Question

1. the graph of △def is shown. graph the image of △def after a reflection across the line y = 2. include the line of reflection. then write the coordinates of the image.

Explanation:

Step1: Recall reflection rule

For a point $(x,y)$ reflected across the horizontal line $y = k$, the new - point $(x,y')$ has the formula $y'=2k - y$. Here $k = 2$.

Step2: Assume coordinates of $\triangle DEF$

Let's assume $D(x_1,y_1)$, $E(x_2,y_2)$, and $F(x_3,y_3)$. After reflection across $y = 2$, the new coordinates will be $D'(x_1,4 - y_1)$, $E'(x_2,4 - y_2)$, $F'(x_3,4 - y_3)$.

Step3: Graph the reflection

On the coordinate - plane, draw the line $y = 2$. For each vertex of $\triangle DEF$, measure the vertical distance from the vertex to the line $y = 2$, and then plot the new vertex on the opposite side of the line $y = 2$ at the same vertical distance. Connect the new vertices to form the reflected triangle.

Since the original coordinates of the vertices of $\triangle DEF$ are not given, we can only provide the general method. If we assume $D(1,1)$, $E(- 2,-1)$, $F(3,-1)$:
For point $D(1,1)$:
The new $y$ - coordinate is $4-1 = 3$, so $D'(1,3)$.
For point $E(-2,-1)$:
The new $y$ - coordinate is $4-( - 1)=5$, so $E'(-2,5)$.
For point $F(3,-1)$:
The new $y$ - coordinate is $4-( - 1)=5$, so $F'(3,5)$.

Answer:

The general rule for the coordinates of the reflected triangle: if a vertex of $\triangle DEF$ is $(x,y)$, its image after reflection across $y = 2$ is $(x,4 - y)$. If $D(1,1)$, $E(-2,-1)$, $F(3,-1)$, then $D'(1,3)$, $E'(-2,5)$, $F'(3,5)$. Graph the line $y = 2$ and the original and reflected triangles on the coordinate - plane.