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# Equations of Parallel Lines

## Use matching slopes to find the equations of parallel lines

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Practice Equations of Parallel Lines
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Equations of Parallel Lines

Suppose a coordinate plane were transposed over the map of a city, and Main Street had the equation y=2x+5\begin{align*}y=2x+5\end{align*}. If Broad Street ran parallel to Main Street, and it passed through the point (3, 14), what would be the equation of Broad Street? How do you know? In this Concept, given the equation of one line, you'll learn how to find the equation of a second line that is parallel to the first line, as long as you know a point that the second line passes through.

### Guidance

In a previous Concept, you learned how to identify parallel lines.

Parallel lines have the same slope.

Each of the graphs below have the same slope, which is 2. According to the definition, all these lines are parallel.

#### Example A

Are y=13x4\begin{align*}y=\frac{1}{3} x-4\end{align*} and 3x+9y=18\begin{align*}-3x+9y=18\end{align*} parallel?

Solution: The slope of the first line is 13\begin{align*}\frac{1}{3}\end{align*}. Any line parallel to this must also have a slope of 13\begin{align*}\frac{1}{3}\end{align*}.

Find the slope of the second equation: A=3\begin{align*}A=-3\end{align*} and B=9\begin{align*}B=9\end{align*}.

slope=AB=3913\begin{align*}slope=\frac{-A}{B}=\frac{3}{9} \rightarrow \frac{1}{3}\end{align*}

These two lines have the same slope so they are parallel.

Writing Equations of Parallel Lines

Sometimes, you will asked to write the equation of a line parallel to a given line that goes through a given point. In the following example, you will see how to do this.

#### Example B

Find the equation parallel to the line y=6x9\begin{align*}y=6x-9\end{align*} passing through (–1, 4).

Solution:

Parallel lines have the same slope, so the slope will be 6. You have a point and the slope, so you can use point-slope form.

yy1y4=m(xx1)=6(x+1)\begin{align*}y-y_1& =m(x-x_1)\\ y-4& =6(x+1)\end{align*}

You could rewrite it in slope-intercept form:

yy=6x+6+4=6x+10\begin{align*}y& =6x+6+4\\ y& =6x+10\end{align*}

#### Example C

Find the equation of the line parallel to the line y5=2(x+3)\begin{align*}y-5=2(x+3)\end{align*} passing through (1, 1).

Solution:

First, we notice that this equation is in point-slope form, so let's use point-slope form to write this equation.

yy1=m(xx1)y1=2(x1)y1=2x2y1+1=2x2+1,y=2x1 Starting with point-slope form. Substituting in the slope and point. Distributing on the left. Rearranging into slope-intercept form.\begin{align*}y-y_1=m(x-x_1)&& \ \text{Starting with point-slope form}. \\ y-1=2(x-1)&& \ \text{Substituting in the slope and point}. \\ y-1=2x-2&& \ \text{Distributing on the left}.\\ y-1+1=2x-2+1, && \ \text{Rearranging into slope-intercept form}.\\ y=2x-1 \end{align*}

### Guided Practice

Find the equation of the line parallel to the line 2x3y=24\begin{align*}2x-3y=24\end{align*} passing through (2, -6).

Solution:

Since this is in standard form, we must first find the slope. For Ax+By=C\begin{align*}Ax+By=C\end{align*}, recall that the slope is m=AB\begin{align*}m=-\frac{A}{B}\end{align*}. Since A=2\begin{align*}A=2\end{align*} and B=3\begin{align*}B=-3\end{align*}:

m=AB=23=23.\begin{align*}m=-\frac{A}{B}=-\frac{2}{-3}=\frac{2}{3}.\end{align*}

Now that we have the slope, we can plug it in:

yy1=m(xx1)y2=23(x+6)y2=4x12y2+2=4x12+2,y=4x10 Starting with point-slope form. Substituting in the slope and point. Distributing on the left. Rearranging into slope-intercept form.\begin{align*}y-y_1=m(x-x_1)&& \ \text{Starting with point-slope form}. \\ y-2=\frac{2}{3}(x+6)&& \ \text{Substituting in the slope and point}. \\ y-2=4x-12&& \ \text{Distributing on the left}.\\ y-2+2=4x-12+2, && \ \text{Rearranging into slope-intercept form}.\\ y=4x-10 \end{align*}

### Practice

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Equations of Parallel and Perpendicular Lines (9:13)

1. Define parallel lines.

Determine the slope of a line parallel to each line given.

1. y=5x+7\begin{align*}y=-5x+7\end{align*}
2. 2x+8y=9\begin{align*}2x+8y=9\end{align*}
3. x=8\begin{align*}x=8\end{align*}
4. y=4.75\begin{align*}y=-4.75\end{align*}
5. y2=15(x+3)\begin{align*}y-2= \frac{1}{5}(x+3)\end{align*}

For the following equations, find the line parallel to it through the given point.

1. y=35x+2;(0,2)\begin{align*}y=-\frac{3}{5}x+2; (0,-2)\end{align*}
2. 5x2y=7;(2,10)\begin{align*}5x-2y=7; (2,-10)\end{align*}
3. x=y;(2,3)\begin{align*}x=y; (2,3)\end{align*}
4. x=5;(2,3)\begin{align*}x=-5; (-2,-3) \end{align*}

Mixed Review

1. Graph the equation 2xy=10\begin{align*}2x-y=10\end{align*}.
2. On a model boat, the stack is 8 inches high. The actual stack is 6 feet tall. How tall is the mast on the model if the actual mast is 40 feet tall?
4. Simplify 112\begin{align*}\sqrt{112}\end{align*}.
5. Simplify 12272\begin{align*}\sqrt{12^2-7^2}\end{align*}.
6. Is 32\begin{align*}\sqrt{3}-\sqrt{2}\end{align*} rational, irrational, or neither? Explain your answer.
7. Solve for s: 15s=6(s+32)\begin{align*}s: \ 15s=6(s+32)\end{align*}.