Question

error analysis your friend incorrectly says that the reflection of \\( \triangle efg \\) to its image \\( \triangle efg \\) is a reflection across the x - axis. what is the correct reflection? what is your friends mistake?
what is the correct description of the reflection?
a. \\( \triangle efg \\) is the image of \\( \triangle efg \\) after a reflection across the line \\( y = - 7 \\).
b. \\( \triangle efg \\) is the image of \\( \triangle efg \\) after a reflection across the line \\( y = - 2 \\).
c. \\( \triangle efg \\) is the image of \\( \triangle efg \\) after a reflection across the x - axis.
d. \\( \triangle efg \\) is the image of \\( \triangle efg \\) after a reflection across the line \\( x = - 2 \\).

Explanation:

Step1: Recall reflection rules

When reflecting a point \((x,y)\) across the line \(x = a\), the new \(x\)-coordinate is \(2a - x\) and \(y\)-coordinate remains the same. When reflecting across the \(x\)-axis \((y = 0)\), the \(x\)-coordinate remains the same and \(y\)-coordinate becomes \(-y\). When reflecting across the line \(y = b\), the new \(y\)-coordinate is \(2b - y\) and \(x\)-coordinate remains the same.

Step2: Analyze the given triangles

Let's assume a general point \(E(- 8,2)\) in \(\triangle EFG\) and its image \(E'(4,2)\) in \(\triangle E'F'G'\). Using the formula for reflection across the line \(x=a\): If \(x\) - coordinate of a point changes from \(x_1=-8\) to \(x_2 = 4\), then \(2a-x_1=x_2\). Substituting \(x_1=-8\) and \(x_2 = 4\) into \(2a-x_1=x_2\), we get \(2a-(-8)=4\), \(2a=4 - 8=-4\), \(a=-2\).

Answer:

D. \(\triangle E'F'G'\) is the image of \(\triangle EFG\) after a reflection across the line \(x=-2\)