Question

draw the image of $\triangle abc$ under a dilation whose center is $p$ and scale factor is 2.

Explanation:

Step1: Recall Dilation Rules

Dilation with center \( P \) and scale factor \( k = 2 \) means for each vertex \( X \) of \( \triangle ABC \), we find the point \( X' \) such that \( \overrightarrow{PX'} = 2 \cdot \overrightarrow{PX} \) (i.e., \( PX' = 2 \cdot PX \) and \( X' \) lies on the line through \( P \) and \( X \)).

Step2: Locate \( A' \)

• Draw the line from \( P \) through \( A \).
• Measure the distance from \( P \) to \( A \) (given as \( 5.0 \)). Multiply by 2: \( 5.0\times2 = 10.0 \). Mark \( A' \) on the line \( PA \) such that \( PA' = 10.0 \) (twice \( PA \)).

Step3: Locate \( B' \)

• Draw the line from \( P \) through \( B \).
• The distance from \( P \) to \( B \) (related to the length \( \approx 5.8 \) from \( B \) to \( C \), but we use the line through \( P \) and \( B \)). Multiply the distance from \( P \) to \( B \) by 2. Mark \( B' \) on line \( PB \) so \( PB' = 2 \cdot PB \).

Step4: Locate \( C' \)

• Draw the line from \( P \) through \( C \).
• The distance from \( P \) to \( C \): multiply by 2. Mark \( C' \) on line \( PC \) so \( PC' = 2 \cdot PC \).

Step5: Draw \( \triangle A'B'C' \)

• Connect \( A' \), \( B' \), and \( C' \) to form the dilated triangle. The sides of \( \triangle A'B'C' \) will be twice the length of the corresponding sides of \( \triangle ABC \) (e.g., \( A'B' = 2 \times AB \approx 12.2 \), \( B'C' = 2 \times BC \approx 11.6 \), \( A'C' = 2 \times AC = 10.0 \)).

To visualize, follow these steps:

1. For each vertex \( A, B, C \), draw a line from \( P \) through the vertex.
2. Extend (or retract, but scale factor \( 2 > 1 \), so extend) the segment from \( P \) to the vertex until its length from \( P \) is doubled.
3. The new points \( A', B', C' \) are the images. Connect them to get \( \triangle A'B'C' \).

(Note: Since this is a drawing task, the key is to apply the dilation rules: center \( P \), scale factor 2, so each vertex's image is along the line from \( P \) to the vertex, at twice the distance from \( P \) as the original vertex.)

Answer:

Step1: Recall Dilation Rules

Dilation with center \( P \) and scale factor \( k = 2 \) means for each vertex \( X \) of \( \triangle ABC \), we find the point \( X' \) such that \( \overrightarrow{PX'} = 2 \cdot \overrightarrow{PX} \) (i.e., \( PX' = 2 \cdot PX \) and \( X' \) lies on the line through \( P \) and \( X \)).

Step2: Locate \( A' \)

• Draw the line from \( P \) through \( A \).
• Measure the distance from \( P \) to \( A \) (given as \( 5.0 \)). Multiply by 2: \( 5.0\times2 = 10.0 \). Mark \( A' \) on the line \( PA \) such that \( PA' = 10.0 \) (twice \( PA \)).

Step3: Locate \( B' \)

• Draw the line from \( P \) through \( B \).
• The distance from \( P \) to \( B \) (related to the length \( \approx 5.8 \) from \( B \) to \( C \), but we use the line through \( P \) and \( B \)). Multiply the distance from \( P \) to \( B \) by 2. Mark \( B' \) on line \( PB \) so \( PB' = 2 \cdot PB \).

Step4: Locate \( C' \)

• Draw the line from \( P \) through \( C \).
• The distance from \( P \) to \( C \): multiply by 2. Mark \( C' \) on line \( PC \) so \( PC' = 2 \cdot PC \).

Step5: Draw \( \triangle A'B'C' \)

• Connect \( A' \), \( B' \), and \( C' \) to form the dilated triangle. The sides of \( \triangle A'B'C' \) will be twice the length of the corresponding sides of \( \triangle ABC \) (e.g., \( A'B' = 2 \times AB \approx 12.2 \), \( B'C' = 2 \times BC \approx 11.6 \), \( A'C' = 2 \times AC = 10.0 \)).

To visualize, follow these steps:

1. For each vertex \( A, B, C \), draw a line from \( P \) through the vertex.
2. Extend (or retract, but scale factor \( 2 > 1 \), so extend) the segment from \( P \) to the vertex until its length from \( P \) is doubled.
3. The new points \( A', B', C' \) are the images. Connect them to get \( \triangle A'B'C' \).

(Note: Since this is a drawing task, the key is to apply the dilation rules: center \( P \), scale factor 2, so each vertex's image is along the line from \( P \) to the vertex, at twice the distance from \( P \) as the original vertex.)