Question

(a) the arrows below show that the coordinates on the left are mapped to the coordinates on the right. fill in the blanks to give the coordinates after the rotation.
original coordinates → final coordinates
d(1, -5) → d(-1, 5)
e(2, 1) → e(\boxed{\quad})
f(4, -7) → f(\boxed{\quad})
(b) choose the general rule below that describes the rotation mapping $\triangle def$ to $\triangle def$.
$\circ\\ (x,y)\to(-x,-y)$
$\circ\\ (x,y)\to(-y,-x)$
$\circ\\ (x,y)\to(y,-x)$
$\circ\\ (x,y)\to(x,-y)$
$\circ\\ (x,y)\to(-y,x)$
$\circ\\ (x,y)\to(-x,y)$
$\circ\\ (x,y)\to(y,x)$

Explanation:

Part (a)

Step 1: Analyze the rotation for point \( E \)

Original coordinates of \( E \) are \( (2, 1) \). Let's assume the rotation rule (we can check with part (b) later, but for now, observe the pattern from \( D \): \( D(1, -5) \to D'(-1, 5) \), which is \( (x,y) \to (-x, -y) \) reversed? Wait, no, \( D(1, -5) \) to \( D'(-1, 5) \): \( x \) becomes \( -x \), \( y \) becomes \( -y \)? Wait \( 1 \to -1 \) (so \( -x \)), \( -5 \to 5 \) (so \( -y \)). So for \( E(2, 1) \), applying \( (x,y) \to (-x, -y) \)? Wait no, \( E(2,1) \) should map to \( E' \). Wait, looking at the graph, \( E \) is at \( (2,1) \), and \( E' \) is at \( (-2, -1) \)? Wait no, the original \( E \) is in the fourth quadrant? Wait the left graph: \( E \) is at \( (-2, -1) \)? Wait no, the grid: let's check the coordinates. Wait the original triangle \( DEF \): \( D \) is at \( (1, -5) \), \( E \) at \( (2, 1) \), \( F \) at \( (4, -7) \). The rotated triangle \( D'E'F' \): \( D' \) is at \( (-1, 5) \), so \( (x,y) \to (-x, -y) \)? Wait \( D(1, -5) \to (-1, 5) \): \( x \) becomes \( -x \), \( y \) becomes \( -y \) (since \( -5 \) becomes \( 5 \), which is \( -(-5) \)). So for \( E(2, 1) \), applying \( (x,y) \to (-x, -y) \), we get \( E'(-2, -1) \)? Wait but the blank is \( E'( \square, \square) \). Wait maybe I misread the graph. Wait the original \( E \) is at \( (2,1) \), and the rotated \( E' \) is at \( (-2, -1) \)? Wait no, let's check the rotation rule from part (b). Let's do part (b) first.

Part (b)

Let's check each option:

1. \( (x,y) \to (-x, -y) \): This is a rotation of \( 180^\circ \) about the origin. Let's test with \( D(1, -5) \): \( (-1, 5) \), which matches \( D'(-1, 5) \). Let's test \( E(2, 1) \): \( (-2, -1) \). Let's test \( F(4, -7) \): \( (-4, 7) \). Let's check the other options:

1. \( (x,y) \to (-y, -x) \): For \( D(1, -5) \), \( (5, -1)

eq (-1,5) \).

1. \( (x,y) \to (y, -x) \): For \( D(1, -5) \), \( (-5, -1)

eq (-1,5) \).

1. \( (x,y) \to (x, -y) \): For \( D(1, -5) \), \( (1, 5)

eq (-1,5) \).

1. \( (x,y) \to (-y, x) \): For \( D(1, -5) \), \( (5, 1)

eq (-1,5) \).

1. \( (x,y) \to (-x, y) \): For \( D(1, -5) \), \( (-1, -5)

eq (-1,5) \).

1. \( (x,y) \to (y, x) \): For \( D(1, -5) \), \( (-5, 1)

eq (-1,5) \).

So the correct rule is \( (x,y) \to (-x, -y) \), which is a \( 180^\circ \) rotation about the origin.

Back to Part (a)

Using the rule \( (x,y) \to (-x, -y) \):

• For \( E(2, 1) \): \( x = 2 \), \( y = 1 \), so \( -x = -2 \), \( -y = -1 \). Thus, \( E'(-2, -1) \).

• For \( F(4, -7) \): \( x = 4 \), \( y = -7 \), so \( -x = -4 \), \( -y = 7 \). Thus, \( F'(-4, 7) \).

Part (a) Answers:

• \( E(2, 1) \to E'(\boldsymbol{-2}, \boldsymbol{-1}) \)
• \( F(4, -7) \to F'(\boldsymbol{-4}, \boldsymbol{7}) \)

Part (b) Answer:

The correct option is \( \boldsymbol{(x,y) \to (-x, -y)} \) (the first option).

Answer:

Part (a)

Step 1: Analyze the rotation for point \( E \)

Original coordinates of \( E \) are \( (2, 1) \). Let's assume the rotation rule (we can check with part (b) later, but for now, observe the pattern from \( D \): \( D(1, -5) \to D'(-1, 5) \), which is \( (x,y) \to (-x, -y) \) reversed? Wait, no, \( D(1, -5) \) to \( D'(-1, 5) \): \( x \) becomes \( -x \), \( y \) becomes \( -y \)? Wait \( 1 \to -1 \) (so \( -x \)), \( -5 \to 5 \) (so \( -y \)). So for \( E(2, 1) \), applying \( (x,y) \to (-x, -y) \)? Wait no, \( E(2,1) \) should map to \( E' \). Wait, looking at the graph, \( E \) is at \( (2,1) \), and \( E' \) is at \( (-2, -1) \)? Wait no, the original \( E \) is in the fourth quadrant? Wait the left graph: \( E \) is at \( (-2, -1) \)? Wait no, the grid: let's check the coordinates. Wait the original triangle \( DEF \): \( D \) is at \( (1, -5) \), \( E \) at \( (2, 1) \), \( F \) at \( (4, -7) \). The rotated triangle \( D'E'F' \): \( D' \) is at \( (-1, 5) \), so \( (x,y) \to (-x, -y) \)? Wait \( D(1, -5) \to (-1, 5) \): \( x \) becomes \( -x \), \( y \) becomes \( -y \) (since \( -5 \) becomes \( 5 \), which is \( -(-5) \)). So for \( E(2, 1) \), applying \( (x,y) \to (-x, -y) \), we get \( E'(-2, -1) \)? Wait but the blank is \( E'( \square, \square) \). Wait maybe I misread the graph. Wait the original \( E \) is at \( (2,1) \), and the rotated \( E' \) is at \( (-2, -1) \)? Wait no, let's check the rotation rule from part (b). Let's do part (b) first.

Part (b)

Let's check each option:

1. \( (x,y) \to (-x, -y) \): This is a rotation of \( 180^\circ \) about the origin. Let's test with \( D(1, -5) \): \( (-1, 5) \), which matches \( D'(-1, 5) \). Let's test \( E(2, 1) \): \( (-2, -1) \). Let's test \( F(4, -7) \): \( (-4, 7) \). Let's check the other options:

1. \( (x,y) \to (-y, -x) \): For \( D(1, -5) \), \( (5, -1)

eq (-1,5) \).

1. \( (x,y) \to (y, -x) \): For \( D(1, -5) \), \( (-5, -1)

eq (-1,5) \).

1. \( (x,y) \to (x, -y) \): For \( D(1, -5) \), \( (1, 5)

eq (-1,5) \).

1. \( (x,y) \to (-y, x) \): For \( D(1, -5) \), \( (5, 1)

eq (-1,5) \).

1. \( (x,y) \to (-x, y) \): For \( D(1, -5) \), \( (-1, -5)

eq (-1,5) \).

1. \( (x,y) \to (y, x) \): For \( D(1, -5) \), \( (-5, 1)

eq (-1,5) \).

So the correct rule is \( (x,y) \to (-x, -y) \), which is a \( 180^\circ \) rotation about the origin.

Back to Part (a)

Using the rule \( (x,y) \to (-x, -y) \):

• For \( E(2, 1) \): \( x = 2 \), \( y = 1 \), so \( -x = -2 \), \( -y = -1 \). Thus, \( E'(-2, -1) \).

• For \( F(4, -7) \): \( x = 4 \), \( y = -7 \), so \( -x = -4 \), \( -y = 7 \). Thus, \( F'(-4, 7) \).

Part (a) Answers:

• \( E(2, 1) \to E'(\boldsymbol{-2}, \boldsymbol{-1}) \)
• \( F(4, -7) \to F'(\boldsymbol{-4}, \boldsymbol{7}) \)

Part (b) Answer:

The correct option is \( \boldsymbol{(x,y) \to (-x, -y)} \) (the first option).