Question

graph the image of rhombus cdef after a reflection over the line x = 1.

Explanation:

Step1: Recall reflection formula

For a point $(x,y)$ reflected over the line $x = a$, the new - $x$ coordinate is $x'=2a - x$ and the $y$ - coordinate remains the same, i.e., $y'=y$. Here $a = 1$.

Step2: Identify original points

Assume the coordinates of the vertices of rhombus $CDEF$ are $C(x_1,y_1)$, $D(x_2,y_2)$, $E(x_3,y_3)$, $F(x_4,y_4)$. Let's say $C(6,1)$, $D(6,6)$, $E(9,9)$, $F(9,4)$ (by observing the graph).

Step3: Calculate new $x$ - coordinates

For point $C$: $x_1'=2\times1 - 6=-4$, $y_1' = 1$. So the new point $C'$ is $(-4,1)$.
For point $D$: $x_2'=2\times1 - 6=-4$, $y_2' = 6$. So the new point $D'$ is $(-4,6)$.
For point $E$: $x_3'=2\times1 - 9=-7$, $y_3' = 9$. So the new point $E'$ is $(-7,9)$.
For point $F$: $x_4'=2\times1 - 9=-7$, $y_4' = 4$. So the new point $F'$ is $(-7,4)$.

Step4: Graph the new rhombus

Plot the points $C'(-4,1)$, $D'(-4,6)$, $E'(-7,9)$, $F'(-7,4)$ on the coordinate - plane and connect them in order to get the reflected rhombus.

Answer:

Graph the points $(-4,1)$, $(-4,6)$, $(-7,9)$, $(-7,4)$ and connect them to form the reflected rhombus.