4.5: Equations of Lines from Graphs
Write the equation, in standard form, of the following graph:
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Khan Academy Slope and y-Intercept Intuition
Guidance
You can determine the equation of a line from a graph by counting. Find the y-intercept (\begin{align*}b\end{align*}
The \begin{align*}y\end{align*}
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 \begin{align*}y\end{align*}
Example A
Determine the equation of the following graph. Write the equation in slope-intercept form.
Solution:
The \begin{align*}y\end{align*}
Example B
Determine the equation, in slope-intercept form of the line shown on the following graph:
Solution:
The \begin{align*}y\end{align*}
\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*}
Example C
Determine the equation, in standard form, for the line on the following graph:
Solution:
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{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\end{align*}
The equation of the line in standard form is \begin{align*}\boxed{3x-5y-12=0}\end{align*}.
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 \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*}.
Vocabulary
- Slope – Intercept Form
- The slope-intercept form is one method for writing the equation of a line. The slope-intercept form is \begin{align*}y = mx + b\end{align*} where \begin{align*}m\end{align*} refers to the slope and \begin{align*}b\end{align*} identifies the \begin{align*}y\end{align*}-intercept.
- Standard Form
- The standard form is another method for writing the equation of a line. The standard form is \begin{align*}Ax + By + C = 0\end{align*} where \begin{align*}A\end{align*} is the coefficient of \begin{align*}x\end{align*}, \begin{align*}B\end{align*} is the coefficient of \begin{align*}y\end{align*} and \begin{align*}C\end{align*} is a constant.
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.
Answers:
1. 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. 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*}
2. 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*}
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.
- \begin{align*}y&=-\frac{4}{5}x+\frac{4}{5}\\ 5y&=-4x+4\\ 4x+5y-4&=0\end{align*}
Practice
For each of the following graphs, write the equation in slope-intercept form:
For each of the following graphs, write the equation in slope-intercept form:
For each of the following graphs, write the equation standard form:
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