Read: Write the Equation of a Linear Function Part I
Learning Objectives
- Define the equation of a line given two points
- Define the equation of a line given the slope and a point
- Write a linear function in standard form
- Identify the equations for vertical and horizontal lines
Point-Slope Formula
We have seen that we can define the slope of a line given two points on the line, and use that information along with the y-intercept to graph the line. If you don't know the y-intercept, or the equation for the line you can use two points to define 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
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.Answer: 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}{l}y-{y}_{1}=m\left(x-{x}_{1}\right)\hfill \\ y - 8=-3\left(x - 4\right)\hfill \\ y - 8=-3x+12\hfill \\ y=-3x+20\hfill \end{array}[/latex]
Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.Example
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.Answer: First, we calculate the slope using the slope formula and two points.
[latex]\begin{array}{l}m\hfill=\dfrac{-3 - 4}{0 - 3}\hfill \\ \hfill =\dfrac{-7}{-3}\hfill \\ \hfill =\dfrac{7}{3}\hfill \end{array}[/latex]
Next, we use the point-slope formula with the slope of [latex]\dfrac{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}{l}y - 4=\dfrac{7}{3}\left(x - 3\right)\hfill \\ y - 4=\dfrac{7}{3}x - 7\hfill&\text{Distribute the }\dfrac{7}{3}.\hfill \\ y=\dfrac{7}{3}x - 3\hfill \end{array}[/latex]
In slope-intercept form, the equation is written as [latex]y=\dfrac{7}{3}x - 3[/latex]. 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}{l}y-\left(-3\right)=\dfrac{7}{3}\left(x - 0\right)\hfill \\ y+3=\dfrac{7}{3}x\hfill \\ y=\dfrac{7}{3}x - 3\hfill \end{array}[/latex]
We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.Standard Form of a Line
Another way that we can represent the equation of a line is in standard form. Standard form is given as[latex]Ax+By=C[/latex]
where [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex] are integers. The x- and y-terms are on one side of the equal sign and the constant term is on the other side.Example
Find the equation of the line with [latex]m=-6[/latex] and passing through the point [latex]\left(\dfrac{1}{4},-2\right)[/latex]. Write the equation in standard form.Answer: We begin using the point-slope formula.
[latex]\begin{array}{l}y-\left(-2\right)=-6\left(x-\dfrac{1}{4}\right)\hfill \\ y+2=-6x+\dfrac{3}{2}\hfill \end{array}[/latex]
From here, we multiply through by [latex]2[/latex], as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right.[latex]\begin{array}{l}2\left(y+2\right)=\left(-6x+\dfrac{3}{2}\right)2\hfill \\ 2y+4=-12x+3\hfill \\ 12x+2y=-1\hfill \end{array}[/latex]
This equation is now written in standard form.Vertical and Horizontal Lines
The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as[latex]x=c[/latex]
where c is a constant. The slope of a vertical line is undefined, and regardless of the y-value of any point on the line, the x-coordinate of the point will be c. Suppose that we want to find the equation of a line containing the following points: [latex]\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right)[/latex], and [latex]\left(-3,5\right)[/latex]. First, we will find the slope.[latex]m=\dfrac{5 - 3}{-3-\left(-3\right)}=\dfrac{2}{0}[/latex]
Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x-coordinates are the same and we find a vertical line through [latex]x=-3[/latex]. The equation of a horizontal line is given as[latex]y=c[/latex]
where c is a constant. The slope of a horizontal line is zero, and for any x-value of a point on the line, the y-coordinate will be c. Suppose we want to find the equation of a line that contains the following set of points: [latex]\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right)[/latex], and [latex]\left(5,-2\right)[/latex]. We can use the point-slope formula. First, we find the slope using any two points on the line.[latex]\begin{array}{l}m=\dfrac{-2-\left(-2\right)}{0-\left(-2\right)}\hfill \\ =\dfrac{0}{2}\hfill \\ =0\hfill \end{array}[/latex]
Use any point for [latex]\left({x}_{1},{y}_{1}\right)[/latex] in the formula, or use the y-intercept.[latex]\begin{array}{l}y-\left(-2\right)=0\left(x - 3\right)\hfill \\ y+2=0\hfill \\ y=-2\hfill \end{array}[/latex]
The graph is a horizontal line through [latex]y=-2[/latex]. Notice that all of the y-coordinates are the same.
The line [latex]x=−3[/latex] is a vertical line. The line [latex]y=−2[/latex] is a horizontal line.Example
Find the equation of the line passing through the given points: [latex]\left(1,-3\right)[/latex] and [latex]\left(1,4\right)[/latex].Answer: The x-coordinate of both points is [latex]1[/latex]. Therefore, we have a vertical line, [latex]x=1[/latex].
Summary
- The slope of a line indicates the direction in which a line slants as well as its steepness. Slope is defined algebraically as:[latex]m=\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[/latex]
- 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]
- Standard form of a line is given as [latex]Ax+By=C[/latex].
- The equation of a vertical line is given as [latex]x=c[/latex]
- The equation of a horizontal line is given as [latex]y=c[/latex]
Licenses & Attributions
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- Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
CC licensed content, Shared previously
- Ex: Find the Equation of a Line in Point Slope and Slope Intercept Form Given the Slope and a Point. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- Ex: Find The Equation of the Line in Point-Slope and Slope Intercept Form Given Two Points. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- College Algebra. Provided by: OpenStax Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution. License terms: Download for free: http://cnx.org/contents/[email protected]:1/Preface.
