Question

graph the image of the figure after a dilation with a scale factor of \\(\frac{1}{4}\\) centered at \\((5, -5)\\). use the polygon tool to graph the triangle by connecting all its vertices. polygon move undo redo reset

Explanation:

Step1: Identify Original Vertices

First, find the coordinates of the original triangle's vertices. From the graph, let's assume the vertices are \( A(x_1, y_1) \), \( B(x_2, y_2) \), \( C(x_3, y_3) \). Let's estimate their coordinates (assuming grid lines are 1 unit each):

• Let's say \( A = (1, 7) \), \( B = (9, 3) \), \( C = (5, -1) \) (wait, no, looking at the graph, maybe better to find actual points. Wait, the center is \( (5, -5) \). Wait, maybe I misread. Let's re-examine: the graph has a triangle with vertices, say, let's pick three points. Let's take the left vertex: x=1, y=7 (since it's at (1,7)), middle-bottom? No, the triangle has three vertices: let's say \( V_1 = (1, 7) \), \( V_2 = (9, 3) \), \( V_3 = (5, -1) \)? Wait, no, the bottom vertex is at (5, -1)? Wait, maybe not. Wait, the grid: the y-axis is up, x-axis right. Let's list the three vertices:

1. Left vertex: x=1, y=7 (so (1,7))
2. Right vertex: x=9, y=3 (so (9,3))
3. Bottom vertex: x=5, y=-1 (so (5,-1))? Wait, no, the bottom vertex is at (5, -1)? Wait, maybe the original vertices are \( (1, 7) \), \( (9, 3) \), \( (5, -1) \). Wait, but the center of dilation is \( (5, -5) \).

Step2: Dilation Formula

The formula for dilation centered at \( (h, k) \) with scale factor \( r \) is:
\( (x', y') = (h + r(x - h), k + r(y - k)) \)

So for each vertex \( (x, y) \), the image \( (x', y') \) is calculated as:
\( x' = 5 + \frac{1}{4}(x - 5) \)
\( y' = -5 + \frac{1}{4}(y - (-5)) = -5 + \frac{1}{4}(y + 5) \)

Let's calculate for each vertex:

Vertex 1: \( (1, 7) \)

\( x' = 5 + \frac{1}{4}(1 - 5) = 5 + \frac{1}{4}(-4) = 5 - 1 = 4 \)
\( y' = -5 + \frac{1}{4}(7 + 5) = -5 + \frac{1}{4}(12) = -5 + 3 = -2 \)
So \( (4, -2) \)

Vertex 2: \( (9, 3) \)

\( x' = 5 + \frac{1}{4}(9 - 5) = 5 + \frac{1}{4}(4) = 5 + 1 = 6 \)
\( y' = -5 + \frac{1}{4}(3 + 5) = -5 + \frac{1}{4}(8) = -5 + 2 = -3 \)
So \( (6, -3) \)

Vertex 3: \( (5, -1) \)

\( x' = 5 + \frac{1}{4}(5 - 5) = 5 + 0 = 5 \)
\( y' = -5 + \frac{1}{4}(-1 + 5) = -5 + \frac{1}{4}(4) = -5 + 1 = -4 \)
So \( (5, -4) \)

Wait, but maybe my original vertices were wrong. Let's recheck the original triangle. Wait, maybe the original vertices are \( (1, 7) \), \( (9, 3) \), and \( (5, -1) \)? Wait, no, the bottom vertex is at (5, -1)? Wait, the center is (5, -5), so the distance from (5, -1) to (5, -5) is 4 units (since y from -1 to -5 is -4, so 4 units down). So scaling by 1/4: the distance from center to vertex should be 1/4 of original.

Wait, let's take a correct vertex. Let's find the three vertices properly. Let's look at the graph:

• Left vertex: x=1, y=7 (so (1,7))
• Right vertex: x=9, y=3 (so (9,3))
• Bottom vertex: x=5, y=-1 (so (5,-1))

Now, center is (5, -5). Let's compute the vector from center to each vertex:

For \( (1,7) \): vector is \( (1 - 5, 7 - (-5)) = (-4, 12) \). Scale by 1/4: \( (-1, 3) \). Then add to center (5, -5): \( (5 -1, -5 + 3) = (4, -2) \).

For \( (9,3) \): vector is \( (9 - 5, 3 - (-5)) = (4, 8) \). Scale by 1/4: \( (1, 2) \). Add to center: \( (5 +1, -5 + 2) = (6, -3) \).

For \( (5,-1) \): vector is \( (5 - 5, -1 - (-5)) = (0, 4) \). Scale by 1/4: \( (0, 1) \). Add to center: \( (5 +0, -5 +1) = (5, -4) \).

So the image vertices are \( (4, -2) \), \( (6, -3) \), \( (5, -4) \).

Step3: Graph the Image

Now, plot these three points \( (4, -2) \), \( (6, -3) \), \( (5, -4) \) and connect them with the polygon tool.

Answer:

To graph the dilated triangle, plot the vertices \( (4, -2) \), \( (6, -3) \), and \( (5, -4) \) (calculated via dilation formula) and connect them using the polygon tool.

(Note: The exact coordinates depend on the original vertices’ accuracy, but the process uses the dilation formula \( (x', y') = (h + r(x - h), k + r(y - k)) \) with \( (h, k) = (5, -5) \) and \( r = \frac{1}{4} \).)