Finding an Equation of the Line Given the Slope and a Point

Finding an equation of a line using the slope–intercept form of the equation works well when you are given the slope and y-intercept or when you read them off a graph. But what happens when you have another point instead of the y-intercept?

We are going to use the slope formula to derive another form of an equation of the line. Suppose we have a line that has slope \(m\) and that contains some specific point \(\left({x}_{1},{y}_{1}\right)\) and some other point, which we will just call \(\left(x,y\right)\). We can write the slope of this line and then change it to a different form.

\(\begin{array}{cccccc}& & & \hfill \phantom{\rule{4em}{0ex}}m& =\hfill & \frac{y-{y}_{1}}{x-{x}_{1}}\hfill \\ \text{Multiply both sides of the equation by}\phantom{\rule{0.2em}{0ex}}x-{x}_{1}.\hfill & & & \hfill \phantom{\rule{4em}{0ex}}m\left(x-{x}_{1}\right)& =\hfill & \left(\frac{y-{y}_{1}}{x-{x}_{1}}\right)\left(x-{x}_{1}\right)\hfill \\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}m\left(x-{x}_{1}\right)& =\hfill & y-{y}_{1}\hfill \\ \text{Rewrite the equation with the}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\text{terms on the left.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}y-{y}_{1}& =\hfill & m\left(x-{x}_{1}\right)\hfill \end{array}\)

This format is called the point–slope form of an equation of a line.

Point–slope Form of an Equation of a Line

The point–slope form of an equation of a line with slope \(m\) and containing the point \(\left({x}_{1},{y}_{1}\right)\) is

We can use the point–slope form of an equation to find an equation of a line when we are given the slope and one point. Then we will rewrite the equation in slope–intercept form. Most applications of linear equations use the the slope–intercept form.

Example: Find an Equation of a Line Given the Slope and a Point

Find an equation of a line with slope \(m=\frac{2}{5}\) that contains the point \(\left(10,3\right)\). Write the equation in slope–intercept form.

Solution

This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. In the first row of the table, the first cell on the left reads: “Step 1. Identify the slope.” The text in the second cell reads: “The slope is given.” The third cell contains the slope of a line, defined as m equals 2 fifths.

In the second row, the first cell reads: “Step 2. Identify the point.” The second cell reads: “The point is given.” The third cell contains the ordered pair (10, 3). A superscript x subscript 1 is written over 10, and a superscript y subscript 1 is written over 3.

In the third row, the first cell reads: “Step 3. Substitute the values into the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses.” The top line of the second cell is left blank. The third cell features the point-slope form written again: y minus y subscript 1 equals m times x minus x subscript 1 in parentheses. Below this is the point-slope form with 10 substituted for x subscript 1, 3 substituted for y subscript 1, and 2 fifths substituted for m: y minus 3 equals 2 fifths times x minus 10 in parentheses. One line down, the instructions in the second cell say: “Simplify.” In the third cell is y minus 3 equals 2 fifths x minus 4.

In the fourth row, the first cell reads: “Write the equation in slope-intercept form.” The second cell is blank. In the third cell is y equals 2 fifths x minus 1.

Find an equation of a line given the slope and a point.

Identify the slope.
Identify the point.
Substitute the values into the point-slope form, \(y-{y}_{1}=m\left(x-{x}_{1}\right)\).
Write the equation in slope–intercept form.

Example

Find an equation of a line with slope \(m=-\frac{1}{3}\) that contains the point \(\left(6,-4\right)\). Write the equation in slope–intercept form.

Solution

Since we are given a point and the slope of the line, we can substitute the needed values into the point–slope form, \(y-{y}_{1}=m\left(x-{x}_{1}\right)\).

Identify the slope.
Identify the point.
Substitute the values into \(y-{y}_{1}=m\left(x-{x}_{1}\right).\)

Simplify.
Write in slope–intercept form.

Example

Find an equation of a horizontal line that contains the point \(\left(-1,2\right)\). Write the equation in slope–intercept form.

Solution

Every horizontal line has slope 0. We can substitute the slope and points into the point–slope form, \(y-{y}_{1}=m\left(x-{x}_{1}\right)\).

Identify the slope.
Identify the point.
Substitute the values into \(y-{y}_{1}=m\left(x-{x}_{1}\right).\)

Simplify.


Write in slope–intercept form.	It is in y-form, but could be written \(y=0x+2\).

Did we end up with the form of a horizontal line, \(y=a\)?

This lesson is part of:

Graphs and Equations

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