Question

line segment xy is dilated to create line segment xy using point t as the center of dilation. what is yt? 15 units, 3 units, 6 units, 12 units

Explanation:

Step1: Find the scale factor of dilation

The length of \( TX \) before dilation: Let's assume the distance from \( T \) to \( X \) is \( TX = 2 + 6=8 \)? Wait, no, looking at the diagram, \( TX' = 6 \) and the segment from \( X \) to \( X' \) is 2? Wait, no, the diagram shows \( X \) and \( X' \) on a vertical line, with \( TX' = 6 \) and the length from \( X \) to \( X' \) is 2? Wait, maybe better: Dilation scale factor \( k=\frac{TX'}{TX} \). Wait, \( TX' = 6 \), and \( TX \) is \( TX' - XX' \)? No, wait, the original segment \( XY \) is dilated to \( X'Y' \), so the scale factor \( k \) is the ratio of corresponding distances from the center \( T \). So \( TX' = 6 \), and let \( TX = x \), then \( XX' = 2 \), so \( TX'=TX - XX' \)? Wait, no, the arrows are upward, so \( T \) is below \( X' \) and \( X \). So \( TX' = 6 \), and \( X \) is above \( X' \) by 2 units, so \( TX = TX' + XX' = 6 + 2 = 8 \)? No, that can't be. Wait, maybe the length from \( T \) to \( X \) is \( TX \), and from \( T \) to \( X' \) is \( TX' = 6 \), and the length \( XX' = 2 \), so \( TX = TX' + XX' = 6 + 2 = 8 \)? No, the dilation scale factor \( k=\frac{TX'}{TX} \). Wait, but \( XY \) and \( X'Y' \) are horizontal, so they are parallel, so triangles \( TYX \) and \( TY'X' \) are similar. So the ratio of \( TX' \) to \( TX \) is equal to the ratio of \( TY' \) to \( TY \). Wait, \( TX' = 6 \), \( TX = TX' + XX' = 6 + 2 = 8 \)? No, maybe \( XX' = 2 \), so \( TX = TX' - XX' = 6 - 2 = 4 \)? Wait, the diagram shows \( X \) is above \( X' \) with a segment of length 2 between them, and \( TX' = 6 \), so \( TX = TX' + 2 = 8 \)? No, this is confusing. Wait, the key is that the scale factor \( k=\frac{TX'}{TX} \). Wait, \( TY' = 9 \), and we need to find \( TY \). Wait, no, the question is \( YT \), which is \( TY \). Wait, maybe the scale factor \( k = \frac{TX'}{TX} \), where \( TX' = 6 \), and \( TX = 6 - 2 = 4 \)? No, the length from \( T \) to \( X' \) is 6, and from \( T \) to \( X \) is \( 6 + 2 = 8 \)? Wait, no, the segment \( XX' \) is 2, so if \( X' \) is closer to \( T \) than \( X \), then \( TX = TX' + XX' = 6 + 2 = 8 \), and \( TX' = 6 \), so scale factor \( k=\frac{TX'}{TX}=\frac{6}{8}=\frac{3}{4} \)? No, but \( TY' = 9 \), so \( TY=\frac{TY'}{k}=\frac{9}{\frac{3}{4}} = 12 \)? No, that's not one of the options. Wait, maybe I got the scale factor reversed. If \( XY \) is dilated to \( X'Y' \), then \( X'Y' \) is the image, so scale factor \( k=\frac{X'Y'}{XY} \), but since \( XY \) and \( X'Y' \) are horizontal, and \( XX' = 2 \), maybe \( XY = X'Y' \)? No, the triangles are similar, so \( \frac{TX'}{TX}=\frac{TY'}{TY} \). Let \( TY = x \), then \( TY' = x - Y'Y \)? No, the diagram shows \( Y' \) is below \( Y \) on the line \( TY \), with \( TY' = 9 \), and \( Y \) is above \( Y' \). Wait, the length \( TY' = 9 \), and we need to find \( TY \). The scale factor \( k \) is \( \frac{TX'}{TX} \). \( TX' = 6 \), \( TX = TX' + XX' = 6 + 2 = 8 \)? No, the options include 3,6,12,15. Wait, maybe \( TX = 2 \), \( TX' = 6 \), so scale factor \( k = \frac{TX'}{TX}=\frac{6}{2}=3 \). Wait, that makes sense! Because \( X \) is 2 units from \( T \)? No, the diagram shows \( X \) is above \( X' \), with \( XX' = 2 \), and \( TX' = 6 \), so \( TX = TX' - XX' = 6 - 2 = 4 \)? No, if \( TX = 2 \), and \( TX' = 6 \), then scale factor \( k = 6/2 = 3 \). Then, since \( TY' = 9 \), and \( TY' = k \times TY \), so \( TY = TY' / k = 9 / 3 = 3 \). Wait, that's one of the options (3 units). Let's check: If scale factor \( k = 3 \), then \( TX' = 3 \times TX \), so \( TX = TX' / 3 = 6 / 3 = 2 \), which matches the \( XX' = 2 \) (since \( X \) is 2 units above \( X' \), so \( TX = 2 \), \( TX' = 6 \)). Then, \( TY' = 9 \), so \( TY = TY' / 3 = 9 / 3 = 3 \). Yes, that works. So \( YT = TY = 3 \) units.

Step2: Confirm the scale factor

Scale factor \( k = \frac{TX'}{TX} = \frac{6}{2} = 3 \). Then, since \( TY' = 9 \), \( TY = \frac{TY'}{k} = \frac{9}{3} = 3 \). So \( YT = 3 \) units.

Answer:

3 units (the option with "3 units")