Parallelogram ABCD is rotated to create image A'B'C'D'. On a coordinate plane, 2 parallelograms are shown. The first parallelogram has points A (2, 5), B (5, 4), C (5, 2), D (2, 3). The second parallelogram has points A prime (5, negative 2), B prime (4, negative 5), C prime (2, negative 5), D prime (3, negative 2). Which rule describes the transformation? (x, y) → (y, –x) (x, y) → (–y, x) (x, y) → (–x, –y) (x, y) → (x, –y)

Respuesta :

Answer:

(x,y) → (y,-x)      

Step-by-step explanation:

We are given two parallelograms.

ABCD with coordinates

A=(2,5)

B=(5,4)

C=(5,2)

D=(2,3)

Another parallelogram A'B'C'D' with coordinates

A'=(5,-2)

B'=(4,-5)

C'=(2,-5)

D'=(3,-2)

Now, A is transformed to A' that is (2,5) is transformed to (5,-2)

We see that the y coordinate becomes the x-coordinate after transformation and the negative of x coordinates becomes y coordinate after transformation.

A similar pattern can be seen for the B, C and D.

Thus, the transformation is given by:

(x,y) → (y,-x)

A translation transformation is a transformation in which all the points on an object are translated in the same direction

The rule that describes the transformation of parallelogram ABCD to parallelogram A'B'C'D' is (x, y) → (y, -x)

The reasons the above selection is correct are as follows;

The coordinates of the vertices of the parallelogram ABCD are;

A(2, 5) , B(5, 4), C(5, 2), and D(2, 3)

The coordinates of the vertices of the parallelogram A'B'C'D' are;

A'(5, -2), B'(4, -5), C'(2, -5), and D'(3, -2)

Required:

To select the rule that determines the transformation

Solution:

By observation, we have;

The x-coordinate and y-coordinate values of the parallelogram ABCD are the same as the y-coordinate and negative x-coordinate values of parallelogram A'B'C'D', respectively

Therefore the rule that describes the transformation is (x, y) → (y, -x)

Which gives;

A(2, 5) [tex]\underset \longrightarrow {(x, \ y) \rightarrow (y, \ -x)}[/tex] A'(5, -2)

B(5, 4) [tex]\underset \longrightarrow {(x, \ y) \rightarrow (y, \ -x)}[/tex] B'(4, -5)

C(5, 2) [tex]\underset \longrightarrow {(x, \ y) \rightarrow (y, \ -x)}[/tex] C'(2, -5)

D(2, 3) [tex]\underset \longrightarrow {(x, \ y) \rightarrow (y, \ -x)}[/tex] D'(3, -2)

The correct option for the rule is (x, y) → (y, -x)

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