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Since A is the center of the dilation, it does not move and we will have A^\prime = A. For vertex B, we know that B^\prime will lie on \overleftrightarrow{AB} because dilations preserve lines through the center of dilation. Since the scale factor is \frac{1}{3} we know that |AB^\prime| = \frac{1}{3}|AB|. From A to B is 4 units to the right and one unit up. Therefore, one third of the way from A to B (along \overline{AB}) will be the point B^\prime = \left(1+ \frac{4}{3}, 1 + \frac{1}{3}\right). Applying this technique to C which is 2 units to the right and 4 units up from A, we find C^\prime = \left(1\frac{2}{3}, 2\frac{1}{3} \right).
The scaled triangle A^\prime B^\prime C^\prime is pictured below:
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Since B is the center of the dilation, it does not move and we will have B^{\prime \prime} = B. For vertex A, we know that A^{\prime \prime} will lie on \overleftrightarrow{AB} because dilations preserve lines through the center of dilation. Since the scale factor is \frac{2}{3} we know that |BA^{\prime \prime}| = \frac{2}{3}|BA|. From B to A is 4 units to the left and one unit down. Therefore, two thirds of the way from B to A (along \overline{AB}) will be the point A^{\prime \prime} = \left(5 - \frac{8}{3}, 2 - \frac{2}{3}\right). Applying this technique to C which is 2 units to the left and 3 units up from B, we find C^{\prime \prime} = \left(3\frac{2}{3}, 4 \right).
The scaled triangle A^{\prime \prime} B^{\prime \prime} C^{\prime \prime} is pictured below:
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The calculations above show that B^\prime = A^{\prime \prime} = \left( 2 \frac{1}{3}, 1 \frac{1}{3} \right). To see why these vertices are the same, note that the sum of the dilation factors \frac{1}{3} and \frac{2}{3} is 1. The dilation about A maps \overline{AB} to a the first third of \overline{AB} while the dilation about B maps \overline{AB} to the second two thirds of this segment. The two scaled triangles with a shared vertex are pictured below: