# 4.5: Determining the Equation of a Line from the Graph

**At Grade**Created by: CK-12

## The Equation of the Line

**Introduction**

In this lesson you will learn how to determine the equation of a line from a graph by counting. The \begin{align*}y-\end{align*}

You will also learn to algebraically determine the equation of a line by using the coordinates of two points on the graph. These two points will be used to calculate the slope of the line by counting. The \begin{align*}y-\end{align*}

**Objectives**

The lesson objectives for Determining the Equation of a Line from the Graph are:

- Understanding the process of locating the \begin{align*}y-\end{align*}
y− intercept on a graph - Counting from the \begin{align*}y-\end{align*}
y− intercept to a second point to determine the equation of the line. - Using two points on the graph of the line to determine the slope of the line by counting and using the slope and one point to algebraically calculate the value of the \begin{align*}y-\end{align*}
y− intercept.

**Introduction**

In previous lessons you learned how to determine the slope of a line by counting between two exact points on the graph. In this lesson, you will use the \begin{align*}y-\end{align*}

The \begin{align*}y-\end{align*}

**Watch This**

Khan Academy Slope and y-Intercept Intuition

**Guidance**

The equation for the graph of a linear function can be determined from the graph by counting. The equation can then be readily written in slope-intercept form.

**Example A**

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

The \begin{align*}y-\end{align*}

\begin{align*}& y=mx+b\\
& \boxed{y=-\frac{2}{5}x+4}\end{align*}

**Example B**

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

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:

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 of the line is five units to the right and three units upward. 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*}

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

**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).

The second step is to count to determine the value of the slope. The slope is 7 units to the right and 4 units upward. The slope of the line 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 5 units to the right and 4 units downward. The slope of the line 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. The first step is to determine the slope of the line.

The slope of the line is 4 units to the right and 3 units upward. 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*}

**Summary**

In this lesson you have learned that the equation of a linear graph can be determined by counting from the \begin{align*}y-\end{align*}intercept to determine the slope of the line. The values of both the \begin{align*}y-\end{align*}intercept and of the slope can be filled into the equation \begin{align*}y=mx+b\end{align*} to write the equation in slope-intercept form. If the \begin{align*}y-\end{align*}intercept of the line is not an exact point, two other points from the graph can be used to determine the slope of the line. The slope of the line can be found by counting. The value of the \begin{align*}y-\end{align*}intercept can then be calculated algebraically. The equation of the line can be written in slope-intercept form using \begin{align*}y=mx+b\end{align*}

You have also learned that 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. 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*} to write the equation in standard form.

**Problem Set**

**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:**

**Answers**

**For each of the following graphs...**

- The \begin{align*}y-\end{align*}intercept is (0, 5). Another point on the line is (1, 2). The value of the slope is 1 unit to the right and 3 units downward. The value of the slope is \begin{align*}\frac{-3}{1}\end{align*}. The equation of the line in slope-intercept form is \begin{align*}\boxed{y=-\frac{3}{1}x+5}\end{align*}

- The \begin{align*}y-\end{align*}intercept is (0, –3). Another point on the line is (4, 2). The value of the slope is 4 units to the right and 5 units upward. The value of the slope is \begin{align*}\frac{5}{4}\end{align*}. The equation of the line in slope-intercept form is \begin{align*}\boxed{y=\frac{5}{4}x-3}\end{align*}

**For each of the following graphs...**

- The \begin{align*}y-\end{align*}intercept is not an exact point. Two points on the line are (–2, –1) and (6, 1). The value of the slope is 8 units to the right and two units upward. The value of the slope is \begin{align*}\frac{2}{8}=\frac{1}{4}\end{align*}. The \begin{align*}y-\end{align*}intercept is:

\begin{align*}y&=mx+b\\ 1&=\frac{1}{4}(6)+b\\ 1&=\frac{6}{4}+b\\ 1-\frac{6}{4}&=\frac{6}{4}-\frac{6}{4}+b\\ \frac{4}{4}-\frac{6}{4}&=b\\ -\frac{2}{4}&=b\\ -\frac{1}{2}&=b\end{align*}

The equation of the line in slope-intercept form is \begin{align*}\boxed{y=\frac{1}{4}x-\frac{1}{2}}\end{align*}

- The \begin{align*}y-\end{align*}intercept is not an exact point. Two points on the line are (–1, 1) and (2, –1). The value of the slope is 3 units to the right and two units downward. The value of the slope is \begin{align*}\frac{-2}{3}\end{align*}. The \begin{align*}y-\end{align*}intercept is:

\begin{align*}y&=mx+b\\ 1&=\frac{-2}{3}(-1)+b\\ 1&=\frac{2}{3}+b\\ 1-\frac{2}{3}&=\frac{2}{3}-\frac{2}{3}+b\\ \frac{3}{3}-\frac{2}{3}&=b\\ \frac{1}{3}&=b\end{align*}

The equation of the line in slope-intercept form is \begin{align*}\boxed{y=-\frac{2}{3}x+\frac{1}{3}}\end{align*}

**For each of the following graphs...**

- The \begin{align*}y-\end{align*}intercept is (0, –4). One point on the line is (2, 6). The slope is 2 units to the right and 10 units upward. The value of the slope is \begin{align*}\frac{10}{2}=\frac{5}{1}\end{align*}.

\begin{align*}y-y_1&=m(x-x_1)\\ y--4&=5(x-0)\\ y+4&=5x-0\\ y+4&=5x\\ y+4-5x&=5x-5x\\ -5x+y+4&=0\\ 5x-y-4&=0\end{align*}

The equation of the line in standard form is \begin{align*}\boxed{5x-y-4=0}\end{align*}

- Two points on the line are (–3, –4) and (8, 6). The slope is 11 units to the right and 10 units upward. The slope is \begin{align*}\frac{10}{11}\end{align*}.

\begin{align*}y-y_1&=m(x-x_1)\\ y-6&=\frac{10}{11}(x-8)\\ y-6&=\frac{10x}{11}-\frac{80}{11}\\ 11(y)-11(6)&=11\left(\frac{10x}{11}\right)-11\left(\frac{80}{11}\right)\\ 11(y)-11(6)&=\cancel{11}\left(\frac{10x}{\cancel{11}}\right)-\cancel{11}\left(\frac{80}{{\cancel{11}}}\right)\\ 11y-66&=10x-80\\ 11y-66+80&=10x-80+80\\ 11y+14&=10x\\ 11y+14-10x&=10x-10x\\ -10x+11y+14&=0\\ 10x-11y-14&=0\end{align*}

The equation of the line in standard form is \begin{align*}\boxed{10x-11y-14=0}\end{align*}

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