If triangle ABC is defined by the coordinates A(-4, -4), B(2, -2), C(0, 4) is dilated by a scale factor of 1 2 , with resulting vertex A' at (-2, -2). What is the center of the dilation? A) (0, 0) B) (0, 2) C) (0, 4) D) (-4, -4)

To find the center of dilation, we can use the information that the image of vertex A after the dilation is A′ with given coordinates. The dilation factor is stated as \( \frac{1}{2} \), suggesting that every point after dilation is halfway (or closer by a factor of \( \frac{1}{2} \)) to the center of dilation compared to its original position.

The original coordinates of A are (-4, -4), and after dilation, its image A' is at (-2, -2). Since the scale factor is \( \frac{1}{2} \), we’re looking for a point (the center of dilation) such that the coordinates of A' are halfway between the coordinates of A and this point.

To find this center, we can use the midpoint formula in reverse. The midpoint \( M \) between two points \( X(x_1,y_1) \) and \( Y(x_2,y_2) \) is given by:

\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

In our case, we want to find the other point Y assuming point A is \( X(-4, -4) \) and \( A' \) is the midpoint \( M(-2, -2) \).

Using the midpoint formula:

\[ -2 = \frac{-4 + x_2}{2} \] \[ -2 = \frac{-4 + y_2}{2} \]

Now solve for \( x_2 \) and \( y_2 \):

\[ -4 = -4 + x_2 \Rightarrow x_2 = 0 \] \[ -4 = -4 + y_2 \Rightarrow y_2 = 0 \]

Thus, the coordinates of the center of dilation are (0, 0).

The correct answer is A) (0, 0).