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Graph of polygon ABCDE with vertices at negative 1 comma negative 4, negative 1 comma negative 1, 3 comma negative 1, 3 comma negative 4, 1 comma negative 6. A second polygon A prime B prime C prime D prime E prime with vertices at negative 11 comma negative 4, negative 11 comma negative 1, negative 15 comma negative 1, negative 15 comma negative 4, negative 13 comma negative 6.

Determine the line of reflection.

Reflection across the x-axis
Reflection across x = −6
Reflection across the y-axis
Reflection across y = −6

User JuicyFruit
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Thanks i was looking for this
User DragonZero
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Answer:

it's clear that the polygon A' B' C' D' E' is obtained by reflecting polygon ABCDE across the x-axis. Thus, the line of reflection is the x-axis.

Explanation:

To determine the line of reflection that maps the polygon ABCDE to its mirror image A' B' C' D' E', we need to find the line of reflection for each vertex individually. Let's consider each option:

Reflection across the x-axis:

The x-axis is a horizontal line with equation y = 0. When reflecting a point across the x-axis, the x-coordinate remains the same, but the y-coordinate changes sign. So, for each vertex (x, y) in ABCDE, the corresponding reflected vertex (x', y') will have the same x-coordinate but a y-coordinate of -y.

For example:

A (-1, -4) is reflected to A' (-1, 4).

B (-1, -1) is reflected to B' (-1, 1).

C (3, -1) is reflected to C' (3, 1).

D (3, -4) is reflected to D' (3, 4).

E (1, -6) is reflected to E' (1, 6).

Reflection across x = -6:

This means we reflect the points across a vertical line with the equation x = -6. When reflecting a point across x = -6, the y-coordinate remains the same, but the x-coordinate changes sign. So, for each vertex (x, y) in ABCDE, the corresponding reflected vertex (x', y') will have the same y-coordinate but an x-coordinate of -2*6 - x = -12 - x.

For example:

A (-1, -4) is reflected to A' (-12 - (-1), -4) => A' (-11, -4).

B (-1, -1) is reflected to B' (-12 - (-1), -1) => B' (-11, -1).

C (3, -1) is reflected to C' (-12 - 3, -1) => C' (-15, -1).

D (3, -4) is reflected to D' (-12 - 3, -4) => D' (-15, -4).

E (1, -6) is reflected to E' (-12 - 1, -6) => E' (-13, -6).

Reflection across the y-axis:

The y-axis is a vertical line with equation x = 0. When reflecting a point across the y-axis, the y-coordinate remains the same, but the x-coordinate changes sign. So, for each vertex (x, y) in ABCDE, the corresponding reflected vertex (x', y') will have the same y-coordinate but an x-coordinate of -x.

For example:

A (-1, -4) is reflected to A' (-(-1), -4) => A' (1, -4).

B (-1, -1) is reflected to B' (-(-1), -1) => B' (1, -1).

C (3, -1) is reflected to C' (-3, -1).

D (3, -4) is reflected to D' (-3, -4).

E (1, -6) is reflected to E' (-1, -6).

Reflection across y = -6:

This means we reflect the points across a horizontal line with the equation y = -6. When reflecting a point across y = -6, the x-coordinate remains the same, but the y-coordinate changes sign. So, for each vertex (x, y) in ABCDE, the corresponding reflected vertex (x', y') will have the same x-coordinate but a y-coordinate of -(-6) - y = 6 - y.

For example:

A (-1, -4) is reflected to A' (-1, 6 - (-4)) => A' (-1, 10).

B (-1, -1) is reflected to B' (-1, 6 - (-1)) => B' (-1, 7).

C (3, -1) is reflected to C' (3, 6 - (-1)) => C' (3, 7).

D (3, -4) is reflected to D' (3, 6 - (-4)) => D' (3, 10).

E (1, -6) is reflected to E' (1, 6 - (-6)) => E' (1, 12).

After analyzing the results, it's clear that the polygon A' B' C' D' E' is obtained by reflecting polygon ABCDE across the x-axis. Thus, the line of reflection is the x-axis.

User Legolas
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