Question

triangle stv was dilated with the origin as the center of dilation to form $\triangle stv$. what is the scale factor of the dilation? $\frac{1}{3}$ $\frac{2}{3}$ $\frac{3}{2}$ $\frac{3}{1}$

Explanation:

Step1: Find the length of corresponding sides

Let's assume the length of a side in \(\triangle STV\) and the corresponding side in \(\triangle S'T'V'\). For example, if we consider the vertical distance from the center (origin - related in dilation) for a vertex. Suppose the \(y -\) coordinate of \(S\) is \(6\) (assuming each grid unit is \(1\)) and the \(y -\) coordinate of \(S'\) is \(4\). But a better approach is using the ratio of side lengths. Let's assume the base - like length. If we consider the horizontal segments (since dilation is uniform). Suppose the length of \(TV\) (horizontal distance between \(T\) and \(V\)) is \(6\) units and the length of \(T'V'\) is \(4\) units.

Step2: Calculate the scale factor

The scale factor \(k\) of a dilation is given by the formula \(k=\frac{\text{length of a side in the image}}{\text{length of the corresponding side in the pre - image}}\). So \(k = \frac{2}{3}\) (if we assume the side length of the smaller triangle (image) is \(2\) and the side length of the larger triangle (pre - image) is \(3\) when measured in terms of grid units. Another way: If we use the distance from the center of dilation (origin) for a point. Let \(S=(0,6)\) and \(S'=(0,4)\). The scale factor \(k=\frac{y_{S'}}{y_S}=\frac{4}{6}=\frac{2}{3}\)

Answer:

\(\frac{2}{3}\)