Question

error analysis describe and correct the error in graphing the image of quadrilateral efgh after the translation (x, y) → (x − 1, y − 2).
8.

error analysis describe and correct the error in graphing the image of △abc after the translation (x, y) → (x − 3, y + 2).
9.

in exercises 10 and 11, write the vertices of quadrilateral pqrs p(−2, 3), q(1, 2), r(3, −1), and s(−2, −1) after the translation

1. (x, y) → (x + 1, y - 3)
2. < 5, − 1 >
3. work backwards. k(-8, 6) is the image of k after a translation along the rule (x, y) → (x − 3, y + 6). what are the coordinates of k?

Explanation:

Problem 10:

Step1: Translate point P

Given \( P(-2, 3) \) and translation \( (x, y) \to (x + 1, y - 3) \).
For \( x \)-coordinate: \( -2 + 1 = -1 \)
For \( y \)-coordinate: \( 3 - 3 = 0 \)
So \( P'(-1, 0) \)

Step2: Translate point Q

Given \( Q(1, 2) \)
For \( x \)-coordinate: \( 1 + 1 = 2 \)
For \( y \)-coordinate: \( 2 - 3 = -1 \)
So \( Q'(2, -1) \)

Step3: Translate point R

Given \( R(3, -1) \)
For \( x \)-coordinate: \( 3 + 1 = 4 \)
For \( y \)-coordinate: \( -1 - 3 = -4 \)
So \( R'(4, -4) \)

Step4: Translate point S

Given \( S(-2, -1) \)
For \( x \)-coordinate: \( -2 + 1 = -1 \)
For \( y \)-coordinate: \( -1 - 3 = -4 \)
So \( S'(-1, -4) \)

Step1: Translate point P

Given \( P(-2, 3) \) and translation \( \langle 5, -1
angle \) (which means \( (x, y) \to (x + 5, y - 1) \))
For \( x \)-coordinate: \( -2 + 5 = 3 \)
For \( y \)-coordinate: \( 3 - 1 = 2 \)
So \( P'(3, 2) \)

Step2: Translate point Q

Given \( Q(1, 2) \)
For \( x \)-coordinate: \( 1 + 5 = 6 \)
For \( y \)-coordinate: \( 2 - 1 = 1 \)
So \( Q'(6, 1) \)

Step3: Translate point R

Given \( R(3, -1) \)
For \( x \)-coordinate: \( 3 + 5 = 8 \)
For \( y \)-coordinate: \( -1 - 1 = -2 \)
So \( R'(8, -2) \)

Step4: Translate point S

Given \( S(-2, -1) \)
For \( x \)-coordinate: \( -2 + 5 = 3 \)
For \( y \)-coordinate: \( -1 - 1 = -2 \)
So \( S'(3, -2) \)

Step1: Let coordinates of K be \( (x, y) \)

Given translation rule \( (x, y) \to (x - 3, y + 6) \) and image \( K'(-8, 6) \)
So we have equations:
\( x - 3 = -8 \) (from \( x \)-coordinate translation)
\( y + 6 = 6 \) (from \( y \)-coordinate translation)

Step2: Solve for x

From \( x - 3 = -8 \), add 3 to both sides:
\( x = -8 + 3 = -5 \)

Step3: Solve for y

From \( y + 6 = 6 \), subtract 6 from both sides:
\( y = 6 - 6 = 0 \)

Answer:

\( P'(-1, 0) \), \( Q'(2, -1) \), \( R'(4, -4) \), \( S'(-1, -4) \)

Problem 11: