Download Solutions to Exercises, Section 2.1

Instructor’s Solutions Manual, Section 2.1
Exercise 1
Solutions to Exercises, Section 2.1
1. What are the coordinates of the unlabeled vertex of the smaller of the
two right triangles in the figure at the beginning of this section?
solution Drawing vertical and horizontal lines from the point in
question to the coordinate axes shows that the coordinates of the point
are (x2 , y1 ).
Instructor’s Solutions Manual, Section 2.1
Exercise 2
2. What are the coordinates of the unlabeled vertex of the larger of the
two right triangles in the figure at the beginning of this section?
solution Drawing vertical and horizontal lines from the point in
question to the coordinate axes shows that the coordinates of the point
are (x4 , y3 ).
Instructor’s Solutions Manual, Section 2.1
Exercise 3
3. Find the slope of the line that contains the points (3, 4) and (7, 13).
solution The line containing the points (3, 4) and (7, 13) has slope
13 − 4
,
7−3
which equals
9
4.
Instructor’s Solutions Manual, Section 2.1
Exercise 4
4. Find the slope of the line that contains the points (2, 11) and (6, −5).
solution The line containing the points (2, 11) and (6, −5) has slope
11 − (−5)
,
2−6
which equals −4.
Instructor’s Solutions Manual, Section 2.1
Exercise 5
5. Find a number t such that the line containing the points (1, t) and
(3, 7) has slope 5.
solution The slope of the line containing the points (1, t) and (3, 7)
equals
7−t
,
3−1
7−t
which equals 2 . We want this slope to equal 5. Thus we must find a
number t such that
7−t
= 5.
2
Solving this equation for t, we get t = −3.
Instructor’s Solutions Manual, Section 2.1
Exercise 6
6. Find a number c such that the line containing the points (c, 4) and
(−2, 9) has slope −3.
solution The slope of the line containing the points (c, 4) and (−2, 9)
equals
4−9
,
c+2
−5
which equals c+2 . We want this slope to equal −3. Thus we must find a
number c such that
−5
= −3.
c+2
1
Solving this equation for c, we get c = − 3 .
Instructor’s Solutions Manual, Section 2.1
Exercise 7
7. Find the equation of the line in the xy-plane with slope 2 that contains
the point (7, 3).
solution If (x, y) denotes a typical point on the line with slope 2 that
contains the point (7, 3), then
y −3
= 2.
x−7
Multiplying both sides of this equation by x − 7 and then adding 3 to
both sides gives the equation
y = 2x − 11.
check The line whose equation is y = 2x − 11 has slope 2. We should
also check that the point (7, 3) is on this line. In other words, we need
to verify the alleged equation
?
3 = 2 · 7 − 11.
Simple arithmetic shows that this is indeed true.
Instructor’s Solutions Manual, Section 2.1
Exercise 8
8. Find the equation of the line in the xy-plane with slope −4 that
contains the point (−5, −2).
solution If (x, y) denotes a typical point on the line with slope −4
that contains the point (−5, −2), then
y +2
= −4.
x+5
Multiplying both sides of this equation by x + 5 and then subtracting 2
from both sides gives the equation
y = −4x − 22.
check The line whose equation is y = −4x − 22 has slope −4. We
should also check that the point (−5, −2) is on this line. In other
words, we need to verify the alleged equation
?
−2 = (−4) · (−5) − 22.
Simple arithmetic shows that this is indeed true.
Instructor’s Solutions Manual, Section 2.1
Exercise 9
9. Find the equation of the line that contains the points (2, −1) and (4, 9).
solution The line that contains the points (2, −1) and (4, 9) has slope
9 − (−1)
,
4−2
which equals 5. Thus if (x, y) denotes a typical point on this line, then
y −9
= 5.
x−4
Multiplying both sides of this equation by x − 4 and then adding 9 to
both sides gives the equation
y = 5x − 11.
check We need to check that both (2, −1) and (4, 9) are on the line
whose equation is y = 5x − 11. In other words, we need to verify the
alleged equations
?
−1 = 5 · 2 − 11
and
?
9 = 5 · 4 − 11.
Simple arithmetic shows that both alleged equations are indeed true.
Instructor’s Solutions Manual, Section 2.1
Exercise 10
10. Find the equation of the line that contains the points (−3, 2) and
(−5, 7).
solution The line that contains the points (−3, 2) and (−5, 7) has
slope
7−2
,
−5 − (−3)
5
which equals − 2 . Thus if (x, y) denotes a typical point on this line,
then
5
y −2
=− .
x+3
2
Multiplying both sides of this equation by x + 3 and then adding 2 to
both sides gives the equation
5
y = −2x −
11
2 .
check We need to check that both (−3, 2) and (−5, 7) are on the line
5
11
whose equation is y = − 2 x − 2 . In other words, we need to verify the
alleged equations
?
5
2 = (− 2 ) · (−3) −
11
2
and
?
5
7 = (− 2 ) · (−5) −
11
2 .
Simple arithmetic shows that both alleged equations are indeed true.
Instructor’s Solutions Manual, Section 2.1
Exercise 11
11. Find a number t such that the point (3, t) is on the line containing the
points (7, 6) and (14, 10).
solution First we find the equation of the line containing the points
(7, 6) and (14, 10). To do this, note that the line containing those two
points has slope
10 − 6
,
14 − 7
which equals
4
7.
Thus if (x, y) denotes a typical point on this line, then
4
y −6
= .
x−7
7
Multiplying both sides of this equation by x − 7 and then adding 6
gives the equation
4
y = 7 x + 2.
Now we can find a number t such that the point (3, t) is on the line
given by the equation above. To do this, in the equation above replace
x by 3 and y by t, getting
t=
4
7
· 3 + 2.
Performing the arithmetic to compute the right side, we get t =
26
7 .
check We should check that all three points (7, 6), (14, 10), and
26 3, 7 are on the line y = 47 x + 2. In other words, we need to verify the
alleged equations
Instructor’s Solutions Manual, Section 2.1
? 4
7
6=
· 7 + 2,
? 4
7
10 =
Exercise 11
· 14 + 2,
26 ? 4
7 = 7
· 3 + 2.
Simple arithmetic shows that all three alleged equations are indeed
true.
Instructor’s Solutions Manual, Section 2.1
Exercise 12
12. Find a number t such that the point (−2, t) is on the line containing the
points (5, −2) and (10, −8).
solution First we find the equation of the line containing the points
(5, −2) and (10, −8). To do this, note that the line containing those two
points has slope
−2 − (−8)
,
5 − 10
which equals − 65 . Thus if (x, y) denotes a typical point on this line,
then
6
y +2
=− .
x−5
5
Multiplying both sides of this equation by x − 5 and then subtracting 2
gives the equation
6
y = − 5 x + 4.
Now we can find a number t such that the point (−2, t) is on the line
given by the equation above. To do this, in the equation above replace
x by −2 and y by t, getting
6
t = (− 5 ) · (−2) + 4.
Performing the arithmetic to compute the right side, we get t =
32
5 .
check We should check that all three points (5, −2), (10, −8), and
32 6
−2, 5 are on the line y = − 5 x + 4. In other words, we need to verify
the alleged equations
Instructor’s Solutions Manual, Section 2.1
?
Exercise 12
6
−2 = − 5 · 5 + 4,
?
−8 = − 65 · 10 + 4,
32 ?
5 =
(− 56 ) · (−2) + 4.
Simple arithmetic shows that all three alleged equations are indeed
true.
Instructor’s Solutions Manual, Section 2.1
Exercise 13
13. Find a number c such that the point (c, 13) is on the line containing the
points (−4, −17) and (6, 33).
solution First we find the equation of the line containing the points
(−4, −17) and (6, 33). To do this, note that the line containing those
two points has slope
33 − (−17)
,
6 − (−4)
which equals 5. Thus if (x, y) denotes a typical point on this line, then
y − 33
= 5.
x−6
Multiplying both sides of this equation by x − 6 and then adding 33
gives the equation
y = 5x + 3.
Now we can find a number c such that the point (c, 13) is on the line
given by the equation above. To do this, in the equation above replace
x by c and y by 13, getting
13 = 5c + 3.
Solving this equation for c, we get c = 2.
check We should check that the three points (−4, −17), (6, 33), and
(2, 13) are all on the line whose equation is y = 5x + 3. In other words,
we need to verify the alleged equations
Instructor’s Solutions Manual, Section 2.1
?
−17 = 5 · (−4) + 3,
Exercise 13
?
33 = 5 · 6 + 3,
?
13 = 5 · 2 + 2.
Simple arithmetic shows that all three alleged equations are indeed
true.
Instructor’s Solutions Manual, Section 2.1
Exercise 14
14. Find a number c such that the point (c, −19) is on the line containing
the points (2, 1) and (4, 9).
solution First we find the equation of the line containing the points
(2, 1) and (4, 9). To do this, note that the line containing those two
points has slope
9−1
,
4−2
which equals 4. Thus if (x, y) denotes a typical point on this line, then
y −1
= 4.
x−2
Multiplying both sides of this equation by x − 2 and then adding 1
gives the equation
y = 4x − 7.
Now we can find a number c such that the point (c, −19) is on the line
given by the equation above. To do this, in the equation above replace
x by c and y by −19, getting
−19 = 4c − 7.
Solving this equation for c, we get c = −3.
check We should check that the three points (2, 1), (4, 9), and
(−3, −19) are all on the line whose equation is y = 4x − 7. In other
words, we need to verify the alleged equations
Instructor’s Solutions Manual, Section 2.1
?
1 = 4 · 2 − 7,
?
9 = 4 · 4 − 7,
Exercise 14
?
−19 = 4 · (−3) − 7.
Simple arithmetic shows that all three alleged equations are indeed
true.
Instructor’s Solutions Manual, Section 2.1
Exercise 15
15. Find a number t such that the point (t, 2t) is on the line containing the
points (3, −7) and (5, −15).
solution First we find the equation of the line containing the points
(3, −7) and (5, −15). To do this, note that the line containing those two
points has slope
−7 − (−15)
,
3−5
which equals −4. Thus if (x, y) denotes a point on this line, then
y − (−7)
= −4.
x−3
Multiplying both sides of this equation by x − 3 and then subtracting 7
gives the equation
y = −4x + 5.
Now we can find a number t such that the point (t, 2t) is on the line
given by the equation above. To do this, in the equation above replace
x by t and y by 2t, getting
2t = −4t + 5.
5
Solving this equation for t, we get t = 6 .
check We should check that the three points (3, −7), (5, −15), and
5
5
6 , 2 · 6 are all on the line whose equation is y = −4x + 5. In other
words, we need to verify the alleged equations
Instructor’s Solutions Manual, Section 2.1
?
−7 = −4 · 3 + 5,
?
−15 = −4 · 5 + 5,
Exercise 15
5 ?
3 =
−4 ·
5
6
+ 5.
Simple arithmetic shows that all three alleged equations are indeed
true.
Instructor’s Solutions Manual, Section 2.1
Exercise 16
16. Find a number t such that the point (t, 2t ) is on the line containing the
points (2, −4) and (−3, −11).
solution First we find the equation of the line containing the points
(2, −4) and (−3, −11). To do this, note that the line containing those
two points has slope
−4 − (−11)
,
2 − (−3)
which equals
7
5.
Thus if (x, y) denotes a typical point on this line, then
7
y +4
= .
x−2
5
Multiplying both sides of this equation by x − 2 and then subtracting 4
gives the equation
7
34
y = 5x − 5 .
Now we can find a number t such that the point (t, 2t ) is on the line
given by the equation above. To do this, in the equation above replace
t
x by t and y by 2 , getting
t
2
= 75 t −
Solving this equation for t, we get t =
34
5 .
68
9 .
check We should check that the three points (2, −4), (−3, −11), and
68 34 7
34
are all on the line whose equation is y = 5 x − 5 . In other
9 , 9
words, we need to verify the alleged equations
Instructor’s Solutions Manual, Section 2.1
Exercise 16
? 7
5
·2−
? 7
5
· (−3) −
−4 =
−11 =
34 ? 7
9 = 5
34
5 ,
· ( 68
9 )−
34
5 ,
34
5 .
Simple arithmetic shows that all three alleged equations are indeed
true.
Instructor’s Solutions Manual, Section 2.1
Exercise 17
17. Let f (x) be the number of seconds in x days. Find a formula for f (x).
solution Each minute has 60 seconds, and each hour has 60 minutes.
Thus each hour has 60 × 60 seconds, or 3600 seconds. Each day has 24
hours; thus each day has 24 × 3600 seconds, or 86400 seconds. Thus
f (x) = 86400x.
Instructor’s Solutions Manual, Section 2.1
Exercise 18
18. Let f (x) be the number of seconds in x weeks. Find a formula for f (x).
solution Each minute has 60 seconds, and each hour has 60 minutes.
Thus each hour has 60 × 60 seconds, or 3600 seconds. Each day has 24
hours; thus each day has 24 × 3600 seconds, or 86400 seconds. Each
week has 7 days; thus each week has 7 × 86400 seconds, or 604800
seconds. Thus
f (x) = 604800x.
Instructor’s Solutions Manual, Section 2.1
Exercise 19
19. Let f (x) be the number of inches in x miles. Find a formula for f (x).
solution Each foot has 12 inches, and each mile has 5280 feet. Thus
each mile has 5280 × 12 inches, or 63360 inches. Thus
f (x) = 63360x.
Instructor’s Solutions Manual, Section 2.1
Exercise 20
20. Let f (x) be the number of miles in x feet. Find a formula for f (x).
solution Each mile has 5280 feet. Thus each foot has length
miles. Thus
x
.
f (x) =
5280
1
5280
Instructor’s Solutions Manual, Section 2.1
Exercise 21
21. Let f (x) be the number of kilometers in x miles. Find a formula for
f (x).
[The exact conversion between the English measurement system and the
metric system is given by the equation 1 inch = 2.54 centimeters.]
solution Multiplying both sides of the equation
1 inch = 2.54 centimeters
by 12 gives
1 foot = 12 × 2.54 centimeters
= 30.48 centimeters.
Multiplying both sides of the equation above by 5280 gives
1 mile = 5280 × 30.48 centimeters
= 160934.4 centimeters
= 1609.344 meters
= 1.609344 kilometers.
Multiplying both sides of the equation above by a number x shows that
x miles = 1.609344x kilometers. In other words,
f (x) = 1.609344x.
Instructor’s Solutions Manual, Section 2.1
Exercise 21
[The formula above is exact. However, the approximation f (x) = 1.61x
is often used.]
Instructor’s Solutions Manual, Section 2.1
Exercise 22
22. Let f (x) be the number of miles in x meters. Find a formula for f (x).
solution In the previous exercise, we saw that
1 mile = 1609.344 meters.
Dividing both sides of this equation by 1609.344 gives the equation
1 meter =
1
miles.
1609.344
Multiplying both sides of the equation above by a number x shows that
x
x meters = 1609.344 miles. In other words,
f (x) =
x
.
1609.344
Instructor’s Solutions Manual, Section 2.1
Exercise 23
23. Let f (x) be the number of inches in x centimeters. Find a formula for
f (x).
solution Dividing both sides of the equation
1 inch = 2.54 centimeters by 2.54 gives
1 centimeter =
1
inches.
2.54
Multiplying both sides of the equation above by a number x shows that
x
x centimeters = 2.54 inches. In other words,
f (x) =
x
.
2.54
Instructor’s Solutions Manual, Section 2.1
Exercise 24
24. Let f (x) be the number of meters in x feet. Find a formula for f (x).
solution Multiplying both sides of the equation
1 inch = 2.54 centimeters
by 12 gives
1 foot = 12 × 2.54 centimeters
= 30.48 centimeters
= 0.3048 meters.
Multiplying both sides of the equation above by a number x shows that
x feet = 0.3048x meters. In other words,
f (x) = 0.3048x.
[The formula above is exact. However, the approximation f (x) =
often used.]
x
3.28
is
Instructor’s Solutions Manual, Section 2.1
Exercise 25
25. Find the equation of the line in the xy-plane that contains the point
(3, 2) and that is parallel to the line y = 4x − 1.
solution The line in the xy-plane whose equation is y = 4x − 1 has
slope 4. Thus each line parallel to it also has slope 4 and hence has the
form
y = 4x + b
for some constant b.
Thus we need to find a constant b such that the point (3, 2) is on the
line given by the equation above. Replacing x by 3 and replacing y by 2
in the equation above, we have
2 = 4 · 3 + b.
Solving this equation for b, we get b = −10. Thus the line that we seek
is described by the equation
y = 4x − 10.
Instructor’s Solutions Manual, Section 2.1
Exercise 26
26. Find the equation of the line in the xy-plane that contains the point
(−4, −5) and that is parallel to the line y = −2x + 3.
solution The line in the xy-plane whose equation is y = −2x + 3 has
slope −2. Thus each line parallel to it also has slope −2 and hence has
the form
y = −2x + b
for some constant b.
Thus we need to find a constant b such that the point (−4, −5) is on
the line given by the equation above. Replacing x by −4 and replacing
y by −5 in the equation above, we have
−5 = (−2) · (−4) + b.
Solving this equation for b, we get b = −13. Thus the line that we seek
is described by the equation
y = −2x − 13.
Instructor’s Solutions Manual, Section 2.1
Exercise 27
27. Find the equation of the line that contains the point (2, 3) and that is
parallel to the line containing the points (7, 1) and (5, 6).
solution The line containing the points (7, 1) and (5, 6) has slope
6−1
,
5−7
5
5
which equals − 2 . Thus each line parallel to it also has slope − 2 and
hence has the form
5
y = −2x + b
for some constant b.
Thus we need to find a constant b such that the point (2, 3) is on the
line given by the equation above. Replacing x by 2 and replacing y by 3
in the equation above, we have
5
3 = − 2 · 2 + b.
Solving this equation for b, we get b = 8. Thus the line that we seek is
described by the equation
5
y = − 2 x + 8.
Instructor’s Solutions Manual, Section 2.1
Exercise 28
28. Find the equation of the line that contains the point (−4, 3) and that is
parallel to the line containing the points (3, −7) and (6, −9).
solution The line containing the points (3, −7) and (6, −9) has slope
−9 − (−7)
,
6−3
2
2
which equals − 3 . Thus each line parallel to it also has slope − 3 and
hence has the form
2
y = −3x + b
for some constant b.
Thus we need to find a constant b such that the point (−4, 3) is on the
line given by the equation above. Replacing x by −4 and replacing y by
3 in the equation above, we have
2
3 = (− 3 ) · (−4) + b.
1
Solving this equation for b, we get b = 3 . Thus the line that we seek is
described by the equation
2
1
y = −3x + 3.
Instructor’s Solutions Manual, Section 2.1
Exercise 29
29. Find a number t such that the line containing the points (t, 2) and
(3, 5) is parallel to the line containing the points (−1, 4) and (−3, −2).
solution The line containing the points (−1, 4) and (−3, −2) has
slope
4 − (−2)
,
−1 − (−3)
which equals 3. Thus each line parallel to it also has slope 3.
The line containing the points (t, 2) and (3, 5) has slope
5−2
,
3−t
3
which equals 3−t . From the paragraph above, we want this slope to
equal 3. In other words, we need to solve the equation
3
= 3.
3−t
Dividing both sides of the equation above by 3 and then multiplying
both sides by 3 − t gives the equation 1 = 3 − t. Thus t = 2.
Instructor’s Solutions Manual, Section 2.1
Exercise 30
30. Find a number t such that the line containing the points (−3, t) and
(2, −4) is parallel to the line containing the points (5, 6) and (−2, 4).
solution The line containing the points (5, 6) and (−2, 4) has slope
6−4
,
5 − (−2)
which equals
2
7.
Thus each line parallel to it also has slope
2
7.
The line containing the points (−3, t) and (2, −4) has slope
t − (−4)
,
−3 − 2
t+4
which equals − 5 . From the paragraph above, we want this slope to
2
equal 7 . In other words, we need to solve the equation
−
2
t+4
= .
5
7
Multiplying both sides of the equation above by −5 and then
38
subtracting 4 shows that t = − 7 .
Instructor’s Solutions Manual, Section 2.1
Exercise 31
31. Find the intersection in the xy-plane of the lines y = 5x + 3 and
y = −2x + 1.
solution Setting the two right sides of the equations above equal to
each other, we get
5x + 3 = −2x + 1.
To solve this equation for x, add 2x to both sides and then subtract 3
2
from both sides, getting 7x = −2. Thus x = − 7 .
To find the value of y at the intersection point, we can plug the value
x = − 27 into either of the equations of the two lines. Choosing the first
2
11
equation, we have y = −5 · 7 + 3, which implies that y = 7 . Thus the
2 11 two lines intersect at the point − 7 , 7 .
2
check As a check, we can substitute the value x = − 7 into the
equation for the second line and see if that also gives the value y =
In other words, we need to verify the alleged equation
11 ?
7 =
−2(− 27 ) + 1.
Simple arithmetic shows that this is true. Thus we indeed have the
correct solution.
11
7 .
Instructor’s Solutions Manual, Section 2.1
Exercise 32
32. Find the intersection in the xy-plane of the lines y = −4x + 5 and
y = 5x − 2.
solution Setting the two right sides of the equations above equal to
each other, we get
−4x + 5 = 5x − 2.
To solve this equation for x, add 4x to both sides and then add 2,
7
getting 9x = 7. Thus x = 9 .
To find the value of y at the intersection point, we can plug the value
x = 79 into either of the equations of the two lines. Choosing the first
7
17
equation, we have y = −4 · 9 + 5, which implies that y = 9 . Thus the
7 17 two lines intersect at the point 9 , 9 .
7
check As a check, we can substitute the value x = 9 into the equation
17
for the second line and see if that also gives the value y = 9 . In other
words, we need to verify the alleged equation
17 ?
9 =
5·
7
9
− 2.
Simple arithmetic shows that this is true. Thus we indeed have the
correct solution.
Instructor’s Solutions Manual, Section 2.1
Exercise 33
33. Find a number b such that the three lines in the xy-plane given by the
equations y = 2x + b, y = 3x − 5, and y = −4x + 6 have a common
intersection point.
solution The unknown b appears in the first equation; thus our first
step will be to find the point of intersection of the last two lines. To do
this, we set the right sides of the last two equations equal to each other,
getting
3x − 5 = −4x + 6.
To solve this equation for x, add 4x to both sides and then add 5 to
11
both sides, getting 7x = 11. Thus x = 7 . Substituting this value of x
into the equation y = 3x − 5, we get
y =3·
11
7
− 5.
Thus y = − 27 .
At this stage, we have shown that the lines given by the equations
11
2
y = 3x − 5 and y = −4x + 6 intersect at the point 7 , − 7 . We want
the line given by the equation y = 2x + b also to contain this point.
11
2
Thus we set x = 7 and y = − 7 in this equation, getting
− 27 = 2 ·
11
7
+ b.
Solving this equation for b, we get b = − 24
7 .
check As a check that the line given by the equation y = −4x + 6
2
11
contains the point 11
7 , − 7 , we can substitute the value x = 7 into the
Instructor’s Solutions Manual, Section 2.1
Exercise 33
equation for that line and see if it gives the value y = − 27 . In other
words, we need to verify the alleged equation
2 ?
− 7 = −4 ·
11
7
+ 6.
Simple arithmetic shows that this is true. Thus we indeed found the
correct point of intersection.
We chose the line whose equation is given by y = −4x + 6 for this
check because the other two lines had been used in direct calculations
in our solution.
Instructor’s Solutions Manual, Section 2.1
Exercise 34
34. Find a number m such that the three lines in the xy-plane given by the
equations y = mx + 3, y = 4x + 1, and y = 5x + 7 have a common
intersection point.
solution The unknown m appears in the first equation; thus our first
step will be to find the point of intersection of the last two lines. To do
this, we set the right sides of the last two equations equal to each other,
getting
4x + 1 = 5x + 7.
To solve this equation for x, subtract 4x from both sides and then
subtract 7, getting x = −6. Substituting this value of x into the
equation y = 4x + 1, we get y = −23.
At this stage, we have shown that the lines given by the equations
y = 4x + 1 and y = 5x + 7 intersect at the point (−6, −23). We want the
line given by the equation y = mx + 3 also to contain this point. Thus
−23 = −6m + 3.
Solving this equation for m, we get m =
13
3 .
check As a check that the line given by the equation y = 5x + 7
contains the point (−6, −23), we can substitute the value x = −6 into
the equation for that line and see if it gives the value y = −23. In other
words, we need to verify the alleged equation
?
−23 = 5 · (−6) + 7.
Instructor’s Solutions Manual, Section 2.1
Exercise 34
Simple arithmetic shows that this is true. Thus we indeed found the
correct point of intersection.
We chose the line whose equation is given by y = 5x + 7 for this check
because the other two lines had been used in direct calculations in our
solution.
Instructor’s Solutions Manual, Section 2.1
Exercise 35
35. Find the equation of the line in the xy-plane that contains the point
(4, 1) and that is perpendicular to the line whose equation is
y = 3x + 5.
solution The line in the xy-plane whose equation is y = 3x + 5 has
1
slope 3. Thus every line perpendicular to it has slope − 3 . Hence the
equation of the line that we seek has the form
1
y = −3x + b
for some constant b. We want the point (4, 1) to be on this line.
Substituting x = 4 and y = 1 into the equation above, we have
1
1 = − 3 · 4 + b.
7
Solving this equation for b, we get b = 3 . Thus the equation of the line
that we seek is
1
7
y = −3x + 3.
Instructor’s Solutions Manual, Section 2.1
Exercise 36
36. Find the equation of the line in the xy-plane that contains the point
(−3, 2) and that is perpendicular to the line whose equation is
y = −5x + 1.
solution The line in the xy-plane whose equation is y = −5x + 1 has
1
slope −5. Thus every line perpendicular to it has slope 5 . Hence the
equation of the line that we seek has the form
1
y = 5x + b
for some constant b. We want the point (−3, 2) to be on this line.
Substituting x = −3 and y = 2 into the equation above, we have
2=
1
5
· (−3) + b.
Solving this equation for b, we get b =
that we seek is
1
y = 5x +
13
5 .
13
5 .
Thus the equation of the line
Instructor’s Solutions Manual, Section 2.1
Exercise 37
37. Find a number t such that the line in the xy-plane containing the
points (t, 4) and (2, −1) is perpendicular to the line y = 6x − 7.
solution The line in the xy-plane whose equation is y = 6x − 7 has
slope 6. Thus every line perpendicular to it has slope − 61 . Thus we
1
want the line containing the points (t, 4) and (2, −1) to have slope − 6 .
In other words, we want
1
4 − (−1)
=− .
t−2
6
Solving this equation for t, we get t = −28.
Instructor’s Solutions Manual, Section 2.1
Exercise 38
38. Find a number t such that the line in the xy-plane containing the
points (−3, t) and (4, 3) is perpendicular to the line y = −5x + 999.
solution The line in the xy-plane whose equation is y = −5x + 999
has slope −5. Thus every line perpendicular to it has slope 15 . Thus we
1
want the line containing the points (−3, t) and (4, 3) to have slope 5 . In
other words, we want
1
t−3
= .
−3 − 4
5
8
Solving this equation for t, we get t = 5 .
Instructor’s Solutions Manual, Section 2.1
Exercise 39
39. Find a number t such that the line containing the points (4, t) and
(−1, 6) is perpendicular to the line that contains the points (3, 5) and
(1, −2).
solution The line containing the points (3, 5) and (1, −2) has slope
5 − (−2)
,
3−1
7
2
which equals 2 . Thus every line perpendicular to it has slope − 7 . Thus
we want the line containing the points (4, t) and (−1, 6) to have slope
2
− 7 . In other words, we want
2
t−6
=− .
4 − (−1)
7
Solving this equation for t, we get t =
32
7 .
Instructor’s Solutions Manual, Section 2.1
Exercise 40
40. Find a number t such that the line containing the points (t, −2) and
(−3, 5) is perpendicular to the line that contains the points (4, 7) and
(1, 11).
solution The line containing the points (4, 7) and (1, 11) has slope
7 − 11
,
4−1
4
3
which equals − 3 . Thus every line perpendicular to it has slope 4 . Thus
we want the line containing the points (t, −2) and (−3, 5) to have slope
3
4 . In other words, we want
3
−2 − 5
= .
t+3
4
Solving this equation for t, we get t = −
37
3 .
Instructor’s Solutions Manual, Section 2.1
Problem 41
Solutions to Problems, Section 2.1
41. Show that the points (−84, −14), (21, 1), and (98, 12) lie on a line.
solution We will compute the slope of the line containing the points
(−84, −14) and (21, 1), and also the slope of the line containing the
points (21, 1) and (98, 12). If these two slopes are equal, then all three
points lie on a line.
The slope of the line containing the points (−84, −14) and (21, 1) is
15
1
1 − (−14)
=
= .
21 − (−84)
105
7
The slope of the line containing the points (21, 1) and (98, 12) is
11
1
12 − 1
=
= .
98 − 21
77
7
Thus the two slopes are equal, and hence all three points lie on a line.
Instructor’s Solutions Manual, Section 2.1
Problem 42
42. Show that the points (−8, −65), (1, 52), and (3, 77) do not lie on a line.
solution We will compute the slope of the line containing the points
(−8, −65) and (1, 52), and also the slope of the line containing the
points (1, 52) and (3, 77). If these two slopes are not equal, then the
three points do not lie on a line.
The slope of the line containing the points (−8, −65) and (1, 52) is
117
52 − (−65)
=
= 13.
1 − (−8)
9
The slope of the line containing the points (1, 52) and (3, 77) is
25
77 − 52
=
.
3−1
2
Thus the two slopes are not equal, and hence the three points do not lie
on a line.
Instructor’s Solutions Manual, Section 2.1
Problem 43
43. Change just one of the six numbers in the problem above so that the
resulting three points do lie on a line.
solution The line containing the first two points in the problem
above has slope 13. In order to make the line containing the last two
points have slope 13 by changing just one number, we replace the last
point (3, 77) in the problem above by (3, 78). Now the slope containing
the points (1, 52) and (3, 78) is
26
78 − 52
=
= 13.
3−1
2
Thus the slope of the line containing the points (−8, −65) and (1, 52)
equals the slope of the line containing the points (1, 52) and (3, 78).
Thus the points (−8, −65), (1, 52), and (3, 78) all lie on a line.
Instructor’s Solutions Manual, Section 2.1
Problem 44
44. Show that for every number t, the point (5 − 3t, 7 − 4t) is on the line
containing the points (2, 3) and (5, 7).
solution The line containing the points (2, 3) and (5, 7) has slope
4
7−3
= .
5−2
3
The line containing the points (5, 7) and (5 − 3t, 7 − 4t) has slope
−4t
4
(7 − 4t) − 7
==
= .
(5 − 3t) − 5
−3t
3
Thus the slope of the line containing the points (2, 3) and (5, 7) equals
the slope of the line containing the points (5, 7) and (5 − 3t, 7 − 4t).
Thus these points all lie on a line.
Instructor’s Solutions Manual, Section 2.1
Problem 45
45. Show that the composition of two linear functions is a linear function.
solution Suppose f and g are linear functions, with
f (x) = ax + b
and g(x) = cx + d
for all real numbers x. Then
(f ◦ g)(x) = f g(x) = f (cx + d) = a(cx + d) + b = acx + (ad + b).
Thus f ◦ g is a linear function.
Instructor’s Solutions Manual, Section 2.1
Problem 46
46. Show that if f and g are linear functions, then the graphs of f ◦ g and
g ◦ f have the same slope.
solution Suppose f and g are functions, with
f (x) = ax + b
and g(x) = cx + d
for all real numbers x. Then, as was shown in the previous problem,
(f ◦ g)(x) = acx + (ad + b).
Similarly,
(g ◦ f )(x) = acx + (bc + d).
Thus the graph of f and the graph of g are both lines with slope ac.
Instructor’s Solutions Manual, Section 2.1
Problem 47
47. Show that a linear function is increasing if and only if the slope of its
graph is positive.
solution Suppose f is a linear function, with
f (x) = mx + c
for every real number x. Thus the slope of the graph of f equals m.
Suppose also that a and b are real numbers with b > a. Then
f (a) = ma + c
and f (b) = mb + c.
Thus f (b) > f (a) if and only if
mb + c > ma + c,
which is equivalent to the inequality
mb > ma,
which is equivalent to the inequality
m(b − a) > 0,
which is equivalent to the inequality m > 0 (obtained from the previous
inequality by dividing both sides by the positive number b − a). Thus f
is increasing if and only if the slope of its graph (which equals m) is
positive.
Instructor’s Solutions Manual, Section 2.1
Problem 48
48. Show that a linear function is decreasing if and only if the slope of its
graph is negative.
solution Suppose f is a linear function, with
f (x) = mx + c
for every real number x. Thus the slope of the graph of f equals m.
Suppose also that a and b are real numbers with b > a. Then
f (a) = ma + c
and f (b) = mb + c.
Thus f (b) < f (a) if and only if
mb + c < ma + c,
which is equivalent to the inequality
mb < ma,
which is equivalent to the inequality
m(b − a) < 0,
which is equivalent to the inequality m < 0 (obtained from the previous
inequality by dividing both sides by the positive number b − a). Thus f
is decreasing if and only if the slope of its graph (which equals m) is
negative.
Instructor’s Solutions Manual, Section 2.1
Problem 49
49. Show that every nonconstant linear function is a one-to-one function.
solution Suppose f is a nonconstant linear function. Thus there
exist numbers m and b, with m = 0, such that
f (x) = mx + b
for every real number b.
Suppose y is any real number. Then the equation f (x) = y implies that
mx + b = y,
which implies that
x=
y −b
.
m
In other words, for each number y there is only one number x (given by
the formula above) such that f (x) = y. Thus f is a one-to-one function.
Instructor’s Solutions Manual, Section 2.1
Problem 50
50. Show that if f is the linear function defined by f (x) = mx + b, where
m = 0, then the inverse function f −1 is defined by the formula
1
b
f −1 (y) = m y − m .
solution Suppose f is the linear function defined by f (x) = mx + b,
where m = 0.
Suppose y is any real number. To find a formula for f −1 (y), we must
find a number x such that f (x) = y. In other words, we need to solve
the equation
mx + b = y
for x. Subtracting b from both sides and then dividing by m gives
x=
Thus f −1 (y) =
1
my
−
b
m.
1
my
−
b
m.
Instructor’s Solutions Manual, Section 2.1
Problem 51
51. Show that the linear function f defined by f (x) = mx + b is an odd
function if and only if b = 0.
solution First suppose b = 0, which means that f (x) = mx. Thus
f (−x) = m(−x) = −(mx) = −f (x),
which implies that f is an odd function.
To prove the implication in the other direction, suppose now that f is
an odd function. Because f (1) = m + b and f (−1) = −m + b and
because f is an odd function, we have
m + b = f (1) = −f (−1) = −(−m + b) = m − b,
which implies that b = 0.
Instructor’s Solutions Manual, Section 2.1
Problem 52
52. Show that the linear function f defined by f (x) = mx + b is an even
function if and only if m = 0.
solution First suppose m = 0, which means that f (x) = b. Thus
f (−x) = b = f (x),
which implies that f is an even function.
To prove the implication in the other direction, suppose now that f is
an even function. Because f (1) = m + b and f (−1) = −m + b and
because f is an even function, we have
m + b = f (1) = f (−1) = −m + b,
which implies that m = 0.
Instructor’s Solutions Manual, Section 2.1
Problem 53
53. We used the similar triangles to show that the product of the slopes of
two perpendicular lines equals −1. The steps below outline an
alternative proof that avoids the use of similar triangles but uses more
algebra instead. Use the figure below, which is the same as the figure
used earlier except that there is now no need to label the angles.
Q
b
a
S
P
c
T
QP is perpendicular to P T .
(a) Apply the Pythagorean Theorem to triangle P SQ to find the length
of the line segment P Q in terms of a and b.
(b) Apply the Pythagorean Theorem to triangle P ST to find the length
of the line segment P T in terms of a and c.
(c) Apply the Pythagorean Theorem to triangle QP T to find the length
of the line segment QT in terms of the lengths of the line segments
of P Q and P T calculated in the first two parts of this problem.
Instructor’s Solutions Manual, Section 2.1
Problem 53
(d) As can be seen from the figure, the length of the line segment QT
equals b + c. Thus set the formula for length of the line segment
QT , as calculated in the previous part of this problem, equal to
b + c, and solve the resulting equation for c in terms of a and b.
(e) Use the result in the previous part of this problem to show that the
slope of the line containing P and Q times the slope of the line
containing P and T equals −1.
solution
(a) By the Pythagorean Theorem, the length of the line segment P Q equals
√
a2 + b 2 .
(b) By the Pythagorean Theorem, the length of the line segment P T equals
√
a2 + c 2 .
(c) By the Pythagorean Theorem, the length of the line segment QT equals
√
u2 + ν 2 , where u is the length of the line segment P Q and ν is the
length of the line segment P T . From the previous two parts of this
problem, we have u2 = a2 + b2 and ν 2 = a2 + c 2 . Thus the length of the
√
line segment QT equals 2a2 + b2 + c 2 .
(d) From the previous part of this problem, the the line segment QT equals
√
2a2 + b2 + c 2 . The figure shows that the line segment QT also equals
b + c. Thus
2a2 + b2 + c 2 = b + c.
Squaring both sides of this equation gives
2a2 + b2 + c 2 = b2 + 2bc + c 2 .
Instructor’s Solutions Manual, Section 2.1
Problem 53
Subtracting b2 + c 2 from both sides and then dividing by 2 gives
a2 = bc,
which implies that
c=
a2
.
b
b
(e) The figure shows that the slope of the line containing P and Q is a
. The
c
figure also shows that the slope of the line containing P and T is − a .
Thus the product of these slopes is
b
b a2
b c
· −
= − 2c = − 2 ·
= −1.
a
a
a
a
b
Instructor’s Solutions Manual, Section 2.1
Problem 54
54. Show that the graphs of two linear functions f and g are perpendicular
if and only if the graph of f ◦ g has slope −1.
solution Suppose f and g are linear functions, with
f (x) = ax + b
and g(x) = cx + d
for all real numbers x. Thus the graph of f has slope a and the graph
of g has slope c. Hence the graphs of f and g are perpendicular if and
only if ac = −1.
Also
(f ◦ g)(x) = f g(x) = f (cx + d) = a(cx + d) + b = acx + (ad + b).
Thus the graph of f ◦ g has slope ac. Hence the graphs of f and g are
perpendicular if and only if the graph of f ◦ g has slope −1.