# 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 $$\left[\begin{array}{c} – 2 \\ 3 \end{array}\right]$$.
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 $$\left[\begin{array}{c} 4 \\ 1 \\ – 1 \end{array}\right]$$.
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.
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.
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.
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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.
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.
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.
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.
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 = 
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 = 
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 = $$\left[\begin{array}{ll} 1 & 4 \\ 0 & 1 \end{array}\right]$$
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 = $$\left[\begin{array}{lll} 2 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 3 \end{array}\right]$$
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 = $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{array}\right]$$
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 = $$\left[\begin{array}{lll} 1 & 3 & 0 \\ 1 & 1 & 1 \\ 0 & 2 & 1 \end{array}\right]$$
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 = $$\left[\begin{array}{ll} 1 & 3 \\ 1 & 2 \end{array}\right]$$
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 = $$\left[\begin{array}{cc} 1 & 2 \\ – 2 & 1 \end{array}\right]$$
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 = $$\left[\begin{array}{ll} 1 & 4 \\ 0 & 1 \end{array}\right]$$
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 = $$\left[\begin{array}{lll} 2 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 3 \end{array}\right]$$
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 = $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{array}\right]$$
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 = $$\left[\begin{array}{lll} 1 & 3 & 0 \\ 1 & 1 & 1 \\ 0 & 2 & 1 \end{array}\right]$$
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 $$\overline{P Q}$$.
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 $$\left[\begin{array}{l} x \\ y \end{array}\right]$$ = $$\left[\begin{array}{l} 2 \\ 1 \end{array}\right] + \left[\begin{array}{l} 0 \\ 4 \end{array}\right] \boldsymbol{t}$$
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.
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.