7. Parallel and Perpendicular Lines
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Chapter 2
7.
Parallel and Perpendicular Lines
Parallel and perpendicular lines are fundamental concepts in geometry. Parallel lines are two coplanar lines that never intersect, while perpendicular lines intersect at a right angle. In a coordinate plane, the relationship between the slopes of these lines helps in determining their parallel or perpendicular nature. For instance, parallel lines have the same slope, whereas the slopes of perpendicular lines are negative reciprocals of each other. These relationships are vital when graphing linear relationships or analyzing geometric figures. Recognizing and understanding these lines can aid in various real-world applications, such as designing community gardens or planning urban infrastructure. For example, in a community garden design scenario, stone paths might be laid out in parallel or perpendicular patterns to create an aesthetically pleasing and functional layout.
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Lesson Settings & Tools
| | 14 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Parallel and Perpendicular Lines
Slide of 14
Two different lines on the same plane can either intersect or not. If the lines do not intersect, then they are parallel. If the lines do intersect, they might be perpendicular. It can be useful to know how to identify parallel and perpendicular lines, which will be explored in this lesson.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Observing Parallel Lines
Consider the following applet showing two parallel lines and their equations in slope-intercept form.
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Discussion
Parallel Lines
Two coplanar lines — lines that are on the same plane — that do not intersect are said to be parallel lines. In a diagram, triangular hatch marks are drawn on lines to denote that they are parallel. The symbol ∥
is used to algebraically denote that two lines are parallel. In the diagram, lines m and l are parallel.
Rule
Slopes of Parallel Lines Theorem
In a coordinate plane, two distinct non-vertical lines are parallel if and only if their slopes are equal.
If l_1 and l_2 are two parallel lines and m_1 and m_2 their respective slopes, then the following statement is true.
l_1 ∥ l_2 ⇔ m_1 = m_2
The slope of a vertical line is not defined. Therefore, this theorem only applies to non-vertical lines. However, any two distinct vertical lines are parallel.
Proof
Since the theorem consists of a biconditional statement, the proof consists of two parts.
- If two distinct non-vertical lines are parallel, then their slopes are equal.
- If the slopes of two distinct non-vertical lines are equal, then the lines are parallel.
Part 1
Consider two distinct non-vertical parallel lines in a coordinate plane. Their equations can be written in slope-intercept form. l_1: y=m_1x+b_1 l_2: y=m_2x+b_2 Suppose that the slopes of the lines are not the same. The system of equations formed by the equations above can be solved by using the Substitution Method.
y=m_1x+b_1 & (I) y=m_2x+b_2 & (II)
Substitute
(I): y= m_2x+b_2
m_2x+b_2=m_1x+b_1 y=m_2x+b_2
x= b_1-b_2m_2-m_1 y=m_2x+b_2
Since m_1≠ m_2, the expression b_1-b_2m_2-m_1 is not undefined because its denominator cannot be zero. To find the value of the y-variable, b_1-b_2m_2-m_1 can be substituted for x in Equation (II).
x= b_1-b_2m_2-m_1 y=m_2x+b_2
Substitute
(II): x= b_1-b_2/m_2-m_1
x= b_1-b_2m_2-m_1 y=m_2( b_1-b_2m_2-m_1)+b_2
The solution to the system formed by the equations was found. Since there is a solution for the system, the lines l_1 and l_2 intersect each other. However, this contradicts the fact that the lines are parallel. Therefore, the assumption that the slopes are different is false. Consequently, the slopes of the lines are equal.
l_1 ∥ l_2 ⇒ m_1 = m_2
Part 2
Now, consider two distinct non-vertical lines l_1 and l_2 that have the same slope m. Their equations can be written in slope-intercept form. l_1: y=mx+b_1 l_2: y=mx+b_2 Since these lies are distinct, b_1 and b_2 are not equal. With this information in mind, suppose that the lines intersect. Solving the system of equations will give the point of intersection. The Substitution Method will be used again.
y=mx+b_1 & (I) y=mx+b_2 & (II)
Substitute
(I): y= mx+b_2
mx+b_2=mx+b_1 y=mx+b_2
SubEqn
(I): LHS-mx=RHS-mx
b_2=b_1 y=mx+b_2
The obtained result contradicts the fact that b_1 and b_2 are different. Therefore, there is no point of intersection between the lines l_1 and l_2. This means that they are parallel lines.
m_1 = m_2 ⇒ l_1 ∥ l_2
Both directions of the biconditional statement have been proved.
l_1 ∥ l_2 ⇔ m_1 = m_2
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Pop Quiz
Identifying Parallel Lines
Consider two lines on the same coordinate plane and their equations in slope-intercept form. Are the lines parallel?
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Example
Finding a Parallel Line Through a Point
Write an equation in slope-intercept form of the line that passes through (- 2,3) and is parallel to y=2x-1.
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Hint
What do parallel lines have in common?
Solution
Consider the given equation of the given line.
y= 2x-1
Parallel lines have the same slope. Therefore, the slope of all lines parallel to the given line is 2. A general equation in slope-intercept form for these lines can be written.
y= 2x+ b
The desired parallel line passes through ( - 2, 3). By substituting the coordinates of this point into the above equation for x and y, the y-intercept b of the parallel line can be determined.
Knowing the y-intercept, the equation of the line parallel to y=2x-1 through (- 2,3) can be written. y= 2x+ 7 The solution can be verified by graphing the line on the same coordinate plane.
As seen above, the graph of y=2x+7 passes through (- 2,3) and is parallel to the graph of y=2x-1.
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Example
Bike Path
Kevin's class is entering a contest to redesign parts of their town. They are using new software to create a model of their ideas for changes to the park. Kevin wants to propose a new bike path in the park. The bike path has to be parallel to an already existing pavement. He visualized the situation on the following graph.
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Hint
Find the slope of the line shown on the graph.
Solution
Two parallel lines have the same slope. To find the equation of the line that corresponds to the bike path, the slope of the line that represents the pavement has to be determined first. This can be done by using the Slope Formula.
m = y_2-y_1/x_2-x_1
In the above formula, (x_1,y_1) and (x_2,y_2) are two points on the line. Therefore, two points on the line corresponding to the pavement have to be identified.
Now (0,3) and (1,1) will be substituted for (x_1,y_1) and (x_2,y_2), respectively.
m = y_2-y_1/x_2-x_1
SubstitutePoints
Substitute ( 0,3) & ( 1,1)
m=1- 3/1- 0
SubTerms
Subtract terms
m=- 2/1
DivByOne
a/1=a
m= - 2
The slope of the pavement is - 2. As mentioned before, parallel lines have the same slope. Therefore, all lines parallel to the line that corresponds to the pavement will have a slope of - 2. A general equation in slope-intercept form for these lines can be written. y= - 2x+ b The line that represents the bike path passes through ( - 4, 3.5). By substituting the coordinates of this point into the above equation for x and y, the y-intercept b of the parallel line can be determined.
Knowing the y-intercept, the equation of the line parallel to the pavement through (- 4,3.5) can be written. y= - 2x+( - 4.5) ⇔ y=- 2x-4.5 The solution will be verified by plotting the line on Kevin's graph.
As seen above, the graph of y=- 2x-4.5 is parallel to the pavement and passes through (- 4,3.5).
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Example
Starry Scene
Zosia wants to propose a new mural to be painted on the side of the planetarium. She starts with a moon and two stars that are already painted on the building.
Zosia wants to place more stars in the line that connects the two existing stars. She also wants to make a second line of stars that is parallel to the first and passes through the moon. The two stars and the moon can be represented on a coordinate plane.
Write the equation of the line of stars that passes through the moon. Give the answer in slope-intercept form.
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Hint
Find the slope of the line that passes through the stars.
Solution
The coordinates of the points that represent the stars are (- 4,- 3) and (3,0.5). Since two points on the line are known, the line's slope can be found using the Slope Formula.
m = y_2-y_1/x_2-x_1
SubstitutePoints
Substitute ( - 4,- 3) & ( 3,0.5)
m=0.5-( - 3)/3-( - 4)
▼
Simplify right-hand side
SubNeg
a-(- b)=a+b
m=0.5+3/3+4
AddTerms
Add terms
m=3.5/7
ExpandFrac
a/b=a * 2/b * 2
m=7/14
ReduceFrac
a/b=.a /7./.b /7.
m= 1/2
The slope of the line that passes through the stars is 12, or 0.5. Parallel lines have the same slope. Therefore, all lines that are parallel to the line that passes through the stars will have a slope of 12. A general equation in slope-intercept form for these lines can be written. y= 1/2x+b The parallel line should pass through the moon, which in the coordinate plane has coordinates ( - 2, 1). These two numbers will be substituted into the above equation. Then, the y-intercept b of the parallel line can be determined.
y=1/2x+b
SubstituteII
x= - 2, y= 1
1=1/2( - 2)+b
▼
Solve for b
MoveRightFacToNumOne
1/b* a = a/b
1=- 2/2+b
MoveNegNumToFrac
Put minus sign in front of fraction
1=-2/2+b
QuotOne
a/a=1
1=- 1+b
AddEqn
LHS+1=RHS+1
2=b
RearrangeEqn
Rearrange equation
b=2
Now, the equation of the parallel line can be completed. y= 1/2x+2 The solution will be verified by graphing the above line on the visual representation of Zosia's proposed mural.
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Observing Perpendicular Lines
The following applet shows two perpendicular lines and their equations in slope-intercept form.
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Discussion
Perpendicular Lines
Two coplanar lines — lines that are on the same plane — that intersect at a right angle are said to be perpendicular lines. The symbol ⊥
is used to algebraically denote that two lines are perpendicular. In the diagram, lines m and l are perpendicular.
Rule
Slopes of Perpendicular Lines Theorem
In a coordinate plane, two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals.
If l_1 and l_2 are two perpendicular lines and m_1 and m_2 their respective slopes, the following relation holds true.
l_1 ⊥ l_2 ⇔ m_1 * m_2=- 1
This theorem does not apply to vertical lines because their slope is undefined. However, vertical lines are always perpendicular to horizontal lines.
Proof
Since the theorem is a biconditional statement, the proof consists of two parts.
- If two non-vertical lines are perpendicular, then the product of their slopes is - 1.
- If the product of the slopes of two non-vertical lines is - 1, then the lines are perpendicular.
Part 1
Let l_1 and l_2 be two perpendicular lines. Therefore, they intersect at one point. For simplicity, the lines will be translated so that the point of intersection is the origin.
Let m_1 and m_2 be the slopes of the lines l_1 and l_2, respectively. Next, consider the vertical line x=1. This line intersects both l_1 and l_2.
Since l_1 and l_2 are assumed to be perpendicular, △ AOC is a right triangle. Using the Distance Formula, the lengths of the sides of this triangle can be found.
| Side | Points | Distance Formula sqrt((x_2-x_1)^2+(y_2-y_1)^2) | Length |
|---|---|---|---|
| AO | A( 1, m_1) & O( 0, 0) | sqrt(( 0- 1)^2+( 0- m_1)^2) | sqrt(1+m_1^2) |
| CO | C( 1, m_2) & O( 0, 0) | sqrt(( 0- 0)^2+( 0- m_2)^2) | sqrt(1+m_2^2) |
| CA | C( 1, m_2) & A( 1, m_1) | sqrt(( 1- 1)^2+( m_1- m_2)^2) | m_1-m_2 |
Since △ AOC is a right triangle, its side lengths satisfy the Pythagorean Equation. AO^2+ CO^2 = CA^2 The next step is to substitute the lengths shown in the table.
AO^2+ CO^2 = CA^2
SubstituteExpressions
Substitute expressions
( sqrt(1+m_1^2) )^2 + ( sqrt(1+m_2^2) )^2 = ( m_1-m_2)^2
▼
Simplify
PowSqrt
( sqrt(a) )^2 = a
1+m_1^2 + 1+m_2^2 = (m_1-m_2)^2
AddTerms
Add terms
2+m_1^2 +m_2^2 = (m_1-m_2)^2
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
2+m_1^2 +m_2^2 = m_1^2-2m_1m_2+m_2^2
SubEqn
LHS-m_1^2=RHS-m_1^2
2+m_2^2 = - 2m_1m_2+m_2^2
SubEqn
LHS-m_2^2=RHS-m_2^2
2 = - 2m_1m_2
DivEqn
.LHS /(- 2).=.RHS /(- 2).
2/- 2 = m_1m_2
MoveNegDenomToFrac
Put minus sign in front of fraction
- 2/2 = m_1m_2
QuotOne
a/a=1
- 1 = m_1m_2
RearrangeEqn
Rearrange equation
m_1* m_2 = - 1
It has been proven that if two lines are perpendicular, then the product of their slopes is - 1.
l_1 ⊥ l_2 ⇒ m_1* m_2 = - 1
Part 2
Here it is assumed that the slopes of two lines l_1 and l_2 are opposite reciprocals. m_1* m_2 =- 1 Consider the steps taken in Part 1. This time, it should be found that △ AOC is a right triangle.
If the lengths of the sides of △ AOC satisfy the Pythagorean Theorem, then the triangle is a right triangle. AO^2+ CO^2 ? = CA^2 The side lengths, which were previously found in Part 1, can be substituted into the above equation.
AO^2+ CO^2 ? = CA^2
SubstituteExpressions
Substitute expressions
(sqrt(1+m_1 ^2))^2+ (sqrt(1+m_2 ^2))^2 ? = (m_1-m_2)^2
▼
Simplify
PowSqrt
( sqrt(a) )^2 = a
1+m_1 ^2 +1+m_2 ^2 ? =(m_1-m_2)^2
AddTerms
Add terms
2+m_1 ^2 +m_2 ^2 ? =(m_1-m_2)^2
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
2+m_1 ^2 +m_2 ^2 ? =m_1^2-2m_1 m_2 +m_2 ^2
SubEqn
LHS-m_1^2=RHS-m_1^2
2+m_2 ^2 ? =- 2m_1 m_2 +m_2 ^2
SubEqn
LHS-m_2^2=RHS-m_2^2
2 ? =- 2m_1 m_2
Substitute
m_1 m_2= - 1
2 ? =- 2( - 1)
MultNegNegOnePar
- a(- b)=a* b
2=2 ✓
Since a true statement was obtained, △ AOC is a right triangle. Therefore, l_1 and l_2 are perpendicular lines. This completes the second part.
m_1* m_2 = - 1 ⇒ l_1 ⊥ l_2
The biconditional statement has been proven.
l_1 ⊥ l_2 ⇔ m_1* m_2 = - 1
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Pop Quiz
Identifying Perpendicular Lines
Consider two lines on the same coordinate plane and their equations in slope-intercept form. Are the lines perpendicular?
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Example
Finding a Perpendicular Line Through a Point
Write an equation in slope-intercept form of the line that passes through (6,- 1) and is perpendicular to y=- 3x+2.
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Hint
What makes two lines perpendicular?
Solution
Consider the given equation.
y= - 3x+2
Two lines are perpendicular if and only if their slopes are negative reciprocals. This means that their product is - 1.
m_1* m_2=- 1
The slope of the given equation is - 3. By substituting this value into the above equation the slope of a perpendicular line m_2 can be found.
All perpendicular lines to the line whose equation is given have a slope of 13. A general equation in slope-intercept form for these lines can be written. y= 1/3x+ b The desired perpendicular line passes through the point with coordinates ( 6, - 1). By substituting these numbers into the obtained equation for x and y, the y-intercept b of the perpendicular line can be determined.
y=1/3x+b
SubstituteII
x= 6, y= - 1
- 1=1/3( 6)+b
▼
Solve for b
MoveRightFacToNumOne
1/b* a = a/b
- 1=6/3+b
CalcQuot
Calculate quotient
- 1=2+b
SubEqn
LHS-2=RHS-2
- 3=b
RearrangeEqn
Rearrange equation
b= - 3
Knowing the y-intercept, the equation of the perpendicular line to y=- 3x+2 through (6,- 1) can now be written. y= 1/3x+( - 3) ⇔ y=1/3x-3 The solution can be verified by graphing this line on the same coordinate plane.
As seen above, the graph of y= 13x-3 is perpendicular to the graph of y=- 3x+2 and passes through (6,- 1).
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Example
Water Supply Network
Zain's class is modeling a neighborhood that is being built outside of town. For extra credit, Zain decides to use the neighborhood's plumbing plan determine where the pipe that connects a new house to the water supply network will be placed. They have the following plan of the network.
Zain knows that the new pipe has to be perpendicular to the main pipe. It also must pass through (3,- 1), which represents the new house. Help Zain find the equation in slope-intercept of the line that represents the new pipe.
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Solution
First, the slope of the line that represents the main pipe will be determined. To do so, the Slope Formula will be used.
m = y_2-y_1/x_2-x_1
Here, (x_1,y_1) and (x_2,y_2) are the coordinates of two points on the line. Therefore, two points on the given line are needed.
The coordinates of two points on the main pipe are (0,2) and (2,3). These will be substituted into the formula.
m = y_2-y_1/x_2-x_1
SubstitutePoints
Substitute ( 0,2) & ( 2,3)
m=3- 2/2- 0
SubTerms
Subtract terms
m= 1/2
The slope of the line that represents the main pipe is 12. Two lines are perpendicular when their slopes are negative reciprocals. This means that the product of the slopes is - 1. m_1* m_2=- 1 The slope of the given line is 12. By substituting this value into the above equation for m_1, the slope of a perpendicular line m_2 can be found.
All lines perpendicular to the line that represents the main pipe have a slope of - 2. A general equation in slope-intercept form for these lines can be written. y= - 2x+ b The line that corresponds to the new pipe should pass through ( 3, - 1). By substituting 3 and - 1 in the general equation for x and y, the y-intercept b of the perpendicular line can be determined.
Knowing the y-intercept, the equation of the new pipe can now be written. y= - 2x+ 5 The solution can be verified by graphing the line on the given plan.
As seen above, the graph of y=- 2x+5 is perpendicular to the given line and passes through (3,- 1). The new pipe is a part of y=- 2x+5.
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Example
Space Planning
Tiffaniqua is designing a new community garden for the class proposal. The entrance to the garden is in its corner. In the opposite corner there is a wooden bench. She plans to place flower beds in another corner of the garden. Consider a plan of the garden made in a specialized software.
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Solution
The stone path that connects the bench with the entrance lies on the line that passes through the points with coordinates (0,4) and (6,0). These can be used to find the line's slope by using the Slope Formula.
m = y_2-y_1/x_2-x_1
SubstitutePoints
Substitute ( 0,4) & ( 6,0)
m=0- 4/6- 0
▼
Evaluate right-hand side
SubTerms
Subtract terms
m=- 4/6
MoveNegNumToFrac
Put minus sign in front of fraction
m=-4/6
ReduceFrac
a/b=.a /2./.b /2.
m=-2/3
The x-coordinate of (0,4) is 0. Therefore, the y-intercept of the line through this point is 4. Knowing the slope an the y-intercept of the line, its equation in slope-intercept form can be written. y=-2/3x+ 4 The path from the bench to the entrance is a part of the line described by the above equation. Note that the whole equation is not actually needed. It would be enough to know only the slope of the line passing through the bench and the entrance to find a line perpendicular to it.
Now, the equation of the line that corresponds to the second path must be found. First, the slope of this line will be determined. The second line is perpendicular to y=- 23x+4. Perpendicular lines have negative reciprocal slopes. This means that their product is - 1. m_1* m_2=- 1 The slope of the known equation is - 23. By substituting this value into the above equation for m_1, the slope of a perpendicular line m_2 can be found.
All lines perpendicular to the line that passes through the bench and the entrance have a slope of 32. A general equation in slope-intercept form for these lines can be written. y= 3/2x+ b The desired line has to allow community members to easily access the flower beds. This means that the line should pass through the point with coordinates ( 6, 4). By substituting them into the above equation for x and y, the y-intercept b of the perpendicular line can be found.
y=3/2x+b
SubstituteII
x= 6, y= 4
4=3/2( 6)+b
▼
Solve for b
MoveRightFacToNum
a/c* b = a* b/c
4=18/2+b
CalcQuot
Calculate quotient
4=9+b
SubEqn
LHS-9=RHS-9
- 5=b
RearrangeEqn
Rearrange equation
b= - 5
Knowing the y-intercept, the equation of the line perpendicular to y=- 23x+4 through (6,4) can be written. y= 3/2x+( - 5) ⇔ y =3/2x-5 The path that will let visitors to the garden access the flowers is a part of the above line. Below it is shown how the solution looks in the specialized software.
It can be seen that the lines are perpendicular and that y= 32x-5 passes through (6,4), which corresponds to the flower beds.
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Closure
Lines in the Three-Dimensional Space
As mentioned at the beginning of the lesson, two lines in a plane are either parallel or intersecting lines. A particular type of intersecting lines are be perpendicular lines.
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Parallel and Perpendicular Lines
Exercise 3.1
Consider the following diagram.
The lines l_1 and l_2 are perpendicular. Find the value of a.
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We are told that l_1 and l_2 are perpendicular. This means that their slopes are negative reciprocals. Let's find the slope m_1 of l_1 first!
Next, we find the slope m_2 of l_2.
Since the lines are perpendicular their slopes are negative reciprocals. This means that their product is equal to -1. We will substitute the values of m_1 and m_2 to write an equation for a.
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Parallel and Perpendicular Lines
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