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what is the correct description of the reflection? a. $\\triangle efg$ …

Question

what is the correct description of the reflection?
a. $\triangle efg$ is the image of $\triangle efg$ after a reflection across the $x$-axis
b. $\triangle efg$ is the image of $\triangle efg$ after a reflection across the line $x = -1$
c. $\triangle efg$ is the image of $\triangle efg$ after a reflection across the line $y = -1$
d. $\triangle efg$ is the image of $\triangle efg$ after a reflection across the line $y = -4$
what is your friends mistake?
a. your friend took the reflection across the $x$-axis instead of the correct line of reflection.
b. your friend took the reflection across the $x$-axis instead of the $y$-axis.
c. your friend took the reflection across the $y$-axis instead of the $x$-axis.

Explanation:

Response
First Question (Reflection Description)
Brief Explanations

To determine the correct reflection, we analyze the line of reflection. For a reflection over a horizontal line \( y = k \), the \( y \)-coordinates of points change symmetrically around \( y = k \). The line \( y=-1 \) is a horizontal line, and reflecting over \( y = - 1 \) would map points such that the distance from each point to \( y=-1 \) is the same as the distance from its image to \( y = - 1 \). Option A (x - axis reflection) would invert \( y \)-coordinates (e.g., \( (x,y)\to(x, - y) \)), which is not the case here. Option B (vertical line \( x=-1 \)) would affect \( x \)-coordinates, and option D (\( y = - 4 \)) is too far. So the correct reflection is over \( y=-1 \).

Brief Explanations

The friend's mistake was choosing a reflection over the \( x \)-axis (which is \( y = 0 \)) instead of the correct line of reflection ( \( y=-1 \) ). Option B suggests a mix - up with the \( y \)-axis, which is not relevant here. Option C has the axes reversed. So the friend took the reflection across the \( x \)-axis instead of the correct line of reflection.

Answer:

C. \( \triangle EFG' \) is the image of \( \triangle EFG \) after a reflection across the line \( y = - 1 \)

Second Question (Friend's Mistake)