Question

1. a figure is transformed by (x, y) → (x + 2, y − 3) followed by (x, y) → (x, 3y). does this sequence of transformations produce a pair of similar figures? explain your reasoning.

Explanation:

Step1: Analyze the first transformation

The first transformation is \((x, y) \to (x + 2, y - 3)\). This is a translation (a rigid transformation) because it shifts the figure horizontally by 2 units and vertically by - 3 units. Rigid transformations preserve the shape and size of the figure, so the figure after translation is congruent (and thus similar, with a scale factor of 1) to the original figure.

Step2: Analyze the second transformation

The second transformation is \((x, y) \to (x, 3y)\). This is a vertical stretch (a non - rigid transformation) with a scale factor of 3. A vertical stretch changes the size of the figure along the y - axis. For two figures to be similar, the ratios of their corresponding side lengths must be equal (the scale factor must be constant for all corresponding sides). In a vertical stretch by a factor of 3, the horizontal side lengths (along the x - axis) remain the same, and the vertical side lengths are multiplied by 3. This means that the ratio of the vertical side lengths of the final figure to the original figure is 3, and the ratio of the horizontal side lengths is 1. Since the scale factor is not constant for all sides, the figure after the vertical stretch is not similar to the figure after the translation? Wait, no. Wait, the first transformation is a translation (rigid, so similar with scale factor 1), the second transformation: let's consider the composition. Let the original figure have points with coordinates \((x,y)\). After the first transformation, the coordinates are \((x + 2,y - 3)\). After the second transformation, the coordinates are \((x + 2,3(y - 3))=(x + 2,3y-9)\).

Now, for similarity, we need to check if the transformation is a similarity transformation (a combination of rigid transformations and dilations). A dilation is a transformation that multiplies all side lengths by a constant scale factor. The first transformation is a translation (rigid), the second transformation: if we look at the effect on the sides. Suppose we have two points \((x_1,y_1)\) and \((x_2,y_2)\) in the original figure. After translation, they are \((x_1 + 2,y_1-3)\) and \((x_2 + 2,y_2 - 3)\). The distance between them is \(\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) (since the translation doesn't change the distance). After the second transformation, the points are \((x_1 + 2,3(y_1-3))\) and \((x_2 + 2,3(y_2 - 3))\). The distance between these two points is \(\sqrt{(x_2 - x_1)^2+[3(y_2 - y_1)]^2}=\sqrt{(x_2 - x_1)^2 + 9(y_2 - y_1)^2}\). This is not a constant multiple of the original distance \(\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) unless \(y_2 - y_1 = 0\) (horizontal lines). So, in general, the composition of a translation and a non - uniform stretch (since the x - coordinate is unchanged and the y - coordinate is multiplied by 3) is not a similarity transformation. Wait, but wait: the first transformation is a translation (rigid, so similar), the second transformation: is it a dilation? No, because it's only stretching in the y - direction. A dilation must stretch (or shrink) all directions by the same scale factor. So the second transformation is a non - uniform scaling. Therefore, the sequence of transformations: translation (which preserves similarity) followed by a non - uniform scaling (which does not preserve similarity) will result in a figure that is not similar to the original? Wait, no, the question is: does the sequence of transformations produce a pair of similar figures? Wait, the "pair of similar figures" - the original figure and the figure after both transformations?

Wait, let's re - express the transformations. Let \(T_1:(x,y)\to(x + 2,y - 3)\) (translation) and \(T_2:(x,y)\to(x,3y)\) (vertical stretch). The composition \(T=T_2\circ T_1\) is \(T(x,y)=(x + 2,3(y - 3))=(x + 2,3y-9)\).

For two figures to be similar, there must exist a similarity transformation (a transformation that is a composition of rigid transformations and dilations) that maps one to the other. A dilation has the form \((x,y)\to(kx,ky)\) where \(k\) is a non - zero constant. Our transformation \(T\) is \((x,y)\to(x + 2,3y-9)\), which is not a dilation (because of the \(x + 2\) and \(-9\) terms) and not a rigid transformation (because the y - coordinate is scaled by 3). The translation \(T_1\) is rigid (so the figure after \(T_1\) is similar to the original with scale factor 1). The transformation \(T_2\) applied to the figure after \(T_1\): the figure after \(T_1\) has the same shape as the original. When we apply \(T_2\) (vertical stretch by 3), the horizontal sides remain the same length, and the vertical sides become 3 times longer. So the ratio of vertical to horizontal side lengths changes. In a similar figure, all corresponding side length ratios must be equal. Since the horizontal sides have a ratio of 1 (compared to the figure after \(T_1\)) and the vertical sides have a ratio of 3, the figure after \(T_2\) is not similar to the figure after \(T_1\). But wait, the question is "Does this sequence of transformations produce a pair of similar figures?" - probably the original figure and the figure after both transformations.

Wait, no, maybe I made a mistake. Let's think about the definition of similar figures: two figures are similar if their corresponding angles are equal and their corresponding side lengths are in proportion (the scale factor is constant).

The first transformation: translation. Translation is a rigid transformation, so the figure after translation is congruent (and thus similar with scale factor 1) to the original figure.

The second transformation: \((x,y)\to(x,3y)\). Let's take a square with vertices \((0,0)\), \((1,0)\), \((1,1)\), \((0,1)\). After translation (say \(x+2,y - 3\)), the vertices are \((2,-3)\), \((3,-3)\), \((3,-2)\), \((2,-2)\). After the second transformation, the vertices are \((2,-9)\), \((3,-9)\), \((3,-6)\), \((2,-6)\). Now, the original square has side lengths 1 (horizontal) and 1 (vertical). The figure after translation has side lengths 1 (horizontal) and 1 (vertical). The figure after the second transformation has horizontal side lengths 1 (distance between \((2,-9)\) and \((3,-9)\) is 1) and vertical side lengths 3 (distance between \((3,-9)\) and \((3,-6)\) is 3). So the figure after the second transformation is a rectangle with length 1 and height 3, while the original is a square. These two figures are not similar because the ratios of corresponding sides are not equal (1/1 = 1 for horizontal, 3/1 = 3 for vertical; 1≠3).

But wait, the question is about the sequence: first translation, then vertical stretch. The translation preserves similarity (scale factor 1), the vertical stretch: if we consider the two figures - the one after translation and the one after the vertical stretch, are they similar? No, because the vertical stretch changes the aspect ratio. But wait, maybe the key is that the first transformation is a rigid transformation (so similar), the second transformation: is it a dilation? No, because it's only in the y - direction. A dilation must be uniform (scale both x and y by the same factor). So the composition of a rigid transformation and a non - uniform dilation is not a similarity transformation, so the final figure is not similar to the original? Wait, no, the original figure and the figure after both transformations: let's check angles. In the original figure, if we have a right angle, after translation, the right angle remains. After vertical stretch, the right angle still remains (because the x and y axes are still perpendicular). Wait, for a rectangle (or square), the angles are right angles. But the side length ratios: original square has side ratio 1:1, after vertical stretch, the rectangle has side ratio 1:3. So the angles are equal, but the side ratios are not equal, so they are not similar.

Wait, but maybe I messed up. Let's recall: a similarity transformation is a transformation that preserves the shape of the figure, which means that all corresponding angles are equal and all corresponding side lengths are in proportion (the scale factor is constant). The first transformation is a translation (rigid, so angles equal, side lengths equal - scale factor 1). The second transformation: when we do \((x,y)\to(x,3y)\), this is a vertical stretch. For a figure, if we have a triangle with base \(b\) (horizontal) and height \(h\) (vertical), after the vertical stretch, the base remains \(b\) and the height becomes \(3h\). The ratio of the new height to the old height is 3, and the ratio of the new base to the old base is 1. So the ratio of the sides (height/base) changes from \(h/b\) to \(3h/b\), which means the shape changes. Therefore, the figure after the vertical stretch is not similar to the figure after the translation (and thus not similar to the original).

But wait, the question is "Does this sequence of transformations produce a pair of similar figures?" - maybe the two figures are the one after the first transformation and the one after the second transformation? The figure after the first transformation (translation) has the same shape as the original. The figure after the second transformation (vertical stretch) has a different shape (since vertical sides are stretched by 3, horizontal sides same). So they are not similar.

Wait, no, let's think again. The first transformation is a translation (rigid, so similar with scale factor 1). The second transformation: a vertical stretch by 3. A stretch in one direction is not a similarity transformation (since similarity transformations require uniform scaling). So the composition of a rigid transformation and a non - uniform stretch is not a similarity transformation, so the final figure is not similar to the original.

Wait, but maybe the answer is yes? Wait, no. Let's take a simple example. Let the original figure be a triangle with vertices at \((0,0)\), \((0,1)\), \((1,0)\) (a right - angled triangle with legs of length 1). After translation \((x + 2,y - 3)\), the vertices are \((2,-3)\), \((2,-2)\), \((3,-3)\). After the second transformation \((x,3y)\), the vertices are \((2,-9)\), \((2,-6)\), \((3,-9)\). Now, the original triangle has legs of length 1 (vertical) and 1 (horizontal), hypotenuse \(\sqrt{2}\). The triangle after translation has legs of length 1 (vertical) and 1 (horizontal), hypotenuse \(\sqrt{2}\). The triangle after the second transformation has vertical leg length \(|-6-(-9)| = 3\), horizontal leg length \(|3 - 2|=1\), hypotenuse \(\sqrt{1^2+3^2}=\sqrt{10}\). Now, the ratio of vertical legs between the final triangle and the translated triangle is \(3/1 = 3\), the ratio of horizontal legs is \(1/1 = 1\), and the ratio of hypotenuses is \(\sqrt{10}/\sqrt{2}=\sqrt{5} eq3\) and \( eq1\). So the corresponding sides are not in proportion, so the triangles are not similar.

So the sequence of transformations (translation followed by vertical stretch) does not produce similar figures because the vertical stretch is not a uniform dilation (it only stretches the y - coordinate, not the x - coordinate), so the scale factor is not constant for all sides, and thus the corresponding side lengths are not in proportion.

Wait, but the first transformation is a rigid transformation (so the figure after translation is similar to the original). The second transformation: when we apply it to the figure after translation, the figure after translation is similar to the original, but the figure after the second transformation is not similar to the figure after translation (because of the non - uniform stretch), so the pair of figures (original and final) are not similar, and the pair (after translation and after second transformation) are not similar.

Answer:

No, the sequence of transformations does not produce a pair of similar figures. The first transformation \((x, y)\to(x + 2, y - 3)\) is a translation (a rigid transformation) that preserves the shape and size of the figure. The second transformation \((x, y)\to(x, 3y)\) is a vertical stretch (a non - uniform transformation) that only scales the \(y\) - coordinate by a factor of 3 while leaving the \(x\) - coordinate unchanged. For two figures to be similar, the ratios of all corresponding side lengths must be equal (a constant scale factor). In this case, the horizontal side lengths remain the same (scale factor of 1) and the vertical side lengths are scaled by 3, so the scale factor is not constant for all sides. Thus, the corresponding side lengths of the figure after both transformations and the original figure (or the figure after the first transformation) are not in proportion, so the figures are not similar.