Tamika is trying to dilate a triangle drawn on a coordinate plane using a magnitude of 4 with reference to the origin. If the original triangle has vertices at coordinate points (12,0) , (5,0) , and (4,3) , what are the coordinate points of the new vertices of the dilated triangle?

Answer 1

To perform a dilation of a figure on the coordinate plane, we use the following formula for dilating the coordinates of a point (x, y) by a scale factor \( k \) with respect to the origin:

\[ (x', y') = (kx, ky) \]

In this case, Tamika is dilating the triangle with a scale factor \( k = 4 \) and the original vertices of the triangle are located at points \( (12, 0) \), \( (5, 0) \), and \( (4, 3) \).

Let's dilate each vertex of the triangle:

1. For the vertex (12, 0): \[ (x', y') = (4 \cdot 12, 4 \cdot 0) = (48, 0) \]

2. For the vertex (5, 0): \[ (x', y') = (4 \cdot 5, 4 \cdot 0) = (20, 0) \]

3. For the vertex (4, 3): \[ (x', y') = (4 \cdot 4, 4 \cdot 3) = (16, 12) \]

Now, we can summarize the coordinates of the new vertices of the dilated triangle:

• The new vertices after dilation are:
  • \( (48, 0) \)
  • \( (20, 0) \)
  • \( (16, 12) \)

Thus, the coordinate points of the new vertices of the dilated triangle are (48, 0), (20, 0), and (16, 12).