Question

the triangle tuv is a dilation of the triangle tuv. what is the scale factor of the dilation? simplify your answer and write it as a proper fraction, an improper fraction, or a whole number.

Explanation:

Step1: Find coordinates of original and dilated points

Original triangle \( TUV \): \( T(-3, 3) \), \( U(0, 3) \), \( V(-3, -9) \) (assuming grid units, checking the graph: \( T \) is at \( x=-3, y=3 \), \( U \) at \( x=0, y=3 \), \( V \) at \( x=-3, y=-9 \)). Dilated triangle \( T'U'V' \): \( T'(-1, 1) \), \( U'(0, 1) \), \( V'(-1, -3) \).

Step2: Calculate scale factor using distance or coordinates

Scale factor \( k \) is the ratio of corresponding side lengths (or coordinates, since dilation is about origin? Wait, check the x-coordinates: \( T \) is \( -3 \), \( T' \) is \( -1 \). So \( k=\frac{-1}{-3}=\frac{1}{3} \)? Wait, no, wait \( U \) is \( (0,3) \), \( U' \) is \( (0,1) \). So \( y \)-coordinate of \( U \) is 3, \( U' \) is 1. So \( k = \frac{1}{3} \). Alternatively, length of \( TU \): distance between \( T(-3,3) \) and \( U(0,3) \) is \( |0 - (-3)| = 3 \). Length of \( T'U' \): distance between \( T'(-1,1) \) and \( U'(0,1) \) is \( |0 - (-1)| = 1 \). So scale factor \( k = \frac{1}{3} \).

Answer:

\(\frac{1}{3}\)