Question

graph the image of the triangle below using a scale factor of k = 4.
image of grid with triangle
write the numerical answer (ie if you get 2 for your answer, type \2\ not \two\)
s ( type your answer... , type your answer... )
t ( type your answer... , type your answer... )
u ( type your answer... , type your answer... )

Explanation:

To solve this, we first need the coordinates of the original triangle's vertices \( S \), \( T \), and \( U \) from the graph. Let's assume (from typical grid problems) the original coordinates are:

• \( S(0, 0) \) (on the origin),
• \( U(1, 1) \) (1 unit right, 1 unit up from \( S \)),
• \( T(2, 2) \) (2 units right, 2 units up from \( S \)) (adjust as per actual graph, but we'll use these for demonstration).

Step 1: Recall the scaling rule

When scaling a figure with a scale factor \( k \), the new coordinates \( (x', y') \) of a vertex \( (x, y) \) are given by:
\[ x' = k \cdot x, \quad y' = k \cdot y \]

Step 2: Scale \( S(0, 0) \)

For \( S(0, 0) \):
\[ x' = 4 \cdot 0 = 0, \quad y' = 4 \cdot 0 = 0 \]
Thus, \( S'(0, 0) \).

Step 3: Scale \( U(1, 1) \)

For \( U(1, 1) \):
\[ x' = 4 \cdot 1 = 4, \quad y' = 4 \cdot 1 = 4 \]
Thus, \( U'(4, 4) \).

Step 4: Scale \( T(2, 2) \)

For \( T(2, 2) \):
\[ x' = 4 \cdot 2 = 8, \quad y' = 4 \cdot 2 = 8 \]
Thus, \( T'(8, 8) \).

(Note: If the original coordinates differ, adjust the scaling accordingly. For example, if \( S \) is \( (1, 0) \), \( U \) is \( (2, 1) \), \( T \) is \( (3, 2) \), then:

• \( S'(4, 0) \), \( U'(8, 4) \), \( T'(12, 8) \), etc.

Assuming the initial coordinates (from the grid) are \( S(0, 0) \), \( U(1, 1) \), \( T(2, 2) \), the scaled coordinates are:

\( S'(0, 0) \), \( T'(8, 8) \), \( U'(4, 4) \)

Final Answer (Example, adjust for actual coordinates):

• \( S'(0, 0) \)
• \( T'(8, 8) \)
• \( U'(4, 4) \)

(Replace with actual coordinates from the graph. If the original \( S \) is \( (1, 0) \), \( U(2, 1) \), \( T(3, 2) \), then:
\( S'(4, 0) \), \( U'(8, 4) \), \( T'(12, 8) \))

Answer:

To solve this, we first need the coordinates of the original triangle's vertices \( S \), \( T \), and \( U \) from the graph. Let's assume (from typical grid problems) the original coordinates are:

• \( S(0, 0) \) (on the origin),
• \( U(1, 1) \) (1 unit right, 1 unit up from \( S \)),
• \( T(2, 2) \) (2 units right, 2 units up from \( S \)) (adjust as per actual graph, but we'll use these for demonstration).

Step 1: Recall the scaling rule

When scaling a figure with a scale factor \( k \), the new coordinates \( (x', y') \) of a vertex \( (x, y) \) are given by:
\[ x' = k \cdot x, \quad y' = k \cdot y \]

Step 2: Scale \( S(0, 0) \)

For \( S(0, 0) \):
\[ x' = 4 \cdot 0 = 0, \quad y' = 4 \cdot 0 = 0 \]
Thus, \( S'(0, 0) \).

Step 3: Scale \( U(1, 1) \)

For \( U(1, 1) \):
\[ x' = 4 \cdot 1 = 4, \quad y' = 4 \cdot 1 = 4 \]
Thus, \( U'(4, 4) \).

Step 4: Scale \( T(2, 2) \)

For \( T(2, 2) \):
\[ x' = 4 \cdot 2 = 8, \quad y' = 4 \cdot 2 = 8 \]
Thus, \( T'(8, 8) \).

(Note: If the original coordinates differ, adjust the scaling accordingly. For example, if \( S \) is \( (1, 0) \), \( U \) is \( (2, 1) \), \( T \) is \( (3, 2) \), then:

• \( S'(4, 0) \), \( U'(8, 4) \), \( T'(12, 8) \), etc.

Assuming the initial coordinates (from the grid) are \( S(0, 0) \), \( U(1, 1) \), \( T(2, 2) \), the scaled coordinates are:

\( S'(0, 0) \), \( T'(8, 8) \), \( U'(4, 4) \)

Final Answer (Example, adjust for actual coordinates):

• \( S'(0, 0) \)
• \( T'(8, 8) \)
• \( U'(4, 4) \)

(Replace with actual coordinates from the graph. If the original \( S \) is \( (1, 0) \), \( U(2, 1) \), \( T(3, 2) \), then:
\( S'(4, 0) \), \( U'(8, 4) \), \( T'(12, 8) \))