Question

the figure below is dilated by a factor of \\(\frac{4}{3}\\) centered at the origin. plot the resulting image. click twice to plot a segment. click a segment to delete it.

Explanation:

Step1: Identify coordinates of points

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

• \( O \) is at \( (3, 0) \) (wait, no, looking at the graph, the origin - centered dilation, and the original points: Let's re - check. Wait, the point \( O \) (maybe a typo, let's see the graph: the bottom vertex is at \( (3,0) \)? Wait, no, looking at the axes, the x - axis has marks from - 10 to 10, y - axis from - 10 to 10. Let's find the coordinates:
• Point \( Q \): From the graph, it seems to be at \( (- 3,6) \) (since x=-3, y = 6)
• Point \( P \): At \( (3,3) \) (x = 3, y=3)
• Point \( O \) (the bottom - right vertex): At \( (3,0) \)
• Point \( N \): At \( (-3,-6) \)

Step2: Apply dilation formula

The rule for dilation centered at the origin with a scale factor \( k \) is \( (x,y)\to(kx,ky) \). Here, \( k=\frac{4}{3} \).

For point \( Q(-3,6) \):

New \( x \) - coordinate: \( \frac{4}{3}\times(-3)=-4 \)
New \( y \) - coordinate: \( \frac{4}{3}\times6 = 8 \)
So, \( Q'(-4,8) \)

For point \( P(3,3) \):

New \( x \) - coordinate: \( \frac{4}{3}\times3 = 4 \)
New \( y \) - coordinate: \( \frac{4}{3}\times3=4 \)
So, \( P'(4,4) \)

For point \( O(3,0) \):

New \( x \) - coordinate: \( \frac{4}{3}\times3 = 4 \)
New \( y \) - coordinate: \( \frac{4}{3}\times0 = 0 \)
So, \( O'(4,0) \)

For point \( N(-3,-6) \):

New \( x \) - coordinate: \( \frac{4}{3}\times(-3)=-4 \)
New \( y \) - coordinate: \( \frac{4}{3}\times(-6)=-8 \)
So, \( N'(-4,-8) \)

Step3: Plot the new points

Plot the points \( Q'(-4,8) \), \( P'(4,4) \), \( O'(4,0) \), and \( N'(-4,-8) \) and connect them in the same order as the original figure.

Answer:

The dilated points are \( Q'(-4,8) \), \( P'(4,4) \), \( O'(4,0) \), and \( N'(-4,-8) \). Plot these points and connect them to get the dilated figure.