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# Linear Equations in Point-Slope Form

## Solving point-slope form in terms of y

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Linear Equations in Point-Slope Form

Suppose that the cost of a wedding was a function of the number of guests attending. If you knew the slope of the function and you also knew how much the wedding would cost if 150 guests attended, could you write a linear equation representing this situation? If so, what form of the equation would be easiest to use? In this Concept, you'll learn about the point-slope form of a linear equation so that you can answer questions like these.

### Objectives

You will be able to...

1. Write the equation of a line in point-slope form given two points
2. Write the equation of a line in point-slope form given a point and a slope
3. Determine a point and slope from an equation in point-slope form
4. Write the equation of a line in point-slope form given the graph of the line
5. Graph a line given an equation in point-slope form

### Key Vocabulary & Concepts

• If a point is on a line, we say that the contains "contains the point."

### Overview

Equations can be written in many forms. Though you are probably more familiar with slope-intercept form (y = mx + b), this page will provide a second way to write an equation of a line: point-slope form.

The equation of the line between any two points (x1,y1)\begin{align*}(x_1,y_1)\end{align*} and (x2,y2)\begin{align*}(x_2,y_2)\end{align*} can be written in the following form: yy1=m(xx1)\begin{align*}y-y_1=m(x-x_1)\end{align*}.

To write an equation in point-slope form, you need two things:

1. A point on the line
2. The slope of the line

### Examples

Objective 1:  Write the equation of a line in point-slope form given two points

Example 1 Write an equation for a line containing (9, 3) and (4, 5).

Solution: Begin by finding the slope.

slope=y2y1x2x1=5349=25\begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}=\frac{5-3}{4-9}=-\frac{2}{5}\end{align*}

Instead of trying to find b\begin{align*}b\end{align*} (the y\begin{align*}y-\end{align*}intercept), you will use the point-slope formula.

yy1y3=m(xx1)=25(x9)\begin{align*}y-y_1& =m(x-x_1)\\ y-3& = \frac{-2}{5}(x-9)\end{align*}

It doesn't matter which point you use.

You could also use the other ordered pair to write the equation:

y5=25(x4)\begin{align*}y-5= \frac{-2}{5}(x-4)\end{align*}

These equations may look completely different, but if you rewrite the equations in slope-intercept form (by solving each one for y\begin{align*}y\end{align*}), you can compare the slope-intercept form to check your answer.

y3yy5yy=25(x9)y=25x+185+3=25x+335=25(x4)=25x+85+5=25x+335\begin{align*}y-3& = \frac{-2}{5} (x-9) \Rightarrow y=\frac{-2}{5} x+\frac{18}{5}+3\\ y& =\frac{-2}{5} x+\frac{33}{5}\\ y-5& =\frac{-2}{5} (x-4) \\ y& =\frac{-2}{5} x+\frac{8}{5}+5\\ y& =\frac{-2}{5} x+\frac{33}{5}\end{align*}

Objective 2:  Write the equation of a line in point-slope form given a point and a slope

Example 2: Write the equation of the linear function in point-slope form.

m=9.8\begin{align*}m=9.8\end{align*} and f(5.5)=12.5\begin{align*}f(5.5)=12.5\end{align*}

Solution: This function has a slope of 9.8 and contains the ordered pair (5.5, 12.5). Substituting the appropriate values into point-slope form, we get the following:

y12.5=9.8(x5.5)\begin{align*}y-12.5=9.8(x-5.5)\end{align*}

Replacing yy1\begin{align*}y-y_1\end{align*} with f(x)y1\begin{align*}f(x)-y_1\end{align*}, the equation in point-slope form is:

f(x)12.5f(x)12.5=9.8x53.9f(x)=9.8x41.4=9.8(x5.5)\begin{align*}f(x)-12.5& =9.8(x-5.5)\\ f(x)-12.5 =9.8x-53.9\\ f(x) =9.8x - 41.4 \end{align*}

where the last equation is in slope-intercept form.

Objective 3:  Determine a point and slope from an equation in point-slope form

Example 3:   y5=25(x+1)\begin{align*}y-5= \frac{2}{5}(x+1)\end{align*}

Solution:  Begin by rewriting the equation to make it point-slope form:  y5=25(x(1))\begin{align*}y-5= \frac{2}{5}(x-(-1))\end{align*}Now we see that point (5, -1) is on the line and that the slope = 25\begin{align*}\frac{2}{5}\end{align*} .

Example 4:  y+6=43(x+7)\begin{align*}y+6= \frac{-4}{3}(x+7)\end{align*} .

Solution:  Begin by rewriting the equation to make it point-slope form: y(6)=43(x(7))\begin{align*}y-(-6)= \frac{-4}{3}(x-(-7))\end{align*} Now we see that point (–6, -7) is on the line and that the slope =43\begin{align*}\frac{-4}{3}\end{align*} .

Objective 4:  Write the equation of a line in point-slope form given the graph of the line

If the graph of a line does not have a "nice" y-intercept or doesn't show a y-intercept, then writing an equation in point-slope form will be easier and more accurate.

Example 5:  Given the graph below, write an equation for the line in point-slope form.

Identify any two points on the line. Below, I have highlighted points A (-5, 4) and B(-3, 3).

Count or calculate the slope between A and B.  The slope is 12\begin{align*}\frac{-1}{2}\end{align*}.  Now use the coordinates of either point as your x1 and y1 in the formula.  Let's use A(-5, 4).  Plugging slope, x1, and y1 into the equation gives us: y4=12(x(5))\begin{align*}y-4=\frac{-1}{2}(x-(-5))\end{align*}, which simplifies to

y4=12(x+5)\begin{align*}y-4=\frac{-1}{2}(x+5)\end{align*}

Objective 5:  Graph a line given an equation in point-slope form

If you are given an equation in point-slope form, it is not necessary to re-write it in slope-intercept form in order to graph it. The point-slope form of the equation gives you enough information so you can graph the line.

Example 6:  Make a graph of the line given by the equation y2=23(x+2)\begin{align*}y-2=\frac{2}{3}(x+2)\end{align*}.

Solution:  Begin by rewriting the equation to make it point-slope form: y2=23(x(2))\begin{align*}y-2= \frac{2}{3}(x-(-2))\end{align*} Now we see that point (–2, 2) is on the line and that the slope=23\begin{align*}\text{slope}=\frac{2}{3}\end{align*}. First plot point (–2, 2) on the graph.

A slope of 23\begin{align*}\frac{2}{3}\end{align*} tells you that from your point you should move 2 units up and 3 units to the right and draw another point.

Now draw a line through the two points and extend the line in both directions.

### [Figure3] License: CC BY-NC 3.0  Common Errors

• sign errors:  In the equation, the number next to y is the opposite of the y-coordinate
• sign errors:  In the equation, the number next to x is the opposite of the x-coordinate
• x1 and y1 switched:  In the equation, x1 goes next to x, y1 goes next to y

### Guided Practice

Rewrite y5=3(x2)\begin{align*}y-5=3(x-2)\end{align*} in slope-intercept form.

Solution: Use the Distributive Property to simplify the right side of the equation:

y5=3x6\begin{align*}y-5=3x-6\end{align*}

Solve for y\begin{align*}y\end{align*}:

y5+5y=3x6+5=3x1\begin{align*}y-5+5& =3x-6+5\\ y& =3x-1\end{align*}

### Practice

Sample explanations for some of the practice problems below are available by viewing the following video.  The problems that have video solutions are labeled, along with the time that they are presented in the video.The problem number in the video and the number in the practice set will not always match. (i.e., the first problem from the video might be practice problem #6).  However, the answers for all practice problems can be found at the bottom of this page. CK-12 Basic Algebra: Linear Equations in Point-Slope Form(9:38).

1. What is the equation for a line containing the points (x1,y1)\begin{align*}(x_1,y_1)\end{align*} and (x2,y2)\begin{align*}(x_2,y_2)\end{align*} in point-slope form?
2. In what ways is it easier to use point-slope form rather than slope-intercept form?

Objective 1:  Write the equation of a line in point-slope form given two points

Write the equation, in point slope form, of each line being described.

1. The line contains the points (–2, 3) and (–1, –2).
2. The line contains the points (0, 0) and (1, 2).
3. The line contains the points (10, 12) and (5, 25).   (video solution @4:03)
4. The line contains the points (2, 3) and (0, 3).
5. The line contains the points (–4, –2) and (8, 12).
6. f(7)=5\begin{align*}f(-7)=5\end{align*} and f(3)=4\begin{align*}f(3)=-4\end{align*}  (video solution @7:40)
7. f(6)=0\begin{align*}f(6)=0\end{align*} and f(0)=6\begin{align*}f(0)=6\end{align*}
8. f(32)=0\begin{align*}f(32)=0\end{align*} and f(77)=25\begin{align*}f(77)=25\end{align*}

Objective 2:  Write the equation of a line in point-slope form given a point and a slope
Write the equation, in point slope form, of each line being described.

1. The slope is 13\begin{align*}\frac{1}{3}\end{align*}; the y\begin{align*}y-\end{align*}intercept is –4.
2. The slope is 110\begin{align*}-\frac{1}{10}\end{align*} and contains the point (10, 2). (video solution @0:00)
3. The slope is –75 and contains the point (0, 125).
4. The slope is 10 and contains the point (8, –2).
5. The line has a slope of 35\begin{align*}\frac{3}{5}\end{align*} and a y\begin{align*}y-\end{align*}intercept of –3.  (video solution @5:27)
6. The line has a slope of –6 and a y\begin{align*}y-\end{align*}intercept of 0.5.
7. m=15\begin{align*}m=-\frac{1}{5}\end{align*} and f(0)=7\begin{align*}f(0)=7\end{align*}
8. m=12\begin{align*}m=-12\end{align*} and \begin{align*}f(-2)=5\end{align*}
9. \begin{align*}m=3\end{align*} and \begin{align*}f(2)=-9\end{align*}
10. \begin{align*}m=-\frac{9}{5}\end{align*} and \begin{align*}f(0)=32\end{align*}
11. \begin{align*}m=25\end{align*} and \begin{align*}f(0)=250\end{align*}

Objective 3:  Determine a point and slope from an equation in point-slope form
Use the equation to determine the coordinates of a point on the line and the slope of the line.

1. \begin{align*}y-5=3(x-2)\end{align*}
2. \begin{align*}y-1=-5(x+7)\end{align*}
3. \begin{align*}y+1=\frac{-1}{6}(x-12)\end{align*}
4. \begin{align*}y+15=\frac{8}{9}(x+43)\end{align*}

Objective 4:  Write the equation of a line in point-slope form given the graph of the line

Write the equation, in point slope form, of the line whose graph is shown.

Objective 5:  Graph a line given an equation in point-slope form

Graph each line.

1. \begin{align*}y-5=3(x-2)\end{align*}
2. \begin{align*}y-8=-5(x+7)\end{align*}
3. \begin{align*}y+1=\frac{-5}{6}(x+7)\end{align*}

Other problems

Rewrite each equation in point slope form.

1. \begin{align*}y-2=3(x-1)\end{align*}
2. \begin{align*}y+4=\frac{-2}{3}(x+6)\end{align*}
3. \begin{align*}0=x+5\end{align*}
4. \begin{align*}y=\frac{1}{4}(x-24)\end{align*}

Answers to all the practice problems above are located in this document (pdf).

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