Question

pentagon abcde is dilated according to the rule \\(d_{0.3}(x,y)\\) to create the image pentagon abcde, which is shown on the graph.

what are the coordinates of point a of the pre-image?

\\((-1, 1)\\)
\\((-1, 2)\\)
\\((-9, 6)\\)
\\((-9, 18)\\)

Explanation:

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<reasoning>

Identify the coordinates of point A' from the graph

\[
A' = (-3, 2)
\]

Apply the inverse dilation rule to find the pre-image point A

\[

$$\begin{aligned} &D_{0.3}(x, y) = (0.3x, 0.3y) = A'\\ &(0.3x, 0.3y) = (-3, 2)\\ &x = \frac{-3}{0.3} = -10\\ &y = \frac{2}{0.3} = \frac{20}{3} \approx 6.67 \end{aligned}$$

\]
Re-evaluating the coordinates of \(A'\) from the grid:
Looking closely at the grid:
The x-axis has labels: \(-1, -2, -3, -4, -5\). Point \(A'\) is at \(x = -3\).
The y-axis has labels: \(2, 4, 6, 8\).
Point \(A'\) is at \(y = 6\).
Let's recalculate with \(A' = (-3, 6)\):
\[

$$\begin{aligned} &x = \frac{-3}{0.3} = -10\\ &y = \frac{6}{0.3} = 20 \end{aligned}$$

\]
Let's re-examine the grid coordinates of \(A'\) very carefully:
The vertical axis has grid lines. The label \(2\) is at 2 units up. \(4\) is at 4 units up. \(6\) is at 6 units up. \(8\) is at 8 units up.
Point \(B'\) is at \((0, 8)\) or \((-1, 8)\)? It is at \((-1, 8)\).
Point \(A'\) is at \(x = -3\), and its height is at the grid line for \(y = 6\) or \(y = 5.4\)?
If \(A = (-9, 18)\), then \(A' = D_{0.3}(-9, 18) = (-2.7, 5.4)\).
If \(A = (-10, 20)\), that is not an option.
Let's check the options:

• \((-1, 1)\)
• \((-1, 2)\)
• \((-9, 6)\)
• \((-9, 18)\)

If \(A = (-9, 18)\):
\[
A' = D_{0.3}(-9, 18) = (-9 \times 0.3, 18 \times 0.3) = (-2.7, 5.4)
\]
Looking at the graph, \(A'\) is slightly to the right of \(-3\) (which is \(-2.7\)) and slightly below \(6\) (which is \(5.4\)). This perfectly matches the visual representation.

Calculate the pre-image coordinates

\[

$$\begin{aligned} &A = (-9, 18) \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \((-1, 1)\)</mcq-option>
<mcq-option>(B) \((-1, 2)\)</mcq-option>
<mcq-option>(C) \((-9, 6)\)</mcq-option>
<mcq-correct>(D) \((-9, 18)\)</mcq-correct>
</answer>

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Answer:

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<reasoning>

Identify the coordinates of point A' from the graph

\[
A' = (-3, 2)
\]

Apply the inverse dilation rule to find the pre-image point A

\[

$$\begin{aligned} &D_{0.3}(x, y) = (0.3x, 0.3y) = A'\\ &(0.3x, 0.3y) = (-3, 2)\\ &x = \frac{-3}{0.3} = -10\\ &y = \frac{2}{0.3} = \frac{20}{3} \approx 6.67 \end{aligned}$$

\]
Re-evaluating the coordinates of \(A'\) from the grid:
Looking closely at the grid:
The x-axis has labels: \(-1, -2, -3, -4, -5\). Point \(A'\) is at \(x = -3\).
The y-axis has labels: \(2, 4, 6, 8\).
Point \(A'\) is at \(y = 6\).
Let's recalculate with \(A' = (-3, 6)\):
\[

$$\begin{aligned} &x = \frac{-3}{0.3} = -10\\ &y = \frac{6}{0.3} = 20 \end{aligned}$$

\]
Let's re-examine the grid coordinates of \(A'\) very carefully:
The vertical axis has grid lines. The label \(2\) is at 2 units up. \(4\) is at 4 units up. \(6\) is at 6 units up. \(8\) is at 8 units up.
Point \(B'\) is at \((0, 8)\) or \((-1, 8)\)? It is at \((-1, 8)\).
Point \(A'\) is at \(x = -3\), and its height is at the grid line for \(y = 6\) or \(y = 5.4\)?
If \(A = (-9, 18)\), then \(A' = D_{0.3}(-9, 18) = (-2.7, 5.4)\).
If \(A = (-10, 20)\), that is not an option.
Let's check the options:

• \((-1, 1)\)
• \((-1, 2)\)
• \((-9, 6)\)
• \((-9, 18)\)

If \(A = (-9, 18)\):
\[
A' = D_{0.3}(-9, 18) = (-9 \times 0.3, 18 \times 0.3) = (-2.7, 5.4)
\]
Looking at the graph, \(A'\) is slightly to the right of \(-3\) (which is \(-2.7\)) and slightly below \(6\) (which is \(5.4\)). This perfectly matches the visual representation.

Calculate the pre-image coordinates

\[

$$\begin{aligned} &A = (-9, 18) \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \((-1, 1)\)</mcq-option>
<mcq-option>(B) \((-1, 2)\)</mcq-option>
<mcq-option>(C) \((-9, 6)\)</mcq-option>
<mcq-correct>(D) \((-9, 18)\)</mcq-correct>
</answer>

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