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Cartesian Coordinate System and Functions

Solutions to Exercises

8.1. Solutions: I’ll just use equation (1).
(a) P(0, 0) and Q(3,−4)

(b) P(−3, 4) and Q(−1,−1)

(c) P(−4, 2) and Q(5, 2)

(d) P(−1, 5) and Q(7, 9)

8.2. Solutions:
(a) Given P(0, 0), Q(1, 1) and R(2, 0), does PQR form a right triangle?
Solution: Verifythe fol lowing calculation :
d(P,Q) =
d(P,R) = 2
d(Q,R) =

Note that is the longest side and
d(P,R)2 = d(P,Q)2 + d(Q,R)2
Thus, the points P, Q, and R form a right triangle.

(b) Given P(6,−7), Q(11,−3) and R(2,−2), does PQR form a right
triangle?
Solution: Verifythe following calculations:
d(P,Q) =
d(P,R) =
d(Q,R) =
The side is the longest side. Note that
d(Q,R)2 = d(P,Q)2 + d(P,R)2
Thus, the points P, Q, and R form a right triangle.

(c) P(1, 2), Q(−3, 4) and R(4,−2)
Solution: Verify the following calculations:
d(P,Q) =
d(P,R) = 5
d(Q,R) =
The side is the longest side. Note that
d(Q,R)2 = 85 ≠ 45 = 20+25 = d(P,Q)2 + d(P,R)2
The square of the longest side does not equal to the sum of the
squares of the other two sides; therefore, P, Q and R do not
form a right triangle.

8.3. Solutions:
(a) Given P(1, 3), Q(2, 5) and R(4, 9), is Q between P and R?
Solution:
d(P,Q) =
d(P,R) =
d(Q,R) =
(Verifythese calculations.) Is it true that

They are equal! Indeed, Q does lie between P and R.

(b) Given P(−1, 10), Q(2,−5), and R(5,−12), is Q between P and
R?
Solution:
d(P,Q) =
d(P,R) =
d(Q,R) =
(Verifythese calculations.) Is it true that

Make a calculator calculation to see that

In this case, the left hand side is not equal to the right hand
side. Conclusion: Q is not between P and R.

8.4. Solution: The perimeter of a triangle is the sum of the lengths
of its sides.
Verifythe following calculations, and e-mail if I am in error.
d(P,Q) =
d(Q,R) =
d(R, P) =
The perimeter is

Question. Is this triangle a right-triangle?
(a) Yes
(b) No

8.5. Solution: I’ll take my own hint—I hope you did too.
Given: P(1 + t, 3 − t) and O(0, 0)

We want d(P,O) = 4, therefore,
Solve for t :

Square both sides of the equation, expand and combine .

square both sides
expand
combine
divide both sides by 2
subtract 18 from both sides

We now solve the equation t 2 − 2t − 3 = 0. We could use the Quadratic
Formula; however, this is an equation that can be solved by
factoring:
t2 − 2t − 3 = 0
(t − 3)(t + 1) = 0

Therefore,
t = −1, 3
Presentation of Solution:
There are actually two times at which the particle is exactly4 units
away from the origin.

8.6. Demonst ration : Since the points are vertically oriented, x1 = x2:

 
since x1 = x2
by (1) of Lesson 2

8.7. Solution: We use equation (3) throughout these solutions.
(a) Given: P(1, 2) and Q(1, 9), compute d(P,Q). The first coordinates
are equal; therefore, these are verticallyorien ted points.
The distance between them is the upper most minus the lower
most:
d(P,Q) = |2 − 9| = 9−2 =

(b) Given: P(−3, 3) and Q(−3,−4), compute d(P,Q). These points
are vertically oriented because the x-coordinates are equal.
d(P,Q) = |3 − (−4)| =

(c) Given: P(4, 3) and Q(−3, 3), compute d(P,Q). Here, the second
coordinates are equal; these are horizontally oriented points.
d(P,Q) = |4 − (−3)| =

(d) Given: P(π,−5) and Q(π,−2), compute d(P,Q). Vertically oriented
points.
d(P,Q) = | − 5 − (−2)| =

What do you know about that, they’re all 7 units apart! Oops! All
but one—how did that one get in there!

8.8. Solutions: Hopefully you used formula (4).
(a) Find the midpoint between P(−1, 3) and Q(5, 7).

(b) Find the midpoint between P(2, 4) and Q(2,−5).

Did you note that these two points were vertically oriented? You
did, didn’t you.

(c) Find the midpoint between P(5,−3) and Q(12,−3).

These were two horizontallyorien ted points.

(d) Find the midpoint between P(−1,−1) and Q(4, 2).

8.9. Solutions:
(a) Find the center and radius of the circle containing the points
P(1, 2) and Q(−3,−1), which are diametrically opposite.
Calculation of the Center: The radius is the midpoint of the line
segment PQ, since the points lie of the same diameter.

Radius Calculation: The radius is one-half the diameter. Thus,
from the distance formula, we have

Presentation of Answer:
Center:
Radius:

(b) Find the center and radius of the circle containing the points
P(−1, 4) and Q(0, 0), which are diametrically opposite.
Calculation of the Center: Again, the center is midway between
the endpoints of any of its diameters.

Calculation of the radius: The radius is one-half the diameter.
By the distance formula, we have

Presentation of Answers:
Center:
Radius:

8.10. Solutions: Just use the replacement technique. Be sure to verify
these calculations.
(a) Given f(x) = 2x2 − 3x; evaluate f(2), f(−2), f(−1/2).
f(2) = 2 · 22 − 3 · 2 = 2
f(−2) = 2(−2)2 − 3(−2) = 14
f(−1/2) = 2(−1/2)2 − 3(−1/2) = 2

(b) Given g(s) = s(s + 1)(s + 2); evaluate g(0), g(1), g(−1), g(−3).

(c) Given evaluate h(1), h(−2), h(1/2), h(−1/2).

(d) Given D(w) = evaluate D(9); D( 1/9 ).

8.11. Solution to: (a) Find the domain of f(x) = |x − 1|. Here, the
natural domain is all of R, the real number line .
Dom(f) = R.
This is because there are no constraints on the value of x. For any x
we can calculate x − 1 and then calculate its absolute value |x − 1|.

Solution to (b) Find the domain of In this problem,
we just must make sure the denominator is never equal to zero .

To identifythese x’s, we find where x2 + 2x − 8 = 0.
Solve:

given
factor
the solutions

Presentation of Solution:

initial description
set notation
interval notation

Solution to (c) Find the domain of Based on the
strategy, we see there are a number of constraints on the values of x:
x ≠ 0
x ≠ −2
x(x + 2) ≥ 0

These three can be summarized by a single inequality:

We use the Sign Chart Method to analyze x(x + 2).
The Sign Chart of x(x + 2)

Presentation of Solution:

initial description
set notation
interval notation

Solution to (d) Find the domain of From the basic
strategy, we see that

Realizing that x2 −1 ≠ 0 is equivalent to x ≠ −1 and x ≠ 1 is see . . .

Begin by doing a Sign Chart Analysis on
The first step is to factor completely.

It is to this expression that we now apply the method .

The Sign Chart of

Taking the blue line as our solution, and keeping in mind that x ≠ −1
and x ≠ 1 we get

initial description
set notation
interval notation

8.12. Solutions:
(a) f(x) = 4x − 1

 
substitute
add 1 to both sides
divide by 4


Presentation of Answer: The x- intercept is

(b) f(x) = x2 − 3x + 2

 
substitute
factor
solved!


Presentation of Answer: The x-intercepts are

(c) f(x) = x3 + 6x2 ++8x

 
substitute
factor
solved!


Presentation of Answer: The x-intercepts are

(d) f(x) = x2 + x + 1

 
substitute ( quadratic equation )
negative discriminant


No Solutions
Presentation of Answer: The function does not cross the x-axis.
Solutions to Exercises

(e) f(x) = 2x2 − x − 1

 
substitute
positive discriminant
quadratic formula
simplify
 


Note: This equation could have also been solved byfactoring
the left-hand side.
Presentation of Answer: The x-intercepts are

(f) f(x) = x2 + x − 3

 
substitute
positive discriminant
quadratic formula


Presentation of Answer: The x-intercepts are

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