The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin.

Slope

Slope is a measure of the steepness of a line. A line can have positive, negative, zero (horizontal), or undefined (vertical) slope. The slope of a line can be found by calculating “rise over run” or “the change in the $y$ over the change in the $x$.” The symbol for slope is $m$

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