Learning Objectives
In this section you will:
2.4.1 – Finding a Linear Equation
The Point-Slope Formula
Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.
[latex]$$y-{y}_{1}=m\left(x-{x}_{1}\right)$$[/latex]
This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.
The Point-Slope Formula
Given one point and the slope, the point-slope formula will lead to the equation of a line:
[latex]$$y-{y}_{1}=m\left(x-{x}_{1}\right)$$[/latex]
Example 1 – Finding the Equation of a Line Given the Slope and One Point
Write the equation of the line with slope [latex]\,m=-3\,[/latex] and passing through the point [latex]\,\left(4,8\right).\,[/latex] Write the final equation in slope-intercept form.
Using the point-slope formula, substitute [latex]\,-3\,[/latex] for m and the point [latex]\,\left(4,8\right)\,[/latex] for [latex]\,\left({x}_{1},{y}_{1}\right).[/latex]
[latex]$$\begin{array}{ccc}\hfill y-{y}_{1}& =& m\left(x-{x}_{1}\right)\hfill \\ \hfill y-8& =& -3\left(x-4\right)\hfill \\ \hfill y-8& =& -3x+12\hfill \\ \hfill y& =& -3x+20\hfill \end{array}$$[/latex]
Analysis
Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.
Try It
Given [latex]\,m=4,[/latex] find the equation of the line in slope-intercept form passing through the point [latex]\,\left(2,5\right).[/latex]
Show answer
[latex]y=4x-3[/latex]
Example 2 – Finding the Equation of a Line Passing Through Two Given Points
Find the equation of the line passing through the points [latex]\,\left(3,4\right)\,[/latex] and [latex]\,\left(0,-3\right).\,[/latex] Write the final equation in slope-intercept form.
First, we calculate the slope using the slope formula and two points.
[latex]$$\begin{array}{ccc}\hfill m& =& \frac{-3-4}{0-3}\hfill \\ & =& \frac{-7}{-3}\hfill \\ & =& \frac{7}{3}\hfill \end{array}$$[/latex]
Next, we use the point-slope formula with the slope of [latex]\,\frac{7}{3},[/latex] and either point. Let’s pick the point [latex]\,\left(3,4\right)\,[/latex] for [latex]\,\left({x}_{1},{y}_{1}\right).[/latex]
[latex]$$\begin{array}{ccc}\hfill y-4& =& \frac{7}{3}\left(x-3\right)\hfill \\ \hfill y-4& =& \frac{7}{3}x-7\phantom{\rule{2em}{0ex}}\text{Distribute the }\frac{7}{3}.\hfill \\ \hfill y& =& \frac{7}{3}x-3\hfill \end{array}$$[/latex]
In slope-intercept form, the equation is written as [latex]\,y=\frac{7}{3}x-3.[/latex]
Analysis
To prove that either point can be used, let us use the second point [latex]\,\left(0,-3\right)\,[/latex] and see if we get the same equation.
[latex]$$\begin{array}{ccc}\hfill y-\left(-3\right)& =& \frac{7}{3}\left(x-0\right)\hfill \\ \hfill y+3& =& \frac{7}