Question

if you dilate triangle XYZ 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?

Answer 1

To dilate triangle XYZ with a center of dilation at (2, 0) and a magnitude of 2, we will use the following formula for dilation:

\[
X' = C_x + k(X - C_x)
\]
\[
Y' = C_y + k(Y - C_y)
\]

where \((C_x, C_y)\) is the center of dilation, \(k\) is the magnitude of dilation, and \((X, Y)\) are the coordinates of the points of triangle XYZ.

Given:
- Center of dilation: \((C_x, C_y) = (2, 0)\)
- Magnitude of dilation: \(k = 2\)

Let's say the original coordinates of the points of triangle XYZ are:
- \(X = (x_1, y_1)\)
- \(Y = (x_2, y_2)\)
- \(Z = (x_3, y_3)\)

After applying the dilation, the coordinates of the dilated triangle \(X'Y'Z'\) will be given by:

\[
X' = (2) + 2(x_1 - 2) = 2 + 2x_1 - 4 = 2x_1 - 2
\]
\[
Y' = (0) + 2(y_1 - 0) = 2y_1
\]

\[
X' = (2) + 2(x_2 - 2) = 2 + 2x_2 - 4 = 2x_2 - 2
\]
\[
Y' = (0) + 2(y_2 - 0) = 2y_2
\]

\[
X' = (2) + 2(x_3 - 2) = 2 + 2x_3 - 4 = 2x_3 - 2
\]
\[
Y' = (0) + 2(y_3 - 0) = 2y_3
\]

So the dilated coordinates will be:
- For point \(X\): \(X' = (2x_1 - 2, 2y_1)\)
- For point \(Y\): \(Y' = (2x_2 - 2, 2y_2)\)
- For point \(Z\): \(Z' = (2x_3 - 2, 2y_3)\)

If you provide the original coordinates of points \(X\), \(Y\), and \(Z\), we can calculate the exact coordinates of \(X'\), \(Y'\), and \(Z'\).