\( X Y \) has endpoints \( X(-3,1) \) and \( Y(4,-5) \). What is the midpoint of its image after a \( 90^{\circ} \) rotation about the origin?

To find the midpoint of the line segment \( XY \) after a \( 90^{\circ} \) rotation about the origin, let's first determine the original midpoint. The formula for the midpoint \( M \) of a line segment with endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] Substituting the coordinates of points \( X(-3,1) \) and \( Y(4,-5) \): \[ M = \left( \frac{-3 + 4}{2}, \frac{1 + (-5)}{2} \right) = \left( \frac{1}{2}, \frac{-4}{2} \right) = \left( \frac{1}{2}, -2 \right) \] Next, we rotate the midpoint \( M \left( \frac{1}{2}, -2 \right) \) by \( 90^{\circ} \) counterclockwise around the origin. The formula for rotating a point \( (x, y) \) by \( 90^{\circ} \) counterclockwise is: \[ (x', y') = (-y, x) \] Applying this to our midpoint: \[ (x', y') = (-(-2), \frac{1}{2}) = (2, \frac{1}{2}) \] Therefore, the image of the midpoint after a \( 90^{\circ} \) rotation about the origin is: \[ (2, \frac{1}{2}) \]