contestada

What set of reflections would carry hexagon ABCDEF onto itself? Hexagon ABCDEF on the coordinate plane with point
A at 0, 1,
point B at negative 1, 0,
point C at negative 2, 1,
point D at negative 2, 3,
point E at negative 1, 4,
and point F at 0, 3.
A. x-axis, y=x, x-axis, y=x
B. y-axis, x-axis, y-axis
C. x-axis, y-axis, y-axis
D. y=x, x-axis, y=x, y-axis

Respuesta :

sqdancefan
You need an even number of reflections to get the original figure, leaving only A and D as potentially viable choices.

The reflection across the x-axis makes the tranformation (x, y) ⇒ (x, -y).
The reflection across the line y=x makse the transformation (x, y) ⇒ (y, x).
The two pairs of transformations of A give
  (x, y) ⇒ (x, -y) ⇒ (-y, x) ⇒ (-y, -x) ⇒ (-x, -y) . . . . . not the original

The reflection across the y-axis makes the transformation (x, y) ⇒ (-x, y).
The two pairs of transformations of D give
  (x, y) ⇒ (y, x) ⇒ (y, -x) ⇒ (-x, y) ⇒ (x, y) . . . . . . the original point

Selection D is appropriate.
MrRoyal

When a point is reflected, it must be reflected across a line. The reflection of ABCDEF that carries it onto itself is:

D. y=x, x-axis, y=x, y-axis

Of the given options, (D) is correct.

The proof is as follows:

Using coordinate point A.

We have:

[tex]A = (0,1)[/tex]

The first reflection in (D) is [tex]y = x[/tex]

This means that:

[tex](x,y) \to (y,x)[/tex]

So, we have:

[tex](0,1) \to (1,0)[/tex]

The next is across the x-axis.

The rule of this reflection is: [tex](x,y) \to (x,-y)[/tex]

So, we have:

[tex](1,0) \to (1,0)[/tex]

The next is [tex]y = x[/tex]

This means that: [tex](x,y) \to (y,x)[/tex]

So, we have:

[tex](1,0) \to (0,1)[/tex]

The last is across the y-axis.

The rule of this reflection is: [tex](x,y) \to (-x,y)[/tex]

So, we have:

[tex](0,1) \to (0,1)[/tex]

Compare the end point to point A;

We can see that both points are the same i.e. (0,1)

Hence, the set of reflections would carry hexagon ABCDEF onto itself is (d)

Read more about reflections at:

https://brainly.com/question/17983440

Otras preguntas