# 5.1: Determining the Equation of a Line

**At Grade**Created by: CK-12

**Practice**Determining the Equation of a Line

What if you were given the slope of a line and either its *y*-intercept or one of its points? Or what if you were given two of its points? How could you write the equation of that line? After completing this Concept, you'll be able to write and graph equations from such information.

### Watch This

CK-12 Foundation: 0501S Linear Equations (H264)

### Try This

Another applet at http://www.cut-the-knot.org/Curriculum/Calculus/StraightLine.shtml lets you create multiple lines and see how they intersect. Each line is defined by two points; you can change the slope of a line by moving either of the points, or just drag the whole line around without changing its slope. To create another line, just click Duplicate and then drag one of the lines that are already there.

### Guidance

We saw in the last chapter that many real-world situations can be described with linear graphs and equations. In this chapter, we’ll see how to find those equations in a variety of situations.

**Write an Equation Given Slope and \begin{align*}y-\end{align*} y−Intercept**

You’ve already learned how to write an equation in slope–intercept form: simply start with the general equation for the slope-intercept form of a line, \begin{align*}y=mx+b\end{align*}

If you are given just the graph of a line, you can read off the slope and \begin{align*}y-\end{align*}

**Write an Equation Given the Slope and a Point**

Often, we don’t know the value of the \begin{align*}y-\end{align*}**point-slope form.** An equation in point-slope form is written as \begin{align*}y-y_0=m(x-x_0)\end{align*}

#### Example A

*A line has a slope of \begin{align*}\frac{3}{5}\end{align*}, and the point (2, 6) is on the line. Write the equation of the line in point-slope form.*

**Solution**

Start with the formula \begin{align*}y-y_0=m(x-x_0)\end{align*}.

Plug in \begin{align*}\frac{3}{5}\end{align*} for \begin{align*}m\end{align*}, 2 for \begin{align*}x_0\end{align*} and 6 for \begin{align*}y_0\end{align*}.

**The equation in point-slope form is** \begin{align*}y-6=\frac{3}{5}(x-2)\end{align*}.

Notice that the equation in point-slope form is not solved for \begin{align*}y\end{align*}. If we did solve it for \begin{align*}y\end{align*}, we’d have it in \begin{align*}y-\end{align*}intercept form. To do that, we would just need to distribute the \begin{align*}\frac{3}{5}\end{align*} and add 6 to both sides. That means that the equation of this line in slope-intercept form is \begin{align*}y=\frac{3}{5}x-\frac{6}{5}+6\end{align*}, or simply \begin{align*}y=\frac{3}{5}x+\frac{24}{5}\end{align*}.

**Write an Equation Given Two Points**

Point-slope form also comes in useful when we need to find an equation given just two points on a line.

For example, suppose we are told that the line passes through the points (-2, 3) and (5, 2). To find the equation of the line, we can start by finding the slope.

Starting with the slope formula, \begin{align*}m=\frac{y_2-y_1}{x_2-x_1}\end{align*}, we plug in the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}values of the two points to get \begin{align*}m=\frac{2-3}{5-(-2)}=\frac{-1}{7}\end{align*}. We can plug that value of \begin{align*}m\end{align*} into the point-slope formula to get \begin{align*}y-y_0=-\frac{1}{7}(x-x_0)\end{align*}.

Now we just need to pick one of the two points to plug into the formula. Let’s use (5, 2); that gives us \begin{align*}y-2=-\frac{1}{7}(x-5)\end{align*}.

What if we’d picked the other point instead? Then we’d have ended up with the equation \begin{align*}y-3=-\frac{1}{7}(x+2)\end{align*}, which doesn’t look the same. That’s because there’s more than one way to write an equation for a given line in point-slope form. But let’s see what happens if we solve each of those equations for \begin{align*}y\end{align*}.

Starting with \begin{align*}y-2=-\frac{1}{7}(x-5)\end{align*}, we distribute the \begin{align*}-\frac{1}{7}\end{align*} and add 2 to both sides. That gives us \begin{align*}y=-\frac{1}{7} x+\frac{5}{7}+2\end{align*}, or \begin{align*}y=-\frac{1}{7}x+\frac{19}{7}\end{align*}.

On the other hand, if we start with \begin{align*}y-3=-\frac{1}{7}(x+2)\end{align*}, we need to distribute the \begin{align*}-\frac{1}{7}\end{align*} and add 3 to both sides. That gives us \begin{align*}y=-\frac{1}{7}x-\frac{2}{7}+3\end{align*}, which also simplifies to \begin{align*}y=-\frac{1}{7}x+\frac{19}{7}\end{align*}.

So whichever point we choose to get an equation in point-slope form, the equation is still mathematically the same, and we can see this when we convert it to \begin{align*}y-\end{align*}intercept form.

#### Example B

*A line contains the points (3, 2) and (-2, 4). Write an equation for the line in point-slope form; then write an equation in \begin{align*}y-\end{align*}intercept form.*

**Solution**

Find the slope of the line: \begin{align*}m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-2}{-2-3}=-\frac{2}{5}\end{align*}

Plug in the value of the slope: \begin{align*}y-y_0=-\frac{2}{5}(x-x_0)\end{align*}.

Plug point (3, 2) into the equation: \begin{align*}y-2=-\frac{2}{5}(x-3)\end{align*}.

**The equation in point-slope form is** \begin{align*}y-2=-\frac{2}{5}(x-3)\end{align*}.

To convert to \begin{align*}y-\end{align*}intercept form, simply solve for \begin{align*}y\end{align*}:

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

**The equation in \begin{align*}y-\end{align*}intercept form is \begin{align*}y=-\frac{2}{5}x+3\frac{1}{5}\end{align*}.**

**Graph an Equation in Point-Slope Form**

Another useful thing about point-slope form is that you can use it to graph an equation without having to convert it to slope-intercept form. From the equation \begin{align*}y-y_0=m(x-x_0)\end{align*}, you can just read off the slope \begin{align*}m\end{align*} and the point \begin{align*}(x_0, y_0)\end{align*}. To draw the graph, all you have to do is plot the point, and then use the slope to figure out how many units up and over you should move to find another point on the line.

#### Example C

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

**Solution**

To read off the right values, we need to rewrite the equation slightly: \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 is \begin{align*}\frac{2}{3}\end{align*}.

First plot point (2, -2) on the graph:

A slope of \begin{align*}\frac{2}{3}\end{align*} tells you that from that 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 it in both directions:

Watch this video for help with the Examples above.

CK-12 Foundation: Linear Equations

### Vocabulary

- Often, we don’t know the value of the \begin{align*}y-\end{align*}intercept, but we know the value of \begin{align*}y\end{align*} for a non-zero value of \begin{align*}x\end{align*}. In this case, it’s often easier to write an equation of the line in
**point-slope form.**An equation in point-slope form is written as \begin{align*}y-y_0=m(x-x_0)\end{align*}, where \begin{align*}m\end{align*} is the slope and \begin{align*}(x_0, y_0)\end{align*} is a point on the line.

### Guided Practice

*A line contains the points (1, -2) and (0, 0). Write an equation for the line in point-slope form; then write an equation in \begin{align*}y-\end{align*}intercept form.*

**Solution**

Find the slope of the line: \begin{align*}m=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-0}{1-0}=\frac{-2}{1}=-2\end{align*}

Plug in the value of the slope: \begin{align*}y-y_0=-2(x-x_0)\end{align*}.

Plug point (1, -2) into the equation: \begin{align*}y-(-2)=-2(x-1)\end{align*}.

**The equation in point-slope form is** \begin{align*}y+2=-2(x-1)\end{align*}.

To convert to \begin{align*}y-\end{align*}intercept form, simply solve for \begin{align*}y\end{align*}:

\begin{align*}y+2=-2(x-1) &\rightarrow y+2=-2x+2\\ &\rightarrow y=-2x+2-2\\ &\rightarrow y=-2x.\end{align*}

**The equation in \begin{align*}y-\end{align*}intercept form is \begin{align*}y=-2x\end{align*}.**

### Practice

Find the equation of each line in slope–intercept form.

- The line has a slope of 7 and a \begin{align*}y-\end{align*}intercept of -2.
- The line has a slope of -5 and a \begin{align*}y-\end{align*}intercept of 6.
- The line has a slope of \begin{align*}-\frac{1}{4}\end{align*} and contains the point (4, -1).
- The line contains points (3, 5) and (-3, 0).
- The line contains points (10, 15) and (12, 20).

Write the equation of each line in slope-intercept form.

Find the equation of each linear function in slope–intercept form.

- \begin{align*}m=5, f(0)=-3\end{align*}
- \begin{align*}m=-7, f(2)=-1\end{align*}
- \begin{align*}m=\frac{1}{3}, f(-1)=\frac{2}{3}\end{align*}
- \begin{align*}m=4.2, f(-3)=7.1\end{align*}
- \begin{align*}f \left(\frac{1}{4}\right)=\frac{3}{4}, f(0)=\frac{5}{4}\end{align*}
- \begin{align*}f(1.5)=-3, f(-1)=2\end{align*}

Write the equation of each line in point-slope form.

- The line has slope \begin{align*}-\frac{1}{10}\end{align*} and goes through the point (10, 2).
- The line has slope -75 and goes through the point (0, 125).
- The line has slope 10 and goes through the point (8, -2).
- The line goes through the points (-2, 3) and (-1, -2).
- The line contains the points (10, 12) and (5, 25).
- The line goes through the points (2, 3) and (0, 3).
- The line has a slope of \begin{align*}\frac{3}{5}\end{align*} and a \begin{align*}y-\end{align*}intercept of -3.
- The line has a slope of -6 and a \begin{align*}y-\end{align*}intercept of 0.5.

Write the equation of each linear function in point-slope form.

- \begin{align*}m=-\frac{1}{5}\end{align*} and \begin{align*}f(0)=7\end{align*}
- \begin{align*}m=-12\end{align*} and \begin{align*}f(-2)=5\end{align*}
- \begin{align*}f(-7)=5\end{align*} and \begin{align*}f(3)=-4\end{align*}
- \begin{align*}f(6)=0\end{align*} and \begin{align*}f(0)=6\end{align*}
- \begin{align*}m=3\end{align*} and \begin{align*}f(2)=-9\end{align*}
- \begin{align*}m=-\frac{9}{5}\end{align*} and \begin{align*}f(0)=32\end{align*}

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### Image Attributions

Here you'll learn how to write the equations of lines given their slope and @$\begin{align*}y\end{align*}@$-intercept or two of their points.