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6

A
(1, 6)
5
C = (6, 4)
4
3
2
1
B = (3, 0)
0
1
2
3
4
5
6
7
8
-1
If you dilate triangle ABC to form X'Y'Z' using a magnitude of 2 and a center of dilation at (2,0), what are the coordinates of the dilated triangle?



(1 point)
Responses

A' (1.5,3), B' (2.5,0), C' (4,2)
A' (1.5,3), B' (2.5,0), C' (4,2)

A' (0,12), B' (4,0), C' (10,8)
A' (0,12), B' (4,0), C' (10,8)

A' (.5,3), B' (1.5,0), C' (3,2)
A' (.5,3), B' (1.5,0), C' (3,2)

A' (2,12), B' (6,0), C' (12,8)

1 answer

To find the coordinates of the dilated triangle \( A'B'C' \) from triangle \( ABC \) with a center of dilation at \( (2,0) \) and a magnitude of \( 2 \), we use the formula for dilation:

\[ P' = C + k(P - C) \]

where:

  • \( P \) are the original coordinates of the points,
  • \( C \) is the center of dilation,
  • \( k \) is the magnitude of dilation,
  • \( P' \) are the new coordinates after dilation.

Given points:

  • \( A = (1, 6) \)
  • \( B = (3, 0) \)
  • \( C = (6, 4) \)

Center of dilation:

  • \( C = (2, 0) \)

Magnitude of dilation:

  • \( k = 2 \)

Now we perform the dilation for each point.

For point A:

\[ A' = (2, 0) + 2 \left( (1, 6) - (2, 0) \right) \] \[ = (2, 0) + 2 \left( (-1, 6) \right) \] \[ = (2, 0) + (-2, 12) \] \[ = (0, 12) \]

For point B:

\[ B' = (2, 0) + 2 \left( (3, 0) - (2, 0) \right) \] \[ = (2, 0) + 2 \left( (1, 0) \right) \] \[ = (2, 0) + (2, 0) \] \[ = (4, 0) \]

For point C:

\[ C' = (2, 0) + 2 \left( (6, 4) - (2, 0) \right) \] \[ = (2, 0) + 2 \left( (4, 4) \right) \] \[ = (2, 0) + (8, 8) \] \[ = (10, 8) \]

Final dilated points:

  • \( A' = (0, 12) \)
  • \( B' = (4, 0) \)
  • \( C' = (10, 8) \)

Thus, the coordinates of the dilated triangle \( A'B'C' \) are:

  • \( A' (0, 12), B' (4, 0), C' (10, 8) \)

The correct response is: A' (0,12), B' (4,0), C' (10,8).