Question

write the coordinates of the vertices after a dilation with a scale factor of 4, centered at the origin.

Explanation:

Step1: Find original coordinates

From the graph:
- \( E(-2, -2) \)
- \( F(1, -2) \) (Wait, no, looking at the grid, F should be (1? Wait, no, the grid lines: E is at x=-2, y=-2; F is at x=1? Wait, no, the horizontal line from E to F: E is (-2, -2), F is (1? Wait, no, the grid: each square is 1 unit. Let's check again. The points: E is at (-2, -2), F is at (1? No, wait, the vertical line from F to G: G is at (1, 2)? Wait, no, the original points: H is (-2, 2), G is (1, 2)? Wait, no, the horizontal line from H to G: H is (-2, 2), G is (1, 2)? Wait, no, the grid: x-axis, y-axis. Let's see: E is (-2, -2), F is (1, -2)? No, wait, the distance between E and F: from x=-2 to x=1? No, maybe I misread. Wait, the graph: E is at (-2, -2), F is at (1, -2)? No, wait, the vertical line from F to G: G is at (1, 2), and H is at (-2, 2). Wait, no, the horizontal line from H to G: H(-2,2), G(1,2)? No, the grid lines: x from -10 to 10, y from -10 to 10. Let's check the coordinates again. Looking at the points:

- E: x=-2, y=-2 (so (-2, -2))
- F: x=1? No, wait, the vertical line from F to G: G is at (1, 2)? Wait, no, the original G is at (1, 2)? Wait, no, the user's graph: H is at (-2, 2), G is at (1, 2)? No, maybe the original points are E(-2, -2), F(1, -2), G(1, 2), H(-2, 2)? Wait, no, the horizontal line from H to G: H is (-2, 2), G is (1, 2)? Then the distance between H and G is 3 units? But maybe I made a mistake. Wait, the grid: each square is 1 unit. Let's look at the coordinates:

Wait, the problem's graph: E is at (-2, -2), F is at (1, -2)? No, wait, the x-coordinate of F: from the origin (0,0), moving right 1 unit? No, maybe F is at (1, -2)? Wait, no, the vertical line from F to G: G is at (1, 2), so F is (1, -2), G is (1, 2), H is (-2, 2), E is (-2, -2). Yes, that makes a rectangle.

Step2: Apply dilation (scale factor 4, center origin)

Dilation rule: \((x, y) \to (4x, 4y)\)

- For \( E(-2, -2) \):
\( x' = 4 \times (-2) = -8 \)
\( y' = 4 \times (-2) = -8 \)
So \( E'(-8, -8) \)

- For \( F(1, -2) \): Wait, no, wait, maybe F is at (1, -2)? Wait, no, looking at the grid again: the x-coordinate of F: from the origin, moving right 1 unit? Wait, no, the original F: let's check the x-axis. The grid lines: E is at x=-2, y=-2; F is at x=1? No, maybe F is at (1, -2)? Wait, no, the vertical line from F to G: G is at (1, 2), so F is (1, -2). Then:

- \( F(1, -2) \):
\( x' = 4 \times 1 = 4 \)
\( y' = 4 \times (-2) = -8 \)
So \( F'(4, -8) \)

- \( G(1, 2) \):
\( x' = 4 \times 1 = 4 \)
\( y' = 4 \times 2 = 8 \)
So \( G'(4, 8) \)

- \( H(-2, 2) \):
\( x' = 4 \times (-2) = -8 \)
\( y' = 4 \times 2 = 8 \)
So \( H'(-8, 8) \)

Wait, but maybe I misread the original coordinates. Let's check again. The original points: E is at (-2, -2), F is at (1, -2)? No, wait, the horizontal line from E to F: E is (-2, -2), F is (1, -2)? Then the length is 3. But maybe the original F is at (1, -2)? Wait, no, the grid: each square is 1 unit. Let's see the x-coordinate of F: from the origin (0,0), moving right 1 unit? So x=1, y=-2. Then G is (1, 2), H is (-2, 2), E is (-2, -2). Yes.

So applying dilation with scale factor 4, center origin: multiply each coordinate by 4.

So:

- E(-2, -2) → (4*(-2), 4*(-2)) = (-8, -8)
- F(1, -2) → (4*1, 4*(-2)) = (4, -8)
- G(1, 2) → (4*1, 4*2) = (4, 8)
- H(-2, 2) → (4*(-2), 4*2) = (-8, 8)

Wait, but maybe the original F is at (1, -2)? Wait, no, maybe I made a mistake in the original x-coordinate of F. Let's check the graph again. The vertical line from F to G: G is at (1, 2), so F is (1, -2). Yes. So the original coordinates are E(-2, -2), F(1, -2), G(1, 2), H(-2, 2). Then after dilation:

E'(-8, -8), F'(4, -8), G'(4, 8), H'(-8, 8)

Answer:

The coordinates of the vertices after dilation are \( E'(-8, -8) \), \( F'(4, -8) \), \( G'(4, 8) \), and \( H'(-8, 8) \).