Question

Quadrilateral ABCD is graphed on a coordinate plane with the vertices A(-9,6), 3(-4,-6), C(-3,2), and D(-3,-2)

Answer 1

A. The translation rule is (x , y) → (x - 2 , y + 6)

B. The coordinates of the vertices of the image are A' (-11 , 12) , B' (-6 , 0) , C' (-5 , 8) , D' (-5 , 4)

Step-by-step explanation:

Let us revise the translation of a point

• If the point (x , y) translated horizontally to the right by h units  then its image is (x + h , y)
• If the point (x , y) translated horizontally to the left by h units  then its image is (x - h , y)
• If the point (x , y) translated vertically up by k units then its image is (x , y + k)
• If the point (x , y) translated vertically down by k units then its image is (x , y - k)

A.

∵ Quadrilateral ABCD is translated to the left 2 units and up 6 units

- That means subtract 2 from each x-coordinates and add 6

  to each y-coordinates

∵ (x , y) is moved to the left 2 units ad up 6 units

∴ Its image is (x - 2 , y + 6)

The translation rule is (x , y) → (x - 2 , y + 6)

B.

∵ The vertices of the quadrilateral ABCD are

A (-9 , 6) , B (-4 , -6) , C (-3 , 2) , D (-3 , -2)

∵ Quadrilateral ABCD is translated to the left 2 units and up 6 units

- Use the rule in part A to find the image of each vertex

∵ A' = (-9 - 2 , 6 + 6)

∴ A' = (-11 , 12)

∵ B' = (-4 - 2 , -6 + 6)

∴ B' = (-6 , 0)

∵ C' = (-3 - 2 , 2 + 6)

∴ C' = (-5 , 8)

∵ D' = (-3 - 2 , -2 + 6)

∴ D' = (-5 , 4)

The coordinates of the vertices of the image are A' (-11 , 12) , B' (-6 , 0) , C' (-5 , 8) , D' (-5 , 4)

Learn more:

You can learn more about the translation in brainly.com/question/2451812

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