Question

graph the image of rectangle rstu after a dilation with a scale factor of \\(\frac{1}{3}\\), centered at the origin.

Explanation:

Step1: Identify Coordinates

First, find the coordinates of the vertices of rectangle \( RSTU \). From the graph:
- \( R(-6, -3) \)
- \( S(9, -3) \) (Wait, no, looking at the grid, \( S \) is at \( (9, -3) \)? Wait, no, the x-axis: let's check again. The original points: \( U \) is at \( (-6, 6) \), \( T \) at \( (9, 6) \)? Wait, no, the grid lines: each square is 1 unit. Let's re-express:
Looking at the graph:
- \( R \): x = -6, y = -3 (since it's 6 left on x, 3 down on y)
- \( S \): x = 9? Wait, no, the x-axis has marks at -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10. So \( S \) is at (9, -3)? Wait, no, the vertical line for \( S \) and \( T \): \( T \) is at (9, 6)? Wait, no, the y-coordinate for \( T \) is 6, x is 9? Wait, no, the horizontal line from \( U \) to \( T \): \( U \) is at (-6, 6), \( T \) is at (9, 6)? Wait, the distance between \( U \) and \( T \) on x-axis: from -6 to 9? No, that can't be. Wait, maybe I misread. Let's check the grid again. The x-axis: the leftmost point \( U \) is at x = -6 (since the grid line at -6), y = 6. Then \( T \) is at x = 9? No, the grid has up to 10, so \( T \) is at (9, 6)? Wait, no, the horizontal line from \( U \) to \( T \): the x-coordinate of \( U \) is -6, \( T \) is at x = 9? Wait, that would be a length of 15, but dilation scale factor 1/3. Wait, maybe I made a mistake. Wait, looking at the graph, the x-coordinate of \( S \) and \( T \): the vertical line for \( S \) and \( T \) is at x = 9? No, the grid lines: each square is 1 unit. Let's list the correct coordinates:
- \( U(-6, 6) \)
- \( T(9, 6) \)? No, that can't be. Wait, the distance between \( U \) and \( T \) on the x-axis: from -6 to 9 is 15 units. But maybe the correct coordinates are:
Wait, the original rectangle: \( R(-6, -3) \), \( S(9, -3) \), \( T(9, 6) \), \( U(-6, 6) \). Wait, but let's check the y-coordinates: \( R \) and \( S \) are at y = -3, \( U \) and \( T \) at y = 6. So the height is 6 - (-3) = 9, width is 9 - (-6) = 15? No, that seems too big. Wait, maybe the x-coordinate of \( S \) is 9? No, maybe I misread the x-axis. Wait, the grid lines: the x-axis has marks at -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10. So the distance between -6 and 9 is 15, but dilation with scale factor 1/3 would make it 5. Wait, maybe the correct coordinates are:
Wait, maybe \( S \) is at (9, -3)? No, that's not right. Wait, perhaps the original coordinates are:
- \( R(-6, -3) \)
- \( S(9, -3) \)
- \( T(9, 6) \)
- \( U(-6, 6) \)

Wait, but let's confirm with the grid. The vertical line for \( R \) and \( U \) is x = -6, vertical line for \( S \) and \( T \) is x = 9? No, that's a long rectangle. Alternatively, maybe \( S \) is at (9, -3)? Wait, no, the y-coordinate for \( S \) and \( R \) is -3, \( U \) and \( T \) is 6. So the height is 6 - (-3) = 9, width is 9 - (-6) = 15. Then dilation with scale factor 1/3: each coordinate is multiplied by 1/3.

Wait, maybe I made a mistake. Let's re-express the coordinates correctly. Let's look at the graph again:

- \( R \): x = -6, y = -3 (since it's on the grid line at x=-6, y=-3)
- \( S \): x = 9, y = -3 (grid line x=9, y=-3)
- \( T \): x = 9, y = 6 (grid line x=9, y=6)
- \( U \): x = -6, y = 6 (grid line x=-6, y=6)

Yes, that makes sense. So the four vertices are:
\( R(-6, -3) \), \( S(9, -3) \), \( T(9, 6) \), \( U(-6, 6) \)

Step2: Apply Dilation

Dilation centered at the origin with scale factor \( \frac{1}{3} \) means we multiply each coordinate by \( \frac{1}{3} \).

For \( R(-6, -3) \):
New x-coordinate: \( -6 \times \frac{1}{3} = -2 \)
New y-coordinate: \( -3 \times \frac{1}{3} = -1 \)
So \( R'(-2, -1) \)

For \( S(9, -3) \):
New x-coordinate: \( 9 \times \frac{1}{3} = 3 \)
New y-coordinate: \( -3 \times \frac{1}{3} = -1 \)
So \( S'(3, -1) \)

For \( T(9, 6) \):
New x-coordinate: \( 9 \times \frac{1}{3} = 3 \)
New y-coordinate: \( 6 \times \frac{1}{3} = 2 \)
So \( T'(3, 2) \)

For \( U(-6, 6) \):
New x-coordinate: \( -6 \times \frac{1}{3} = -2 \)
New y-coordinate: \( 6 \times \frac{1}{3} = 2 \)
So \( U'(-2, 2) \)

Step3: Graph the New Points

Now, plot the new points \( R'(-2, -1) \), \( S'(3, -1) \), \( T'(3, 2) \), \( U'(-2, 2) \) on the coordinate plane. Connect them to form the dilated rectangle.

Answer:

The dilated rectangle has vertices at \( R'(-2, -1) \), \( S'(3, -1) \), \( T'(3, 2) \), and \( U'(-2, 2) \). To graph it, plot these points and connect them in order.