G-GPE Scaling a Triangle in the Coordinate Plane
Below is a picture of \(\triangle ABC\) with vertices lying on grid points:
- Draw the image of \(\triangle ABC\) when it is scaled with a scale factor of \(\frac{1}{3}\) about the vertex \(A\): label this triangle \(A^\prime B^\prime C^\prime\) and find the coordinates of the points \(A^\prime\), \(B^\prime\), and \(C^\prime\).
- Draw the image of \(\triangle ABC\) when it is scaled with a scale factor of \(\frac{2}{3}\) about the vertex \(B\): label this triangle \(A^{\prime \prime}B^{\prime \prime}C^{\prime \prime}\) and find the coordinates of the points \(A^{\prime \prime}\), \(B^{\prime \prime}\), and \(C^{\prime \prime}\).
- How does \(A^{\prime \prime}\) compare \(B^{\prime}\)? Why?
Solutions
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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 \(A^{\prime \prime} = \left(4 - \frac{8}{3}, 2 - \frac{2}{3}\right)\). 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:
Commentary
The goal of this task is to apply dilations to a triangle in the coordinate plane. Students will need to find the coordinates of a point which cuts a given segment into one third (part a) and into two thirds (part b). Because the centers for the two dilations are vertices of \(\triangle ABC\) and the sum of the dilation factors (one third and two thirds) is one, the two dilated triangles share a vertex. A more challenging question, in place of (b) and (c), would be to ask students which dilations they can apply about \(B\) so that the image will share a vertex with the triangle constructed in part (a): there are two possibilities, namely \(\frac{2}{3}\) and 1.