(5 points) Triangle ABC has its vertices at the following coordinates: $$ \mathrm{A}(0,3), \mathrm{B}(-4,-2), \mathrm{C}(5,-1) $$. Give the coordinates of the image triangle $$ \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}^{\prime} $$ after a $$ 90^{\circ} $$ counterclockwise rotation about the origin.

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The coordinates of the image triangle $$ A'B'C' $$ after the $$ 90^{\circ} $$ counterclockwise rotation are $$ A'(-3, 0), B'(2, -4), C'(1, 5) $$. To perform a $$ 90^{\circ} $$ counterclockwise rotation about the origin, we can use the following transformation for a point $$(x, y)$$: $$ (x, y) \rightarrow (-y, x) $$ Now, we will apply this transformation to each vertex of triangle ABC. 1. For point $$ A(0, 3) $$: $$ A' = (-3, 0) $$ 2. For point $$ B(-4, -2) $$: $$ B' = (2, -4) $$ 3. For point $$ C(5, -1) $$: $$ C' = (1, 5) $$ Thus, the coordinates of the image triangle $$ A'B'C' $$ after the $$ 90^{\circ} $$ counterclockwise rotation about the origin are: $$ A'(-3, 0), \quad B'(2, -4), \quad C'(1, 5) $$

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