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Point-Slope Form of a Line
> ... Math > Slopes, Intercepts, Equation of a Straight Line > Point-Slope Form of a Line

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 \(12\). (
Solution
x
Solution: \({\text{y}} = 12{\text{x}} - 8\)
)
2. Find the equation of the line that passes through the point \((1, 4) \) and has a slope of \(2\). (
Solution
x
Solution: \({\text{y}} = 2{\text{x}} + 2\)

Details:
Use the point-slope form to find the equation of the line. Point-slope form is the following:

\(y-y_1=m(x-x_1)\)

Use the point \(({\color{Red}1}, {\color{Purple}4})\) and the slope of \({\color{DarkOrange}2}\) and substitute them in to \(y-{\color{Purple}y_1}={\color{DarkOrange}m}(x-{\color{Red}x_1})\):

\(y-{\color{Purple}4}={\color{DarkOrange}2}(x-{\color{Red}1})\)

Then distribute on the right side:

\(y-{\color{Purple}4}={\color{DarkOrange}2}x-{\color{DarkOrange}2}(1)\)

Which simplifies to the following:

\(y-4=2x-2\)

Add 4 to both sides:

\(y-4{\color{Red}+4} = 2x-2{\color{Red}+4}\)

Which gives you the equation:

\(y = 2x +{\color{Red}2}\)
)
3. Find the equation of the line that passes through the point \((27, 4)\) and has a slope of \(-\dfrac{2}{9}\). (
Solution
x
Solution: \(y=-\dfrac{2}{9}x+10\)
)
4. Find the equation of the line that passes through the point \((-11,2)\) and has a slope of \(-\dfrac{5}{11}\). (
Solution
x
Solution: \({\text{y}}=-\dfrac{5}{11}{\text{x}}-3\)
)
5. Find the equation of the line that passes through the point \((10, 6)\) and has a slope of \(\dfrac{1}{5}\). What is the y-intercept of the line? (
Solution
x
Solution: \((0,4)\)

Details:
Use the point-slope form to find the equation of the line. Point-slope form is the following:

\({\text{y}}{-}{\text{y}_{1}}=\text{m}\left ( {\text{x}}{-}{\text{x}_{1}} \right )\)

Use the point \(({\color{Red}10}, {\color{Purple}6})\) and the slope of \({\color{DarkOrange} \dfrac{1}{5}}\) and substitute them in to the following:

\({\text{y}}{-}{\color{Purple} {\text{y}_{1}}}={\color{DarkOrange} \text{m}}\left ( {\text{x}}{-}{\color{Red} {\text{x}_{1}}} \right )\)

Then distribute on the right side:

\({\text{y}}{-}{\color{Purple}6}={\color{Orange}\dfrac{1}{5}}{\text{x}}-{\color{Orange}\dfrac{1}{5}}(10)\)

Which simplifies to the following:

\({\text{y}}{-}6={\color{Orange}\dfrac{1}{5}}{\text{x}}-2\)

Add 6 to both sides:

\({\text{y}}{-}6{\color{Red}+6}={\color{Orange}\dfrac{1}{5}}{\text{x}}-2{\color{Red}+6}\)

Which gives you the equation:

\(\text{y}={\color{DarkOrange} \dfrac{1}{5}}\text{x}+{\color{Red} 4}\)

The equation is in slope-intercept form, \(y=mx+b\), so you can see that the y-intercept is at \(4\) or the point \((0,4)\).
)
6. Find the equation of the line that passes through the point \((3, 29)\) and has a slope of \(6\). What is the y-intercept of the line? (
Solution
x
Solution: \((0,11)\)
)

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