Question

draw the image of △abc under a dilation whose center is p and scale factor is 3.

Explanation:

Step1: Recall Dilation Rules

Dilation with center \( P \) and scale factor \( k \) means for each point \( X \) in \( \triangle ABC \), the image \( X' \) satisfies \( \overrightarrow{PX'} = k \cdot \overrightarrow{PX} \). So we need to find the direction and distance from \( P \) to each vertex \( A, B, C \), then scale that distance by 3.

Step2: Analyze Distances (Visual/Grid)

• For vertex \( A \): Find the vector from \( P \) to \( A \). Then multiply the length of \( \overrightarrow{PA} \) by 3 to get \( \overrightarrow{PA'} \), so \( A' \) is on the line \( PA \) such that \( PA' = 3 \cdot PA \).
• For vertex \( B \): Similarly, find \( \overrightarrow{PB} \), scale its length by 3 to get \( \overrightarrow{PB'} \), so \( B' \) is on line \( PB \) with \( PB' = 3 \cdot PB \).
• For vertex \( C \): Find \( \overrightarrow{PC} \), scale its length by 3 to get \( \overrightarrow{PC'} \), so \( C' \) is on line \( PC \) with \( PC' = 3 \cdot PC \).

Step3: Draw the Image

1. Draw line segments from \( P \) through \( A, B, C \) respectively.
2. Measure the distance from \( P \) to \( A \), multiply by 3, mark \( A' \) at that new distance on line \( PA \).
3. Do the same for \( B \) (get \( B' \)) and \( C \) (get \( C' \)).
4. Connect \( A', B', C' \) to form \( \triangle A'B'C' \), the dilated image.

(Note: Since this is a drawing problem, the key is to apply the dilation rules: center at \( P \), scale factor 3, so each vertex's image is 3 times as far from \( P \) as the original vertex, along the line from \( P \) through the original vertex.)

Answer:

To draw the image of \( \triangle ABC \) under dilation with center \( P \) and scale factor 3:

1. For each vertex \( A, B, C \), draw a line from \( P \) through the vertex.
2. Extend (or create) the segment from \( P \) to the vertex so that the new length from \( P \) to the image vertex (\( A', B', C' \)) is 3 times the original length from \( P \) to \( A, B, C \) (respectively).
3. Connect \( A', B', C' \) to form \( \triangle A'B'C' \), the dilated triangle.

(Visually, if \( PA = d \), then \( PA' = 3d \); same for \( PB \to PB' \) and \( PC \to PC' \), with \( A', B', C' \) collinear with \( P \) and their original vertices.)