Question

4.1 reflections · spiral review
graph the image of each figure under the given reflection. determine the coordinates of the image.

1. △abc in the line y = x
2. trapezoid defg in the line x = -1
3. parallelogram rstu in the line y = x
4. square klmn in the line y = -2

Explanation:

Step1: Analyze Reflection over \( y = x \)

For a point \((a, b)\), reflection over \( y = x \) swaps \( x \) and \( y \)-coordinates, giving \((b, a)\). First, find coordinates of \( \triangle ABC \) vertices. From graph: \( A(-3, 2) \), \( B(0, 1) \), \( C(-2, -2) \).

Step2: Apply Reflection to Each Vertex

• Reflect \( A(-3, 2) \): Swap \( x \) and \( y \), get \( A'(2, -3) \).
• Reflect \( B(0, 1) \): Swap \( x \) and \( y \), get \( B'(1, 0) \).
• Reflect \( C(-2, -2) \): Swap \( x \) and \( y \), get \( C'(-2, -2) \)? Wait, no: \( (-2, -2) \) swapped is \( (-2, -2) \)? Wait, no, \( C \) is \((-2, -2)\)? Wait, original graph: Let's recheck. \( A \) is at \( x=-3, y=2 \); \( B \) at \( x=0, y=1 \); \( C \) at \( x=-2, y=-2 \)? Wait, no, looking at the first graph, \( C \) is at \( (-2, -3) \)? Maybe I misread. Let's correct: Assume \( A(-3, 2) \), \( B(0, 1) \), \( C(-2, -3) \). Then reflection over \( y=x \): \( A'(2, -3) \), \( B'(1, 0) \), \( C'(-3, -2) \). Then plot these points and connect to get the reflected triangle.

(Note: For other problems, similar steps: identify vertex coordinates, apply reflection rule (over \( x=-1 \): distance from \( x \)-coordinate to \( -1 \) is \( |x - (-1)| = |x + 1| \), so new \( x \)-coordinate is \( -1 - (x + 1) = -x - 2 \); over \( y=-2 \): new \( y \)-coordinate is \( -2 - (y + 2) = -y - 4 \), etc. Then plot the image.)

Answer:

To solve each reflection problem:

1. \(\triangle ABC\) over \( y = x \):

• Find vertices: \( A(-3, 2) \), \( B(0, 1) \), \( C(-2, -3) \) (approx from graph).
• Reflect: \( A'(2, -3) \), \( B'(1, 0) \), \( C'(-3, -2) \). Plot and connect.

1. Trapezoid \( DEFG \) over \( x = -1 \):

• Let \( D(0, -2) \), \( E(1, 2) \), \( F(3, 2) \), \( G(4, -2) \) (approx). For \( x = -1 \), new \( x = -1 - (x - (-1)) = -x - 2 \).
• \( D'(-2, -2) \), \( E'(-3, 2) \), \( F'(-5, 2) \), \( G'(-6, -2) \)? Wait, no, distance from \( x \) to \( -1 \) is \( |x - (-1)| \), so new \( x = -1 - (x - (-1)) = -1 - x - 1 = -x - 2 \). So \( D(0, -2) \to x=-0 -2=-2 \), \( y=-2 \): \( D'(-2, -2) \); \( E(1, 2) \to x=-1 -2=-3 \), \( y=2 \): \( E'(-3, 2) \); \( F(3, 2) \to x=-3 -2=-5 \), \( y=2 \): \( F'(-5, 2) \); \( G(4, -2) \to x=-4 -2=-6 \), \( y=-2 \): \( G'(-6, -2) \). Plot these.

1. Parallelogram \( RSTU \) over \( y = x \):

• Find vertices: \( R(-2, 2) \), \( S(2, 3) \), \( T(2, -2) \), \( U(-2, -3) \) (approx). Reflect over \( y=x \): \( R'(2, -2) \), \( S'(3, 2) \), \( T'(-2, 2) \), \( U'(-3, -2) \). Plot and connect.

1. Square \( KLMN \) over \( y = -2 \):

• Let \( K(-1, 0) \), \( L(-2, 2) \), \( M(1, 3) \), \( N(2, 1) \) (approx). Reflect over \( y=-2 \): new \( y = -2 - (y - (-2)) = -2 - y - 2 = -y - 4 \).
• \( K'(-1, -4) \), \( L'(-2, -8) \)? Wait, no: \( y \)-coordinate of \( L \) is \( 2 \), so distance to \( y=-2 \) is \( |2 - (-2)| = 4 \), so new \( y = -2 - 4 = -6 \). Correct: \( K(-1, 0) \to y=-2 - (0 + 2) = -4 \), so \( K'(-1, -4) \); \( L(-2, 2) \to y=-2 - (2 + 2) = -6 \), \( L'(-2, -6) \); \( M(1, 3) \to y=-2 - (3 + 2) = -7 \), \( M'(1, -7) \); \( N(2, 1) \to y=-2 - (1 + 2) = -5 \), \( N'(2, -5) \). Plot these.

(The key is identifying vertex coordinates, applying the reflection rule (swap \( x \) and \( y \) for \( y=x \); \( x \to -x - 2 \) for \( x=-1 \); \( y \to -y - 4 \) for \( y=-2 \)), then plotting the image vertices and connecting them.)