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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 email if I am in error.
d(P,Q) =
d(Q,R) =
d(R, P) =
The perimeter is
Question. Is this triangle a righttriangle?
(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:
t^{2} − 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, x_{1} = x_{2}:
since x_{1} = x_{2}  
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 xcoordinates 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 onehalf 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 onehalf 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) = 2x^{2} − 3x; evaluate f(2), f(−2), f(−1/2).
f(2) = 2 · 2^{2} − 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 x^{2} + 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 x^{2} −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) = x^{2} − 3x + 2
substitute  
factor  
solved! 
Presentation of Answer: The xintercepts are
(c) f(x) = x^{3} + 6x^{2} ++8x
substitute  
factor  
solved! 
Presentation of Answer: The xintercepts are
(d) f(x) = x^{2} + x + 1
substitute ( quadratic equation )  
negative discriminant 
No Solutions
Presentation of Answer: The function does not cross the xaxis.
Solutions to Exercises
(e) f(x) = 2x^{2} − x − 1
substitute  
positive discriminant  
quadratic formula  
simplify  
Note: This equation could have also been solved byfactoring
the lefthand side.
Presentation of Answer: The xintercepts are
(f) f(x) = x^{2} + x − 3
substitute  
positive discriminant  
quadratic formula 
Presentation of Answer: The xintercepts are
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