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Translate quadrilateral ADCB 5 units to the left and 2 units down. What are the coordinates of new quadrilateral A′D′C′B′ ? (1 point) Responses A′(−1,0), D′(−1,5), C′(4,2), B′(2,−1) upper A prime left parenthesis negative 1 comma 0 right parenthesis , upper D prime left parenthesis negative 1 comma 5 right parenthesis , upper C prime left parenthesis 4 comma 2 right parenthesis , upper B prime left parenthesis 2 comma negative 1 right parenthesis A′(9,−4), D′(9,1), C′(13,−2), B′(12,−5) upper A prime left parenthesis 9 comma negative 4 right parenthesis , upper D prime left parenthesis 9 comma 1 right parenthesis , upper C prime left parenthesis 13 comma negative 2 right parenthesis , upper B prime left parenthesis 12 comma negative 5 right parenthesis A′(−1,−4), D′(−1,1), C′(4,−2), B′(2,−5) upper A prime left parenthesis negative 1 comma negative 4 right parenthesis , upper D prime left parenthesis negative 1 comma 1 right parenthesis , upper C prime left parenthesis 4 comma negative 2 right parenthesis , upper B prime left parenthesis 2 comma negative 5 right parenthesis A′(9,0), D′(9,5), C′(13,2), B′(12,−1).

5 answers

To translate the quadrilateral ADCB 5 units to the left and 2 units down, you will subtract 5 from the x-coordinates of each vertex and subtract 2 from the y-coordinates of each vertex.

Let's denote the original coordinates of points A, D, C, and B as follows:

  • A(x₁, y₁)
  • D(x₂, y₂)
  • C(x₃, y₃)
  • B(x₄, y₄)

The new coordinates after translation will be:

  • A′(x₁ - 5, y₁ - 2)
  • D′(x₂ - 5, y₂ - 2)
  • C′(x₃ - 5, y₃ - 2)
  • B′(x₄ - 5, y₄ - 2)

Without the original coordinates of the vertices A, D, C, and B, I cannot provide the exact new coordinates. However, you can apply the translation as described above to convert the original coordinates to the new coordinates.

You would then match your results with the provided response options to identify which set of coordinates is correct after the translation. If you provide the original coordinates, I can help you compute the new positions.

What do you think would be the new coordinates?

I need the original coordinates of quadrilateral ADCB in order to accurately translate it. However, without knowing the original coordinates, I can only explain how to apply the translation.

Let's assume that the original coordinates are:

  • A(x₁, y₁)
  • D(x₂, y₂)
  • C(x₃, y₃)
  • B(x₄, y₄)

You would apply the translation as follows:

  • A′(x₁ - 5, y₁ - 2)
  • D′(x₂ - 5, y₂ - 2)
  • C′(x₃ - 5, y₃ - 2)
  • B′(x₄ - 5, y₄ - 2)

For example, let's say the coordinates were:

  • A(4, 2)
  • D(6, 7)
  • C(9, 4)
  • B(7, 1)

After applying the translation of 5 units left and 2 units down, the new coordinates would be:

  • A′(4 - 5, 2 - 2) = A′(-1, 0)
  • D′(6 - 5, 7 - 2) = D′(1, 5)
  • C′(9 - 5, 4 - 2) = C′(4, 2)
  • B′(7 - 5, 1 - 2) = B′(2, -1)

So, the new coordinates in this example would be:

  • A′(-1, 0)
  • D′(1, 5)
  • C′(4, 2)
  • B′(2, -1)

You can apply the same process to the actual coordinates of quadrilateral ADCB when you have them. If you can provide the specific original coordinates, I would be able to calculate the exact new coordinates for you.

The original coordinates are, A is, (4,-2) B is, (7,-3) C is, (9,0) D is, (4,3)

Thank you for providing the original coordinates. Let's translate the quadrilateral ADCB with the following original coordinates:

  • A(4, -2)
  • B(7, -3)
  • C(9, 0)
  • D(4, 3)

To translate the quadrilateral 5 units to the left and 2 units down, we will subtract 5 from the x-coordinates and 2 from the y-coordinates:

  1. For point A(4, -2):

    • A′(4 - 5, -2 - 2) = A′(-1, -4)

  2. For point B(7, -3):

    • B′(7 - 5, -3 - 2) = B′(2, -5)

  3. For point C(9, 0):

    • C′(9 - 5, 0 - 2) = C′(4, -2)

  4. For point D(4, 3):

    • D′(4 - 5, 3 - 2) = D′(-1, 1)

Now, the new coordinates of the translated quadrilateral A′D′C′B′ are:

  • A′(-1, -4)
  • B′(2, -5)
  • C′(4, -2)
  • D′(-1, 1)

So the final new coordinates are:

  • A′(-1, -4)
  • B′(2, -5)
  • C′(4, -2)
  • D′(-1, 1)

The option that corresponds to this result is: A′(−1,−4), D′(−1,1), C′(4,−2), B′(2,−5).