Question

Triangle FDP is reduced with a scale factor of ½ and a center of (0, 0).  Find the coordinates of the new coordinates of the vertex, P’ (__, __):

Answer 1

P = (-6,0)

Answer 2

To find the coordinates of the new vertex \( P' \) of triangle FDP after it has been reduced by a scale factor of \( \frac{1}{2} \) with respect to the center at \( (0, 0) \), you take the coordinates of vertex \( P \) (let's assume the original coordinates of \( P \) are \( (x_P, y_P) \)) and apply the scale transformation.

The formula to find the new coordinates after scaling is given by:

\[
P' = \left( x_P \cdot \text{scale factor}, y_P \cdot \text{scale factor} \right)
\]

Substituting the scale factor of \( \frac{1}{2} \):

\[
P' = \left( x_P \cdot \frac{1}{2}, y_P \cdot \frac{1}{2} \right)
\]

This means you simply multiply both the x and y coordinates of point \( P \) by \( \frac{1}{2} \).

For example, if the original coordinates of \( P \) are \( (4, 6) \), then the new coordinates \( P' \) will be:

\[
P' = \left( 4 \cdot \frac{1}{2}, 6 \cdot \frac{1}{2} \right) = (2, 3)
\]

To find the specific coordinates for vertex \( P' \), simply apply this process using the actual coordinates of vertex \( P \). If you provide the coordinates of \( P \), I can give you the exact coordinates of \( P' \).

Answer 3

Given the coordinates of vertex \( P \) as \( (-6, 0) \), we will apply the scale factor of \( \frac{1}{2} \) to find the new coordinates \( P' \).

Using the scaling formula:

\[
P' = \left( x_P \cdot \frac{1}{2}, y_P \cdot \frac{1}{2} \right)
\]

We substitute \( x_P = -6 \) and \( y_P = 0 \):

\[
P' = \left( -6 \cdot \frac{1}{2}, 0 \cdot \frac{1}{2} \right)
\]

Calculating these values:

\[
P' = \left( -3, 0 \right)
\]

Thus, the coordinates of the new vertex \( P' \) are \( \boxed{(-3, 0)} \).