To solve this, we'll assume that triangle \( \triangle EFG \) has its vertices located at specific coordinates in a Cartesian coordinate system. Let's define the points \( E, F, \) and \( G \) as follows:
- \( E(x_1, y_1) \)
- \( F(x_2, y_2) \)
- \( G(x_3, y_3) \)
When you rotate a point \( (x, y) \) by \( -90^\circ \) clockwise, the new coordinates \( (x', y') \) are given by:
\[ x' = y \] \[ y' = -x \]
Now, let's apply this rotation formula to the points of \( \triangle EFG \):
-
For point \( E(x_1, y_1) \):
- New coordinates: \( E'(y_1, -x_1) \)
-
For point \( F(x_2, y_2) \):
- New coordinates: \( F'(y_2, -x_2) \)
-
For point \( G(x_3, y_3) \):
- New coordinates: \( G'(y_3, -x_3) \)
Thus, the new vertices of triangle \( \triangle XYZ \) are:
- \( X = E'(y_1, -x_1) \)
- \( Y = F'(y_2, -x_2) \)
- \( Z = G'(y_3, -x_3) \)
If you provide the specific coordinates for points \( E \), \( F \), and \( G \), I can give you the exact coordinates for the new points \( X \), \( Y \), and \( Z \).