Eureka Math Precalculus Module 2 Lesson 22 Answer Key

Engage NY Eureka Math Precalculus Module 2 Lesson 22 Answer Key

Eureka Math Precalculus Module 2 Lesson 22 Exercise Answer Key

Opening Exercise
a. Find parametric equations of the line through point P(1, 1) in the direction of vector \(\left[\begin{array}{c}
– 2 \\
3
\end{array}\right]\).
Answer:
A vector form of the equation is \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
1 \\
1
\end{array}\right] + \left[\begin{array}{c}
– 2 \\
3
\end{array}\right] t\), which gives parametric equations x(t) = 1 – 2t and y(t) = 1 + 3t for any real number t.

b. Find parametric equations of the line through point P(2, 3, 1) in the direction of vector \(\left[\begin{array}{c}
4 \\
1 \\
– 1
\end{array}\right]\).
Answer:
A vector form of the equation is \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
3 \\
1
\end{array}\right] + \left[\begin{array}{c}
4 \\
1 \\
– 1
\end{array}\right] t\), which gives parametric equations x(t) = 2 + 4t and y(t) = 3 + t and z(t) = 1 – t for any real number t.

Exercises
Exercise 1.
Consider points P(2, 1, 4) and Q(3, – 1, 2), and define a linear transformation by L\(\left(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\right)\) = \(\left[\begin{array}{ccc}
1 & 2 & – 1 \\
0 & 1 & 2 \\
3 & – 1 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\). Find parametric equations to describe the image of \(\overleftrightarrow{P Q}\) under the transformation L.
Answer:
Direction vector: \(\left[\begin{array}{c}
3 \\
– 1 \\
2
\end{array}\right] – \left[\begin{array}{l}
2 \\
1 \\
4
\end{array}\right]\) = \(\left[\begin{array}{c}
1 \\
– 2 \\
– 2
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
1 \\
4
\end{array}\right] + \left[\begin{array}{c}
1 \\
– 2 \\
– 2
\end{array}\right]\) for all real numbers t
Parametric Equations: x(t) = 2 + t, y(t) = 1 – 2t, and z(t) = 4 – 2t for all real numbers t

Exercise 2.
The process that we developed for images of lines in R3 also applies to lines in R2. Consider points P(2, 3) and Q( – 1, 4). Define a linear transformation by L\(\left(\left[\begin{array}{l}
x \\
y
\end{array}\right]\right)\) = \(\left[\begin{array}{cc}
1 & 2 \\
3 & – 1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\). Find parametric equations to describe the image of \(\overleftrightarrow{P Q}\) under the transformation L.
Answer:
Direction vector: \(\left[\begin{array}{c}
– 1 \\
4
\end{array}\right] – \left[\begin{array}{l}
2 \\
3
\end{array}\right]\) = \(\left[\begin{array}{c}
– 3 \\
1
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
3
\end{array}\right] + \left[\begin{array}{c}
– 3 \\
1
\end{array}\right] t\) for all real numbers t
Parametric Equations: x(t) = 2 – 3t and y(t) = 3 + t for all real numbers t

Exercise 3.
Not only is the image of a line under a linear transformation another line, but the image of a line segment under a linear transformation is another line segment. Let P, Q, and L be as specified in Exercise 2. Find parametric equations to describe the image of \(\overline{P Q}\) under the transformation L.
Answer:
Direction vector: \(\left[\begin{array}{c}
– 1 \\
4
\end{array}\right] – \left[\begin{array}{l}
2 \\
3
\end{array}\right]\) = \(\left[\begin{array}{c}
– 3 \\
1
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
3
\end{array}\right] + \left[\begin{array}{c}
– 3 \\
1
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric Equations: x(t) = 2 – 3t and y(t) = 3 + t for 0 ≤ t ≤ 1

Eureka Math Precalculus Module 2 Lesson 22 Problem Set Answer Key

Question 1.
Find parametric equations of \(\overleftrightarrow{P Q}\) through points P and Q in the plane.
a. P(1, 3), Q(2, – 5)
Answer:
Direction vector: v = \(\left[\begin{array}{c}
2 \\
– 5
\end{array}\right] – \left[\begin{array}{l}
1 \\
3
\end{array}\right]\) = \(\left[\begin{array}{c}
1 \\
– 8
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
1 \\
3
\end{array}\right] + \left[\begin{array}{c}
1 \\
– 8
\end{array}\right] t\) for all real numbers t
Parametric Equations: x(t) = 1 + t and y(t) = 3 – 8t for all real numbers t

b. P(3, 1), Q(0, 2)
Answer:
Direction vector: v = \(\left[\begin{array}{l}
0 \\
2
\end{array}\right] – \left[\begin{array}{l}
3 \\
1
\end{array}\right]\) = \(\left[\begin{array}{c}
– 3 \\
1
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
3 \\
1
\end{array}\right] + \left[\begin{array}{c}
– 3 \\
1
\end{array}\right] t\) for all real numbers t
Parametric Equations: x(t) = 3 – 3t and y(t) = 1 + t for all real numbers t

c. P( – 2, 2), Q( – 3, – 4)
Answer:
Direction vector: v = \(\left[\begin{array}{l}
– 3 \\
– 4
\end{array}\right] – \left[\begin{array}{c}
– 2 \\
2
\end{array}\right]\) = \(\left[\begin{array}{l}
– 1 \\
– 6
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{c}
– 2 \\
2
\end{array}\right] + \left[\begin{array}{l}
– 1 \\
– 6
\end{array}\right] t\) for all real numbers t
Parametric Equations: x(t) = – 2 – t and y(t) = 2 – 6t for all real numbers t

Question 2.
Find parametric equations of \(\overleftrightarrow{P Q}\) through points P and Q in space.
a. P(1, 0, 2), Q(4, 3, 1)
Answer:
Direction vector: v = \(\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right] – \left[\begin{array}{l}
1 \\
0 \\
2
\end{array}\right]\) = \(\left[\begin{array}{c}
3 \\
3 \\
– 1
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
1 \\
0 \\
2
\end{array}\right] + \left[\begin{array}{c}
3 \\
3 \\
– 1
\end{array}\right] t\) for all real numbers t
Parametric Equations: x(t) = 1 + 3t, y(t) = 3t, and z(t) = 2 – t for all real numbers t

b. P(3, 1, 2), Q(2, 8, 3)
Answer:
Direction vector: v = \(\left[\begin{array}{l}
2 \\
8 \\
3
\end{array}\right] – \left[\begin{array}{l}
3 \\
1 \\
2
\end{array}\right]\) = \(\left[\begin{array}{c}
– 1 \\
7 \\
1
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
3 \\
1 \\
2
\end{array}\right] + \left[\begin{array}{c}
– 1 \\
7 \\
1
\end{array}\right] t\) for all real numbers t
Parametric Equations: x(t) = 3 – t, y(t) = 1 + 7t, and z(t) = 2 + t for all real numbers t

c. P(1, 4, 0), Q( – 2, 1, – 1)
Answer:
Direction vector: v = \(\left[\begin{array}{c}
– 2 \\
1 \\
– 1
\end{array}\right] – \left[\begin{array}{l}
1 \\
4 \\
0
\end{array}\right]\) = \(\left[\begin{array}{l}
– 3 \\
– 3 \\
– 1
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
1 \\
4 \\
0
\end{array}\right] + \left[\begin{array}{l}
– 3 \\
– 3 \\
– 1
\end{array}\right] t\) for all real numbers t.
Parametric Equations: x(t) = 1 – 3t, y(t) = 4 – 3t, and z(t) = – t for all real numbers t.

Question 3.
Find parametric equations of \(\overline{P Q}\) through points P and Q in the plane.
a. P(2, 0), Q(2, 10)
Answer:
Direction vector: v = \(\left[\begin{array}{c}
2 \\
10
\end{array}\right] – \left[\begin{array}{l}
2 \\
0
\end{array}\right]\) = \(\left[\begin{array}{c}
0 \\
10
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
0
\end{array}\right] + \left[\begin{array}{c}
0 \\
10
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric Equations: x(t) = 2 and y(t) = 10t for 0 ≤ t ≤ 1

b. P(1, 6), Q( – 3, 5)
Answer:
Direction vector: v = \(\left[\begin{array}{c}
– 3 \\
5
\end{array}\right] – \left[\begin{array}{l}
1 \\
6
\end{array}\right]\) = \(\left[\begin{array}{l}
– 4 \\
– 1
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
1 \\
6
\end{array}\right] + \left[\begin{array}{l}
– 4 \\
– 1
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric Equations: x(t) = 1 – 4t and y(t) = 6 – t for 0 ≤ t ≤ 1

c. ( – 2, 4), Q(6, 9)
Answer:
Direction vector: v = \(\left[\begin{array}{l}
6 \\
9
\end{array}\right] – \left[\begin{array}{c}
– 2 \\
4
\end{array}\right]\) = \(\left[\begin{array}{l}
8 \\
5
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{c}
– 2 \\
4
\end{array}\right] + \left[\begin{array}{l}
8 \\
5
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric Equations: x(t) = – 2 + 8t and y(t) = 4 + 5t for 0 ≤ t ≤ 1

Question 4.
Find parametric equations of \(\overline{P Q}\) through points P and Q in space.
a. P(1, 1, 1), Q(0, 0, 0)
Answer:
Direction vector: v = \(\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] – \left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
– 1 \\
– 1 \\
– 1
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right] + \left[\begin{array}{l}
– 1 \\
– 1 \\
– 1
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric Equations: x(t) = 1 – t, y(t) = 1 – t and z(t) = 1 – t for 0 ≤ t ≤ 1

b. P(2, 1, – 3), Q(1, 1, 4)
Answer:
Direction vector: v = \(\left[\begin{array}{l}
1 \\
1 \\
4
\end{array}\right] – \left[\begin{array}{c}
2 \\
1 \\
– 3
\end{array}\right]\) = \(\left[\begin{array}{c}
– 1 \\
0 \\
7
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{c}
2 \\
1 \\
– 3
\end{array}\right] + \left[\begin{array}{c}
– 1 \\
0 \\
7
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric Equations: x(t) = 2 – t, y(t) = t and z(t) = – 3 + 7t for 0 ≤ t ≤ 1

c. P(3, 2, 1), Q(1, 2, 3)
Answer:
Direction vector: v = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right] – \left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\) = \(\left[\begin{array}{c}
– 2 \\
0 \\
2
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right] + \left[\begin{array}{c}
– 2 \\
0 \\
2
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric Equations: x(t) = 3 – 2t, y(t) = 2, and z(t) = 1 + 2t for 0 ≤ t ≤ 1

Question 5.
Jeanine claims that the parametric equations x(t) = 3 – t and y(t) = 4 – 3t describe the line through points P(2, 1) and Q(3, 4). Is she correct? Explain how you know.
Answer:
Yes, she is correct. If t = 1, then x(t) = 2 and y(t) = 1, so the line passes through point P. If t = 0, then x(t) = 3 and y(t) = 4, so the line passes through point Q.

Question 6.
Kelvin claims that the parametric equations x(t) = 3 + t and y(t) = 4 + 3t describe the line through points P(2, 1) and Q(3, 4). Is he correct? Explain how you know.
Answer:
Yes, he is correct. If t = – 1, then x(t) = 2 and y(t) = 1, so the line passes through point P. If t = 0, then x(t) = 3 and y(t) = 4, so the line passes through point Q.

Question 7.
LeRoy claims that the parametric equations x(t) = 1 + 3t and y(t) = – 2 + 9t describe the line through points P(2, 1) and Q(3, 4). Is he correct? Explain how you know.
Answer:
Yes, he is correct. If = \(\frac{1}{3}\), then x(t) = 2 and y(t) = 1, so the line passes through point P. If = \(\frac{2}{3}\), then x(t) = 3 and y(t) = 4, so the line passes through point Q.

Question 8.
Miranda claims that the parametric equations x(t) = – 2 + 2t and y(t) = 3 – t describe the line through points P(2, 1) and Q(3, 4). Is she correct? Explain how you know.
Answer:
No, she is not correct. If t = 2, then x(t) = 2 and y(t) = 1, so the line passes through point P. However, when we solve –2 + 2t = 3, we find t = \(\frac{5}{2}\), and when we solve 3 – t = 4, we find that t = – 1. Thus, there is no value of t so that (x(t), y(t)) = (3, 4), so this line does not pass through point Q.

Question 9.
Find parametric equations of the image of \(\overleftrightarrow{P Q}\) under the transformation L\(\left(\left[\begin{array}{l}
x \\
y
\end{array}\right]\right)\) = A\(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) for the given points P, Q, and matrix A.
a. P(2, 4), Q(5, – 1), A = \(\)
Answer:
L(P) = \(\left[\begin{array}{cc}
1 & 2 \\
– 2 & 1
\end{array}\right]\left[\begin{array}{c}
1 \\
– 2
\end{array}\right]\) = \(\left[\begin{array}{l}
– 3 \\
– 4
\end{array}\right]\) and L(Q) = \(\left[\begin{array}{cc}
1 & 2 \\
– 2 & 1
\end{array}\right]\left[\begin{array}{l}
0 \\
0
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0
\end{array}\right]\) so v = \(\left[\begin{array}{l}
0 \\
0
\end{array}\right] – \left[\begin{array}{l}
– 3 \\
– 4
\end{array}\right]\) = \(\left[\begin{array}{l}
3 \\
4
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
– 3 \\
– 4
\end{array}\right] + \left[\begin{array}{l}
3 \\
4
\end{array}\right] t\) for all real numbers t
Parametric equations: x(t) = 14 – 12t and y(t) = 10 – 7t for all real numbers t

b. P(1, – 2), Q(0, 0), A = \(\)
Answer:
L(P) = \(\) = \(\)and L(Q) = \(\) = \(\) so v = \(\) = \(\)
Vector equation: \(\) = \(\) for all real numbers t
Parametric equations: x(t) = – 3 + 3t and y(t) = – 4 + 4t for all real numbers t

c. P(2, 3), Q(1, 10), A = \(\left[\begin{array}{ll}
1 & 4 \\
0 & 1
\end{array}\right]\)
Answer:
L(P) = \(\left[\begin{array}{ll}
1 & 4 \\
0 & 1
\end{array}\right]\left[\begin{array}{l}
2 \\
3
\end{array}\right]\) = \(\left[\begin{array}{c}
14 \\
3
\end{array}\right]\) and L(Q) = \(\left[\begin{array}{ll}
1 & 4 \\
0 & 1
\end{array}\right]\left[\begin{array}{c}
1 \\
10
\end{array}\right]\) = \(\left[\begin{array}{c}
41 \\
10
\end{array}\right]\) so v = \(\left[\begin{array}{l}
41 \\
10
\end{array}\right] – \left[\begin{array}{c}
14 \\
3
\end{array}\right]\) = \(\left[\begin{array}{c}
27 \\
7
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{c}
14 \\
3
\end{array}\right] + \left[\begin{array}{c}
27 \\
7
\end{array}\right] t\) for all real numbers t
Parametric equations: x(t) = 14 + 27t and y(t) = 3 + 7t for all real numbers t

Question 10.
Find parametric equations of the image of \(\overleftrightarrow{P Q}\) under the transformation L\(\left(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\right)\) = A\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) for the given points P, Q, and matrix A.
a. P(1, – 2, 1), Q( – 1, 1, 3), A = \(\left[\begin{array}{lll}
2 & 1 & 0 \\
0 & 1 & 1 \\
1 & 2 & 3
\end{array}\right]\)
Answer:
L(P) = \(\left[\begin{array}{lll}
2 & 1 & 0 \\
0 & 1 & 1 \\
1 & 2 & 3
\end{array}\right]\left[\begin{array}{c}
1 \\
– 2 \\
1
\end{array}\right]\) = \(\left[\begin{array}{c}
0 \\
– 1 \\
0
\end{array}\right]\)and L(Q) = \(\left[\begin{array}{lll}
2 & 1 & 0 \\
0 & 1 & 1 \\
1 & 2 & 3
\end{array}\right]\left[\begin{array}{c}
– 1 \\
1 \\
3
\end{array}\right]\) = \(\left[\begin{array}{c}
– 1 \\
4 \\
10
\end{array}\right]\) so v = \(\left[\begin{array}{c}
– 1 \\
4 \\
10
\end{array}\right] – \left[\begin{array}{c}
0 \\
– 1 \\
0
\end{array}\right]\) = \(\left[\begin{array}{c}
– 1 \\
5 \\
10
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
Z
\end{array}\right]\) = \(\left[\begin{array}{c}
0 \\
– 1 \\
0
\end{array}\right] + \left[\begin{array}{c}
– 1 \\
5 \\
10
\end{array}\right] t\) for all real numbers t
Parametric equations: x(t) = – t and y(t) = – 1 + 5t and z(t) = 10t for all real numbers t

b. P(2, 1, 4), Q(1, – 1, – 3), A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 2 \\
1 & 0 & 1
\end{array}\right]\)
Answer:
L(P) = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 2 \\
1 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
2 \\
1 \\
4
\end{array}\right]\) = \(\left[\begin{array}{c}
7 \\
11 \\
6
\end{array}\right]\) and L(Q) = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 2 \\
1 & 0 & 1
\end{array}\right]\left[\begin{array}{c}
1 \\
– 1 \\
– 3
\end{array}\right]\) = \(\left[\begin{array}{l}
– 3 \\
– 6 \\
– 2
\end{array}\right]\) so v = \(\left[\begin{array}{l}
– 3 \\
– 6 \\
– 2
\end{array}\right] – \left[\begin{array}{c}
7 \\
11 \\
6
\end{array}\right]\) = \(\left[\begin{array}{c}
– 10 \\
– 17 \\
– 8
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{c}
7 \\
11 \\
6
\end{array}\right] + \left[\begin{array}{c}
– 10 \\
– 17 \\
– 8
\end{array}\right] t\) for all real numbers t
Parametric equations: x(t) = 7 – 10t and y(t) = 11 – 17t and z(t) = 6 – 8t for all real numbers t

c. P(0, 0, 1), Q(4, 2, 3), A = \(\left[\begin{array}{lll}
1 & 3 & 0 \\
1 & 1 & 1 \\
0 & 2 & 1
\end{array}\right]\)
Answer:
L(P) = \(\left[\begin{array}{lll}
1 & 3 & 0 \\
1 & 1 & 1 \\
0 & 2 & 1
\end{array}\right]\left[\begin{array}{l}
0 \\
0 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
1 \\
1
\end{array}\right]\) and L(Q) = \(\left[\begin{array}{lll}
1 & 3 & 0 \\
1 & 1 & 1 \\
0 & 2 & 1
\end{array}\right]\left[\begin{array}{l}
4 \\
2 \\
3
\end{array}\right]\) = \(\left[\begin{array}{c}
10 \\
9 \\
7
\end{array}\right]\) so v = \(\left[\begin{array}{c}
10 \\
9 \\
7
\end{array}\right] – \left[\begin{array}{l}
0 \\
1 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
9 \\
8 \\
6
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
1 \\
1
\end{array}\right] + \left[\begin{array}{l}
9 \\
8 \\
6
\end{array}\right] t\) for all real numbers t
Parametric equations: x(t) = 9t, y(t) = 1 + 8t and z(t) = 1 + 6t for all real numbers t

Question 11.
Find parametric equations of the image of \(\overline{P Q}\) under the transformation L\(\left(\left[\begin{array}{l}
x \\
y
\end{array}\right]\right)\) = A\(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) for the given points P, Q, and matrix A.
a. P(2, 1), Q( – 1, – 1), A = \(\left[\begin{array}{ll}
1 & 3 \\
1 & 2
\end{array}\right]\)
Answer:
L(P) = \(\left[\begin{array}{ll}
1 & 3 \\
1 & 2
\end{array}\right]\left[\begin{array}{l}
2 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
5 \\
4
\end{array}\right]\) and L(Q) = \(\left[\begin{array}{ll}
1 & 3 \\
1 & 2
\end{array}\right]\left[\begin{array}{l}
– 1 \\
– 1
\end{array}\right]\) = \(\left[\begin{array}{l}
– 4 \\
– 3
\end{array}\right]\) so v = \(\left[\begin{array}{l}
– 4 \\
– 3
\end{array}\right] – \left[\begin{array}{l}
5 \\
4
\end{array}\right]\) = \(\left[\begin{array}{l}
– 9 \\
– 7
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
5 \\
4
\end{array}\right] + \left[\begin{array}{l}
– 9 \\
– 7
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric equations: x(t) = 5 – 9t and y(t) = 4 – 7t for 0 ≤ t ≤ 1

b. P(0, 0), Q(4, 2), A = \(\left[\begin{array}{cc}
1 & 2 \\
– 2 & 1
\end{array}\right]\)
Answer:
L(P) = \(\left[\begin{array}{cc}
1 & 2 \\
– 2 & 1
\end{array}\right]\left[\begin{array}{l}
0 \\
0
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0
\end{array}\right]\) and L(Q) = \(\left[\begin{array}{cc}
1 & 2 \\
– 2 & 1
\end{array}\right]\left[\begin{array}{l}
4 \\
2
\end{array}\right]\) = \(\left[\begin{array}{c}
8 \\
– 6
\end{array}\right]\) so v = \(\left[\begin{array}{c}
8 \\
– 6
\end{array}\right] – \left[\begin{array}{l}
0 \\
0
\end{array}\right]\) = \(\left[\begin{array}{c}
8 \\
– 6
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0
\end{array}\right] + \left[\begin{array}{c}
8 \\
– 6
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric equations: x(t) = 8t and y(t) = – 6t for 0 ≤ t ≤ 1

c. P(3, 1), Q(1, – 2), A = \(\left[\begin{array}{ll}
1 & 4 \\
0 & 1
\end{array}\right]\)
Answer:
L(P) = \(\left[\begin{array}{ll}
1 & 4 \\
0 & 1
\end{array}\right]\left[\begin{array}{l}
3 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
7 \\
1
\end{array}\right]\) and L(Q) = \(\left[\begin{array}{c}
1 \\
– 2
\end{array}\right]]\) = \(\left[\begin{array}{l}
– 7 \\
– 2
\end{array}\right]\) so v = \(\left[\begin{array}{c}
– 7 \\
– 2
\end{array}\right] – \left[\begin{array}{l}
7 \\
1
\end{array}\right]\) = \(\left[\begin{array}{c}
– 14 \\
– 3
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
7 \\
1
\end{array}\right] + \left[\begin{array}{c}
– 14 \\
– 3
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric equations: x(t) = 7 – 14t and y(t) = 1 – 3t for 0 ≤ t ≤ 1

Question 12.
Find parametric equations of the image of \(\overline{P Q}\) under the transformation L\(\left(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\right)\) = A\(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) for the given points P, Q, and matrix A.
a. P(0, 1, 1), Q( – 1, 1, 2), A = \(\left[\begin{array}{lll}
2 & 1 & 0 \\
0 & 1 & 1 \\
1 & 2 & 3
\end{array}\right]\)
Answer:
L(P) = \(\left[\begin{array}{lll}
2 & 1 & 0 \\
0 & 1 & 1 \\
1 & 2 & 3
\end{array}\right]\left[\begin{array}{l}
0 \\
1 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
1 \\
2 \\
5
\end{array}\right]\) and L(Q) = \(\left[\begin{array}{lll}
2 & 1 & 0 \\
0 & 1 & 1 \\
1 & 2 & 3
\end{array}\right]\left[\begin{array}{c}
– 1 \\
1 \\
2
\end{array}\right]\) = [\(\left[\begin{array}{c}
– 1 \\
3 \\
7
\end{array}\right]\) so v = \(\left[\begin{array}{c}
– 1 \\
3 \\
7
\end{array}\right] – \left[\begin{array}{l}
1 \\
2 \\
5
\end{array}\right]\) = \(\left[\begin{array}{c}
– 2 \\
1 \\
2
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
1 \\
2 \\
5
\end{array}\right] + \left[\begin{array}{c}
– 2 \\
1 \\
2
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric equations: x(t) = 1 – 2t, y(t) = 2 + t, and z(t) = 5 + 2t for 0 ≤ t ≤ 1

b. P(2, 1, 1), Q(1, 1, 2), A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 2 \\
1 & 0 & 1
\end{array}\right]\)
Answer:
L(P) = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 2 \\
1 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
2 \\
1 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
4 \\
5 \\
3
\end{array}\right]\)and L(Q) = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 2 \\
1 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right]\) = \(\left[\begin{array}{l}
4 \\
6 \\
3
\end{array}\right]\) so v = \(\left[\begin{array}{l}
4 \\
6 \\
3
\end{array}\right] – \left[\begin{array}{l}
4 \\
5 \\
3
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
4 \\
5 \\
3
\end{array}\right] + \left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric equations: x(t) = 4, y(t) = 5 + t, and z(t) = 3 for 0 ≤ t ≤ 1

c. P(0, 0, 1), Q(1, 0, 0), A = \(\left[\begin{array}{lll}
1 & 3 & 0 \\
1 & 1 & 1 \\
0 & 2 & 1
\end{array}\right]\)
Answer:
L(P) = \(\left[\begin{array}{lll}
1 & 3 & 0 \\
1 & 1 & 1 \\
0 & 2 & 1
\end{array}\right]\left[\begin{array}{l}
0 \\
0 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
1 \\
1
\end{array}\right]\) and L(Q) = \(\left[\begin{array}{lll}
1 & 3 & 0 \\
1 & 1 & 1 \\
0 & 2 & 1
\end{array}\right]\left[\begin{array}{l}
1 \\
0 \\
0
\end{array}\right]\) = \(\left[\begin{array}{l}
1 \\
1 \\
0
\end{array}\right]\) so v = \(\left[\begin{array}{l}
1 \\
1 \\
0
\end{array}\right] – \left[\begin{array}{l}
0 \\
1 \\
1
\end{array}\right]\) = \(\left[\begin{array}{c}
1 \\
0 \\
– 1
\end{array}\right]\)
Vector equation: \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
1 \\
1
\end{array}\right] + \left[\begin{array}{c}
1 \\
0 \\
– 1
\end{array}\right] t\) for 0 ≤ t ≤ 1
Parametric equations: x(t) = t, y(t) = 1, and z(t) = 1 – t for 0 ≤ t ≤ 1

Eureka Math Precalculus Module 2 Lesson 22 Exit Ticket Answer Key

Question 1.
Consider points P(2, 1) and Q(2, 5). Find parametric equations that describe points on \(\overline{P Q}\).
Answer:
A direction vector v is given by v = \(\left[\begin{array}{l}
2 \\
5
\end{array}\right] – \left[\begin{array}{l}
2 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
4
\end{array}\right]\), so a vector form of \(\overline{P Q}\) is \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
1
\end{array}\right] + \left[\begin{array}{l}
0 \\
4
\end{array}\right] \boldsymbol{t}\)
for 0 ≤ t ≤ 1. This gives the parametric equations x(t) = 2 and y(t) = 1 + 4t for 0 ≤ t ≤ 1.

Question 2.
Suppose that points P(2, 1) and Q(2, 5) are transformed under the linear transformation L\(\left(\left[\begin{array}{l}
x \\
y
\end{array}\right]\right)\) = \(\left[\begin{array}{cc}
1 & – 3 \\
0 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\). Find parametric equations that describe the image of \(\overleftrightarrow{P Q}\) under this transformation.
Answer:
The images of P and Q are
L(P) = \(\left[\begin{array}{cc}
1 & – 3 \\
0 & 1
\end{array}\right]\left[\begin{array}{l}
2 \\
1
\end{array}\right]\) = \(\left[\begin{array}{c}
– 1 \\
1
\end{array}\right]\)
L(Q) = \(\left[\begin{array}{cc}
1 & – 3 \\
0 & 1
\end{array}\right]\left[\begin{array}{l}
2 \\
5
\end{array}\right]\) = \(\left[\begin{array}{c}
– 13 \\
5
\end{array}\right]\)
The direction vector v is then v = \(\left[\begin{array}{c}
– 13 \\
5
\end{array}\right] – \left[\begin{array}{c}
– 1 \\
1
\end{array}\right]\) = \(\left[\begin{array}{c}
– 12 \\
4
\end{array}\right]\), so the vector form of the image of \(\overleftrightarrow{P Q}\) is \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{c}
– 1 \\
1
\end{array}\right] + \left[\begin{array}{c}
– 12 \\
4
\end{array}\right]\) for all real numbers t.
Parametric equations that represent the limit of \(\overleftrightarrow{P Q}\) are x(t) = – 1 – 12t and y(t) = 1 + 4t for all real numbers t.

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