Slope of Parallel and
Perpendicular Lines
CK-12
Kaitlyn Spong
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Printed: March 24, 2016
AUTHORS
CK-12
Kaitlyn Spong
www.ck12.org
C HAPTER
Chapter 1. Slope of Parallel and Perpendicular Lines
1
Slope of Parallel and
Perpendicular Lines
Here you will prove that parallel lines have slopes that are equal and perpendicular lines have slopes that are opposite
reciprocals. You will also practice solving problems involving parallel and perpendicular lines.
Find the equation of the line parallel to y = 2x − 4that passes through the point (2, −3). Then, find the equation
of the line perpendicular to y = 2x − 4 that passes through the point (2, −3). How are the two lines that you found
related?
Slope of Parallel and Perpendicular Lines
Consider two lines. There are three ways that the two lines can interact:
1. They are parallel and so they never intersect.
2. They are perpendicular and so they intersect at a right angle.
3. They intersect, but they are not perpendicular.
Recall that the slope of a line is a measure of its steepness. For a line written in the form y = mx + b, “m” is the slope.
Given two lines, their slopes can help you to determine whether the lines are parallel, perpendicular, or neither.
In the past you learned that two lines are parallel if and only if they have the same slope.
In the past you also learned that two lines are perpendicular if and only if they have slopes that are opposite
reciprocals. This means that if the slope of one line is m, the slope of a line perpendicular to it will be − m1 . Another
way of thinking about this is that the product of the slopes of perpendicular lines will always be -1. (Note that
(m)( − m1 ) = − m
m = −1).
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/73893
Consider two lines
How do you know that the two lines are distinct (not the same line?)
The two lines are distinct because they have different y-intercepts. The first line has a y-intercept at (0, b)and the
second line has a y-intercept at (0, c).
Use algebra to find the point of intersection of the lines. What happens?
You can use substitution to attempt to find the point of intersection.
y = ax + b and y = ax + c
Therefore:
1
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ax + b = ax + c
b=c
This is a contradiction because it was stated that b 6= c. Therefore, these two lines do not have a point of intersection.
This means the lines must be parallel. This proves that if two lines have the same slope, then they are parallel.
←
→
←
→
Consider rectangle ABCD with BC = m, EC = 1 and perpendicular lines AE and BE.
Find the length of
Because it is a rectangle, AD = BC = m. The two triangles are similar because they have congruent angles. Let
6 BEC = θ and label all angles in the picture in terms of θ.
You can see that each of the three triangles in the picture have the same angle measures, so they must all be similar.
In particular, 4ADE is similar to 4ECB.
Use the fact that
Because 4ADE is similar to 4ECB, the following proportion is true:
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Chapter 1. Slope of Parallel and Perpendicular Lines
m DE
=
1
m
Solving this proportion you have that DE = m2 .
←
→
←
→
1
m
The slopes of the lines can be found using rise
run . The slope of line AE is − m2 = − m and the slope of BE is
The product of the slopes is − m1 (m) = − m
m = −1.
m
1
= m.
This proves that if two lines are perpendicular, then their slopes will be opposite reciprocals (the product of the
slopes will be -1).
Examples
Example 1
Earlier, you were asked how were the two lines that you found related.
To find the equation of the line parallel to y = 2x − 4 that passes through the point (2, −3), remember that parallel
lines must have equal slopes. This means that the new line must have a slope of 2 and pass through the point (2, −3).
All you need to do is solve for the y-intercept.
−3 = 2(2) + b
−3 = 4 + b
b = −7
The equation of the line is y = 2x − 7.
To find the equation of the line perpendicular to y = 2x − 4 that passes through the point (2, −3), remember that
perpendicular lines will have opposite reciprocal slopes. This means that the new line must have a slope of − 12 and
pass through the point (2, −3). Again, all you need to do is solve for the y-intercept.
1
−3 = − (2) + b
2
−3 = −1 + b
−4 = b
The equation of the line is y = − 12 x − 4
The two lines that were found (y = 2x − 7 and − 12 x − 4 are also perpendicular. Note that they have opposite
reciprocal slopes.
Example 2
Consider two parallel lines y = ax + b and y = cx + d with b 6= d. Show that a = c.
Suppose a 6= c. You can solve a system of equations to find the point of intersection of the two lines.
y = ax + b and y = cx + d
3
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Therefore:
ax + b = cx + d
x(a − c) = d − b
d −b
x=
a−c
If a 6= c, then this point exists so the lines intersect. This is a contradiction because it was stated that the lines were
parallel. Therefore, a must be equal to c. This proves that if two lines are parallel then they must have the same
slope.
Example 3
Consider two lines intersecting at the origin as shown below. Find the lengths of the legs of each triangle. Then,
show that 4BCO is similar to 4ODA.
The lengths of the legs of the triangles are shown below.
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4BCO ∼ 4ODA with a ratio of m : 1 by SAS ∼.
Chapter 1. Slope of Parallel and Perpendicular Lines
BC
OD
=
m
1
and
CO
DA
=
1
1
m
= m1 . Also 6 C ∼
= 6 D.
Example 4
←
→
←
→
Using the picture from #3, find the slopes of lines AO and BO and verify that their product is -1. Then use the fact
←
→
←
→
that 4BCO is similar to 4ODA to show that AO and BO must be perpendicular.
←
→
←
→
The slope of line AO is − m1 and the slope of BO is m1 . The product of the slopes is − m1 (m) = − m
m = −1.
Because 4BCO is similar to 4ODA, their corresponding angles must be congruent. This means that:
• m6 OCB = m6 DOA
• m6 BOC = m6 DAO
Also, because they are right triangles:
• m6 OCB + m6 BOC = 90◦
• m6 DOA + m6 DAO = 90◦
By substitution, m6 DOA + m6 BOC = 90◦ . Because 6 DOA, 6 BOC and 6 AOB form a straight line, the sum of their
measures must be 180◦ . Therefore, m6 AOB must be 90◦ .
←
→
←
→
Because m6 AOB = 90◦ , AO and BO must be perpendicular. This proves that if two lines have opposite reciprocal
slopes, then they are perpendicular.
Review
1. Describe the three ways that two lines could interact. Draw a picture of each.
2. What does it mean for two lines to be parallel? How are the slopes of parallel lines related?
3. What does it mean for two lines to be perpendicular? How are the slopes of perpendicular lines related?
4. Use algebra to show why the lines y = 3x − 4 and y = 3x + 7 (lines with the same slope) must be parallel.
←
→
←
→
5. Use the method from Example B and Example C to show why the slopes of lines FK and KG must be opposite
reciprocals. Assume that FGHJ is a rectangle.
6. Find the line parallel to y = 3x − 5 that passes through (2, 11).
7. Find the line perpendicular to y = 3x − 5 that passes through (6, 11).
8. Find the line parallel to 3x + 4y = 7 that passes through (4, 2).
9. Find the line perpendicular to 3x + 4y = 7 that passes through (3, 10).
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10. Find the line parallel to y = 5 that passes through (2, 16).
11. Find the line perpendicular to y = 5 that passes through (2, 16)
12. Find the line parallel to y = − 31 x − 4 that passes through (6, 8).
13. Find the line perpendicular to y = − 31 x − 4 that passes through (6, 8).
14. Line a passes through the point (2, 4) and (3, 6). Line b passes through the points (6, 7) and (11, 17). Are lines a
and b parallel, perpendicular, or neither?
15. Line a passes through the point (1, −1) and (6, 14). Line b passes through the points (9, 3) and (−6, 8). Are
lines a and b parallel, perpendicular, or neither?
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 10.4.
References
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