**Introduction**

In this lesson, you will learn how to find the point-slope form of a line.

This video illustrates the lesson material below. Watching the video is optional.

**Point-Slope Form of a Line**

With the slope-intercept form, you know the slope (\(m\)) and the *y-*intercept (\(b\)). For example: \(y=mx+b \rightarrow y=\frac{2}{3}x+3\), where the slope is \(\frac{2}{3}\), and the *y*-intercept is 3.

The point-slope form of a line does not include the *y*-intercept. The point-slope form of a line is \(y - y_{1} = m(x - x_{1}\)), where \(m\) is the slope, and (\(x_{1}, y_{1}\)) and (\(x_{2}, y_{2}\)) give coordinates on the line.

You can still find the equation of a line without the *y-*intercept by doing two methods:

- The substitution method with the slope-intercept form: \(y = mx + b\)

- The substitution method but using point-slope form: \(y - y_{1} = m(x - x_{1})\)

**Example 1**

Find the equation of a line that passes through the point \((4, 5)\) and has a slope of \(\frac{2}{3}\). Simplify to slope-intercept form and fill in the missing value: \(y= \frac{2}{3}x\space+\) _______.

\begin{align*}

y&=mx+b &\color{red}\small\text{Formula for slope-intercept form}\\\\

5&=\frac{2}{3}(4)+b &\color{red}\small\text{Substitute \(m=\frac{2}{3}, x =4, y =5\)}\\\\

5&=\frac{8}{3}+b & \color{red}\small\text{Multiply \(\frac{2}{3}(4)\) is \(\frac{8}{3}\)}\\\\

5 \color{red}\mathbf{-\frac{8}{3}} &=\frac{8}{3}+b \color{red}\mathbf{-\frac{8}{3}} &\color{red}\small\text{Subtract \(\frac{8}{3}\) from both sides}\\\\ 5 \color{red}\mathbf{-\frac{8}{3}} &= b &\color{red}\small\text{Right side: the \(\frac{8}{3}\) will cancel out}\\\\

\frac{7}{3}&= b &\color{red}\small\text{Left side: \(5-\frac{8}{3} = \frac{15}{3} - \frac{8}{3} = \frac{7}{3}\)}\\\\

b&=\frac{7}{3} &\color{red}\small\text{The value of the y-intercept}

\end{align*}

With this information, the slope-intercept form is \(y=\frac{2}{3}x+\frac{7}{3}\)

This equation of a line goes through point \((4, 5)\) and has a slope of \(\frac{2}{3}\), with a *y*-intercept at \(\frac{7}{3}\).

**Example 2**

Repeat Example 1, but use the substitution method using point-slope form

\begin{align*}

y-y_{1}&=m(x-x_{1}) &\color{red}\small\text{Formula for point-slope form}\\\\

y-5&=\frac{2}{3}(x-4) &\color{red}\small\text{Substitute \(m=\frac{2}{3}, x_{1} =4, y_{1}=5\)}\\\\

y-5& = \frac{2}{3}x - \frac{8}{3} &\color{red}\small\text{Right side: distributive property}\\\\

y-5 \color{red}\mathbf{+5} &= \frac{2}{3}x - \frac{8}{3} \color{red}\mathbf{+5} &\color{red}\small\text{Add 5 to both sides}\\\\ y&= \frac{2}{3}x - \frac{8}{3} \color{red}\mathbf{+5} &\color{red}\small\text{Left side: the 5's cancel out}\\\\

y&= \frac{2}{3}x + \frac{7}{3} &\color{red}\small\text{Right side: \(-\frac{8}{3} +5 = -\frac{8}{3} + \frac{15}{3} = \frac{7}{3}\)}

\end{align*}

The final equation is \(y=\frac{2}{3}x+\frac{7}{3}\).

Note how the results from Example 1 and 2 are the same.

**Things to Remember**

- You can use the substitution method or the point-slope form of a line formula to write an equation in slope-intercept form.
- Remember to check your work throughout the process.
- The point-slope form of a line is \(y - y_{1} = m(x - x_{1}\)).

### Practice Problems

1. Find the equation of the line that passes through the point \((1, 4)\) and has a slope of \(12\). (**Need More Help?**

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