Question

dilate the triangle A(5, 5)B(1, 7)C(3, 2) using a magnitude of 1/2 and a center of dilation at (2, 2) to form triangle A'B'C'. what are the coordinates of the dilated triangle?

Answer 1

To dilate the triangle \( ABC \) using a magnitude of \( \frac{1}{2} \) with a center of dilation at the point \( (2, 2) \), we can use the formula for dilation of a point \( (x, y) \) about a center \( (x_c, y_c) \) with a scale factor \( k \):

\[
(x', y') = \left( x_c + k(x - x_c), y_c + k(y - y_c) \right)
\]

Here, \( k = \frac{1}{2} \) and the center of dilation \( (x_c, y_c) = (2, 2) \).

We will apply this formula to each of the points \( A(5, 5) \), \( B(1, 7) \), and \( C(3, 2) \).

### For point \( A(5, 5) \):

\[
x' = 2 + \frac{1}{2}(5 - 2) = 2 + \frac{1}{2} \cdot 3 = 2 + 1.5 = 3.5
\]

\[
y' = 2 + \frac{1}{2}(5 - 2) = 2 + \frac{1}{2} \cdot 3 = 2 + 1.5 = 3.5
\]

So, \( A' = (3.5, 3.5) \).

### For point \( B(1, 7) \):

\[
x' = 2 + \frac{1}{2}(1 - 2) = 2 + \frac{1}{2} \cdot (-1) = 2 - 0.5 = 1.5
\]

\[
y' = 2 + \frac{1}{2}(7 - 2) = 2 + \frac{1}{2} \cdot 5 = 2 + 2.5 = 4.5
\]

So, \( B' = (1.5, 4.5) \).

### For point \( C(3, 2) \):

\[
x' = 2 + \frac{1}{2}(3 - 2) = 2 + \frac{1}{2} \cdot 1 = 2 + 0.5 = 2.5
\]

\[
y' = 2 + \frac{1}{2}(2 - 2) = 2 + \frac{1}{2} \cdot 0 = 2 + 0 = 2
\]

So, \( C' = (2.5, 2) \).

### Summary of the Dilated Triangle Coordinates

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

- \( A' = (3.5, 3.5) \)
- \( B' = (1.5, 4.5) \)
- \( C' = (2.5, 2) \)

Thus, the final coordinates for the dilated triangle \( A'B'C' \) are \( (3.5, 3.5) \), \( (1.5, 4.5) \), and \( (2.5, 2) \).