Question

1. $\triangle efg$ is the image of $\triangle efg$. what reflection produces this image?

Explanation:

Step1: Identify Key Points

First, we look at the coordinates of corresponding points (e.g., \(G\) and \(G'\), \(F\) and \(F'\), \(E\) and \(E'\)). For a reflection over the \(x\)-axis, the rule is \((x,y)\to(x,-y)\). Let's check a point, say \(G\): if \(G\) has coordinates \((-4, 3)\) (estimating from the grid), then \(G'\) should be \((-4, -3)\), which matches the position of \(G'\) (around \((-4, -7)\)? Wait, maybe better to check the \(x\)-axis as the mirror. Wait, looking at the \(y\)-coordinates: original triangle is above the \(x\)-axis, image is below. The \(x\)-coordinates of corresponding points are the same, \(y\)-coordinates are negatives (or symmetric over \(x\)-axis).

Step2: Confirm Reflection Rule

The reflection over the \(x\)-axis transforms a point \((x, y)\) to \((x, -y)\). Let's take point \(F\): original \(F\) is at \((-1, 4)\) (approx), image \(F'\) is at \((-1, -8)\)? Wait, maybe the grid lines: each square is 1 unit. Let's check \(G\): \(G\) is at \((-4, 3)\), \(G'\) is at \((-4, -7)\)? Wait, no, maybe the \(x\)-axis is the line of reflection. Wait, the original triangle is above the \(x\)-axis, the image is below, symmetric with respect to the \(x\)-axis. So the reflection is over the \(x\)-axis.

Answer:

Reflection over the \(x\)-axis (or reflection across the \(x\)-axis)