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# Equations of Perpendicular Lines

## Lines meeting at right angles have slopes that are negative reciprocals

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Equations of Perpendicular Lines

Suppose a coordinate grid were transposed over the blueprint for a house under construction, and two lines on the blueprint were perpendicular to each other. If one of the lines had the equation y=12x+1\begin{align*}y= \frac{1}{2} x + 1\end{align*}, and the other line passed through the point (3, 3), what would be the equation of the second line? After completing this Concept, given the equation of one line, you'll be able to find the equation of a second line that is perpendicular to the first line, as long as you know a point on the second line.

### Guidance

Lines can be parallel, coincident (overlap each other), or intersecting (crossing). Lines that intersect at 90\begin{align*}90^\circ\end{align*} angles have a special name: perpendicular lines. The slopes of perpendicular lines have a special property.

Perpendicular lines form a right angle. The product of their slopes is –1.

m1m2=1\begin{align*}m_1 \cdot m_2=-1\end{align*}

#### Example A

Verify that the following lines are perpendicular.

Line a\begin{align*}a\end{align*}: passes through points (–2, –7) and (1, 5)

Line b\begin{align*}b\end{align*}: passes through points (4, 1) and (–8, 4)

Solution: Find the slopes of each line.

Line a:5(7)1(2)=123=41Line b:4184=312=14\begin{align*}Line \ a: \frac{5-(-7)}{1-(-2)}=\frac{12}{3}=\frac{4}{1} && Line \ b: \frac{4-1}{-8-4}=\frac{3}{-12}=\frac{-1}{4}\end{align*}

To verify that the lines are perpendicular, the product of their slopes must equal –1.

41×14=1\begin{align*}\frac{4}{1} \times \frac{-1}{4}=-1\end{align*}

Because the product of their slopes is 1\begin{align*}-1\end{align*}, lines a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} are perpendicular.

#### Example B

Determine whether the two lines are parallel, perpendicular, or neither:

Line 1: 2x=y10\begin{align*}2x=y-10\end{align*}; Line 2: y=2x+5\begin{align*}y=-2x+5\end{align*}

Solution: Begin by finding the slopes of lines 1 and 2.

2x+102x+10=y10+10=y\begin{align*}2x+10& =y-10+10\\ 2x+10& =y\end{align*}

The slope of the first line is 2.

y=2x+5\begin{align*}y=-2x+5\end{align*}

The slope of the second line is –2.

These slopes are not identical, so these lines are not parallel.

To check if the lines are perpendicular, find the product of the slopes. 2×2=4\begin{align*}2 \times -2=-4\end{align*}. The product of the slopes is not –1, so the lines are not perpendicular.

Lines 1 and 2 are neither parallel nor perpendicular.

Writing Equations of Perpendicular Lines

Writing equations of perpendicular lines is slightly more difficult than writing parallel line equations. The reason is because you must find the slope of the perpendicular line before you can proceed with writing an equation.

#### Example C

Find the equation of the line perpendicular to the line y=3x+5\begin{align*}y=-3x+5\end{align*} that passes through point (2, 6).

Solution: Begin by finding the slopes of the perpendicular line. Using the perpendicular line definition, m1m2=1\begin{align*}m_1 \cdot m_2=-1\end{align*}. The slope of the original line is –3. Substitute that for m1\begin{align*}m_1\end{align*}.

3m2=1\begin{align*}-3 \cdot m_2=-1\end{align*}

Solve for m2\begin{align*}m_2\end{align*}, the slope of the perpendicular line.

3m23m2=13=13\begin{align*}\frac{-3m_2}{-3}& =\frac{-1}{-3}\\ m_2& =\frac{1}{3}\end{align*}

The slope of the line perpendicular to y=3x+5\begin{align*}y=-3x+5\end{align*} is 13\begin{align*}\frac{1}{3}\end{align*}.

You now have the slope and a point. Use point-slope form to write its equation.

y6=13(x2)\begin{align*}y-6= \frac{1}{3}(x-2)\end{align*}

You can rewrite this in slope-intercept form: y=13x23+6\begin{align*}y=\frac{1}{3} x-\frac{2}{3}+6\end{align*}.

y=13x+163\begin{align*}y=\frac{1}{3} x+\frac{16}{3}\end{align*}

Multimedia Link: For more help with writing lines, visit AlgebraLab.

### Guided Practice

Find the equation of the line perpendicular to the line y=5\begin{align*}y=5\end{align*} and passing through (5, 4).

Solution:

The line y=5\begin{align*}y=5\end{align*} is a horizontal line with a slope of zero.

Lines that make a 90\begin{align*}90^\circ\end{align*} angle with a horizontal line are vertical lines.

Vertical lines are in the form x=constant\begin{align*}x=constant\end{align*}.

Since the vertical line must go through (5, 4), the equation is x=5\begin{align*}x=5\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 perpendicular lines.
2. What is true about the slopes of perpendicular lines?

Determine the slope of a line perpendicular 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*}

In 8 – 14, determine whether the lines are parallel, perpendicular, or neither.

1. Line a:\begin{align*}a:\end{align*} passing through points (–1, 4) and (2, 6); Line b:\begin{align*}b:\end{align*} passing through points (2, –3) and (8, 1).
2. Line a:\begin{align*}a:\end{align*} passing through points (4, –3) and (–8, 0); Line b:\begin{align*}b:\end{align*} passing through points (–1, –1) and (–2, 6).
3. Line a:\begin{align*}a:\end{align*} passing through points (–3, 14) and (1, –2); Line b:\begin{align*}b:\end{align*} passing through points (0, –3) and (–2, 5).
4. Line a:\begin{align*}a:\end{align*} passing through points (3, 3) and (–6, –3); Line b:\begin{align*}b:\end{align*} passing through points (2, –8) and (–6, 4).
5. Line 1: 4y+x=8\begin{align*}4y+x=8\end{align*}; Line 2: 12y+3x=1\begin{align*}12y+3x=1\end{align*}
6. Line 1: 5y+3x+1\begin{align*}5y+3x+1\end{align*}; Line 2: 6y+10x=3\begin{align*}6y+10x=-3\end{align*}
7. Line 1: 2y3x+5=0\begin{align*}2y-3x+5=0\end{align*}; Line 2: y+6x=3\begin{align*}y+6x=-3\end{align*}

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

1. \begin{align*}x+4y=12; (-3,-2)\end{align*}
2. \begin{align*}y=\frac{1}{3}x+2; (-3,-1)\end{align*}
3. \begin{align*}y=\frac{3}{5}x-4; (6,-2)\end{align*}
4. \begin{align*}2x+y=5; (2,-2)\end{align*}
5. \begin{align*}y=x-6; (-2,0)\end{align*}
6. \begin{align*}5x-7=3y; (8,-2)\end{align*}
7. \begin{align*}y=\frac{2}{3}x-1; (4,7)\end{align*}

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