5.4: Equations of Parallel and Perpendicular Lines
In a previous lesson, you learned how to identify parallel lines.
Parallel lines have the same slope.
In the following graph all the lines have the same slope. According to the definition, all these lines are parallel.
Example 1: Are \begin{align*}y=\frac{1}{3} x-4\end{align*}
Solution: The slope of the first line is \begin{align*}\frac{1}{3}\end{align*}
Find the slope of the second equation: \begin{align*}A=-3, B=9\end{align*}
\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.
Slopes of Perpendicular Lines
Lines can be parallel, coincident (overlap each other), or intersecting (crossing). Lines that intersect at \begin{align*}90^\circ\end{align*}
Perpendicular lines form a right angle. The product of their slopes is –1.
\begin{align*}m_1 \cdot m_2=-1\end{align*}
Example 2: Verify that the following lines are perpendicular.
Line \begin{align*}a\end{align*}
Line \begin{align*}b\end{align*}
Solution: Find the slopes of each line.
\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.
\begin{align*}\frac{4}{1} \times \frac{-1}{4}=-1\end{align*}
Because the product of their slopes is \begin{align*}-1\end{align*}
Example 3: Determine whether the two lines are parallel, perpendicular, or neither:
Line 1: \begin{align*}2x=y-10\end{align*}
Solution: Begin by finding the slopes of lines 1 and 2.
\begin{align*}2x+10& =y-10+10\\ 2x+10& =y\end{align*}
The slope of the first line is 2.
\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. \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 Parallel Lines
Example 4: Find the equation parallel to the line \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.
\begin{align*}y-y_1& =m(x-x_1)\\ y-4& =6(x+1)\end{align*}
You could rewrite it in slope-intercept form:
\begin{align*}y& =6x+6+4\\ y& =6x+10\end{align*}
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: Find the equation perpendicular to the line \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, \begin{align*}m_1 \cdot m_2=-1\end{align*}. The slope of the original line is –3. Substitute that for \begin{align*}m_1\end{align*}.
\begin{align*}-3 \cdot m_2=-1\end{align*}
Solve for \begin{align*}m_2\end{align*}, the slope of the perpendicular line.
\begin{align*}\frac{-3m_2}{-3}& =\frac{-1}{-3}\\ m_2& =\frac{1}{3}\end{align*}
The slope of the line perpendicular to \begin{align*}y=-3x+5\end{align*} is \begin{align*}\frac{1}{3}\end{align*}.
You now have the slope and a point. Use point-slope form to write its equation.
\begin{align*}y-6= \frac{1}{3}(x-2)\end{align*}
You can rewrite this in slope-intercept form: \begin{align*}y=\frac{1}{3} x-\frac{2}{3}+6\end{align*}.
\begin{align*}y=\frac{1}{3} x+\frac{16}{3}\end{align*}
Example 4: Find the equation of the line perpendicular to the line \begin{align*}y=5\end{align*} and passing through (5, 4).
Solution: The line \begin{align*}y=5\end{align*} is a horizontal line with slope of zero. The only thing that makes a \begin{align*}90^\circ\end{align*} angle with a horizontal line is a vertical line. Vertical lines have undefined slopes.
Since the vertical line must go through (5, 4), the equation is \begin{align*}x=5\end{align*}.
Multimedia Link: For more help with writing lines, visit AlgebraLab.
Families of Lines
A straight line has two very important properties, its slope and its \begin{align*}y-\end{align*}intercept. The slope tells us how steeply the line rises or falls, and the \begin{align*}y-\end{align*}intercept tells us where the line intersects the \begin{align*}y-\end{align*}axis. In this section, we will look at two families of lines.
A family of lines is a set of lines that have something in common with each other. Straight lines can belong to two types of families: where the slope is the same and where the \begin{align*}y-\end{align*}intercept is the same.
Family 1: The slope is the same
Remember that lines with the same slope are parallel. Each line on the Cartesian plane below has an identical slope with different \begin{align*}y-\end{align*}intercepts. All the lines look the same but they are shifted up and down the \begin{align*}y-\end{align*}axis. As \begin{align*}b\end{align*} gets larger the line rises on the \begin{align*}y-\end{align*}axis and as \begin{align*}b\end{align*} gets smaller the line goes lower on the \begin{align*}y-\end{align*}axis. This behavior is often called a vertical shift.
Family 2: The \begin{align*}y-\end{align*}intercept is the same
The graph below shows several lines with the same \begin{align*}y-\end{align*}intercept but varying slopes.
Practice Set
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)
- Define parallel lines.
- Define perpendicular lines.
- What is true about the slopes of perpendicular lines?
- What is a family of lines?
Determine the slope of a line a) parallel and b) perpendicular to each line given.
- \begin{align*}y=-5x+7\end{align*}
- \begin{align*}2x+8y=9\end{align*}
- \begin{align*}x=8\end{align*}
- \begin{align*}y=-4.75\end{align*}
- \begin{align*}y-2= \frac{1}{5}(x+3)\end{align*}
In 10 – 16, determine whether the lines are parallel, perpendicular, or neither.
- Line \begin{align*}a:\end{align*} passing through points (–1, 4) and (2, 6); Line \begin{align*}b:\end{align*} passing through points (2, –3) and (8, 1).
- Line \begin{align*}a:\end{align*} passing through points (4, –3) and (–8, 0); Line \begin{align*}b:\end{align*} passing through points (–1, –1) and (–2, 6).
- Line \begin{align*}a:\end{align*} passing through points (–3, 14) and (1, –2); Line \begin{align*}b:\end{align*} passing through points (0, –3) and (–2, 5).
- Line \begin{align*}a:\end{align*} passing through points (3, 3) and (–6, –3); Line \begin{align*}b:\end{align*} passing through points (2, –8) and (–6, 4).
- Line 1: \begin{align*}4y+x=8\end{align*}; Line 2: \begin{align*}12y+3x=1\end{align*}
- Line 1: \begin{align*}5y+3x=1\end{align*}; Line 2: \begin{align*}6y+10x=-3\end{align*}
- Line 1: \begin{align*}2y-3x+5=0\end{align*}; Line 2: \begin{align*}y+6x=-3\end{align*}
- Find the equation of the line parallel to \begin{align*}5x-2y=2\end{align*} that passes through point (3, –2).
- Find the equation of the line perpendicular to \begin{align*}y=-\frac{2}{5}x-3\end{align*} that passes through point (2, 8).
- Find the equation of the line parallel to \begin{align*}7y+2x-10=0\end{align*} that passes through the point (2, 2).
- Find the equation of the line perpendicular to \begin{align*}y+5=3(x-2)\end{align*} that passes through the point (6, 2).
- Find the equation of the line through (2, –4) perpendicular to \begin{align*}y=\frac{2}{7} x+3\end{align*}.
- Find the equation of the line through (2, 3) parallel to \begin{align*}y=\frac{3}{2} x+5\end{align*}.
In 23 – 26, write the equation of the family of lines satisfying the given condition.
- All lines pass through point (0, 4).
- All lines are perpendicular to \begin{align*}4x+3y-1=0\end{align*}.
- All lines are parallel to \begin{align*}y-3=4x+2\end{align*}.
- All lines pass through point (0, –1).
- Write an equation for a line parallel to the equation graphed below.
- Write an equation for a line perpendicular to the equation graphed below and passing through the ordered pair (0, –1).
Mixed Review
- Graph the equation \begin{align*}2x-y=10\end{align*}.
- 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?
- The amount of money charged for a classified advertisement is directly proportional to the length of the advertisement. If a 50-word advertisement costs $11.50, what is the cost of a 70-word advertisement?
- Simplify \begin{align*}\sqrt{112}\end{align*}.
- Simplify \begin{align*}\sqrt{12^2-7^2}\end{align*}.
- Is \begin{align*}\sqrt{3}-\sqrt{2}\end{align*} rational, irrational, or neither? Explain your answer.
- Solve for \begin{align*}s: \ 15s=6(s+32)\end{align*}.
Quick Quiz
1. Write an equation for a line with slope of \begin{align*}\frac{4}{3}\end{align*} and \begin{align*}y-\end{align*}intercept (0, 8).
2. Write an equation for a line containing (6, 1) and (7, –3).
3. A plumber charges $75 for a 2.5-hour job and $168.75 for a 5-hour job.
Assuming the situation is linear, write an equation to represent the plumber’s charge and use it to predict the cost of a 1-hour job.
4. Rewrite in standard form: \begin{align*}y=\frac{6}{5} x+11\end{align*}.
5. Sasha took tickets for the softball game. Student tickets were $3.00 and adult tickets were $3.75. She collected a total of $337.50 and sold 75 student tickets. How many adult tickets were sold?
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Feb 22, 2012Last Modified:
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