Question

1. shawn plots the image of point m under a dilation with center o and scale factor 0.6 as shown in the diagram.

use the properties of dilations to explain how shawn determines the location of point m.

Explanation:

1. Recall the property of dilations: A dilation with center \( O \) and scale factor \( k \) maps a point \( M \) to \( M' \) such that \( \overrightarrow{OM'} = k \cdot \overrightarrow{OM} \), meaning \( O \), \( M \), and \( M' \) are colinear, and \( \frac{OM'}{OM}=k \).
2. For the horizontal segment \( OL \) (where \( L \) is the projection of \( M \) onto the x - axis), under dilation with scale factor \( 0.6 \), the image \( L' \) of \( L \) satisfies \( OL' = 0.6\times OL \). Shawn can first find \( L' \) by multiplying the length of \( OL \) by \( 0.6 \) (since dilation scales distances from the center).
3. Since dilation preserves collinearity, \( M' \) lies on the line \( OM \). Also, the vertical distance from \( M \) to the x - axis (the y - coordinate related to the grid) will be scaled by \( 0.6 \) as well. So, Shawn can project \( M' \) from \( L' \) vertically (parallel to the original vertical segment from \( L \) to \( M \)) such that the vertical length from \( L' \) to \( M' \) is \( 0.6 \) times the vertical length from \( L \) to \( M \). In other words, since \( M' \) is on \( OM \) and the ratio of \( OM' \) to \( OM \) is \( 0.6 \), by constructing the line \( OM \) and finding the point \( M' \) on \( OM \) such that the distance from \( O \) to \( M' \) is \( 0.6 \) times the distance from \( O \) to \( M \) (using the grid to measure and scale the horizontal and vertical components), Shawn can determine the location of \( M' \).

Answer:

Shawn uses the property of dilations that a dilation with center \( O \) and scale factor \( 0.6 \) maps \( M \) to \( M' \) such that \( O \), \( M \), \( M' \) are collinear and \( \frac{OM'}{OM}=0.6 \). He first finds the projection \( L \) of \( M \) onto the x - axis. Then, he scales the distance \( OL \) by \( 0.6 \) to get \( OL' \) (finding \( L' \)). Since \( M' \) lies on \( OM \), he projects vertically from \( L' \) (parallel to the vertical segment from \( L \) to \( M \)) to a point on \( OM \) such that the vertical distance from \( L' \) to \( M' \) is \( 0.6 \) times the vertical distance from \( L \) to \( M \). This gives the location of \( M' \) as it satisfies the dilation property (scaling distances from \( O \) by \( 0.6 \) and preserving collinearity along \( OM \)).