Question

the figure below is dilated by a factor of 3 centered at the origin. plot the resulting image. click twice to plot a segment. click a segment to delete it.

Explanation:

Step1: Identify original coordinates

First, find the coordinates of points \( P \), \( O \), and \( N \) from the graph.

• Point \( P \): \( (-3, 0) \)
• Point \( O \): \( (3, 0) \)
• Point \( N \): \( (0, -2) \)

Step2: Apply dilation factor

Dilation centered at the origin with factor \( k \) transforms a point \( (x, y) \) to \( (k \cdot x, k \cdot y) \). Here, \( k = 3 \).

• For \( P(-3, 0) \): New coordinates \( (3 \cdot (-3), 3 \cdot 0) = (-9, 0) \)
• For \( O(3, 0) \): New coordinates \( (3 \cdot 3, 3 \cdot 0) = (9, 0) \)
• For \( N(0, -2) \): New coordinates \( (3 \cdot 0, 3 \cdot (-2)) = (0, -6) \)

Step3: Plot the new points

Plot the points \( (-9, 0) \), \( (9, 0) \), and \( (0, -6) \) and connect them to form the dilated triangle.

Answer:

The dilated image has vertices at \( (-9, 0) \), \( (9, 0) \), and \( (0, -6) \). Plot these points and connect \( P'(-9, 0) \) to \( O'(9, 0) \), \( O'(9, 0) \) to \( N'(0, -6) \), and \( N'(0, -6) \) to \( P'(-9, 0) \) to get the dilated figure.