A triangle A(-2,-5)B(0,3)C(6,-3) is drawn on a coordinate plane. dilate the figure with a magnitude of 3 in reference to the origin. what are the coordinates of the new vertices of A'B'C'?

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To dilate a triangle (or any figure) from the origin with a scale factor (magnitude) of 3, you multiply the coordinates of each vertex by the scale factor.

The original vertices of the triangle are:

  • A(-2, -5)
  • B(0, 3)
  • C(6, -3)

Now, applying the dilation with a scale factor of 3:

  1. For vertex A: \[ A' = (3 \times -2, 3 \times -5) = (-6, -15) \]

  2. For vertex B: \[ B' = (3 \times 0, 3 \times 3) = (0, 9) \]

  3. For vertex C: \[ C' = (3 \times 6, 3 \times -3) = (18, -9) \]

Thus, the coordinates of the new vertices after the dilation are:

  • A'(-6, -15)
  • B'(0, 9)
  • C'(18, -9)

So, the new vertices of triangle A'B'C' are:

  • A'(-6, -15)
  • B'(0, 9)
  • C'(18, -9)
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