Question

Use the image to answer the question.

A coordinate plane's x-axis ranges from negative 6 to 2 and its y-axis ranges from negative 3 to 2, both by 1-unit increments. 4 points forming a parallelogram are plotted, labeled upper A, upper B, upper C, and upper D, respectively.

Suppose parallelogram ABCD
is dilated using a magnitude of 2 and a center of dilation at (−2,−1)
. What would be the corresponding ordered pair at point A
?

(1 point)
Responses

(−4,3)
left parenthesis negative 4 comma negative 3 right parenthesis

(−5,0)
left parenthesis negative 5 comma 0 right parenthesis

(−3,0)
left parenthesis negative 3 comma 0 right parenthesis

(−1,2)

Answer 1

To find the corresponding ordered pair for point A after dilation, we can use the formula for dilation:

\[
P' = C + k(P - C)
\]

where:
- \(P\) is the original point,
- \(P'\) is the dilated point,
- \(C\) is the center of dilation,
- \(k\) is the magnitude of the dilation.

Suppose point A has coordinates \((x_A, y_A)\). In this case, the center of dilation \(C\) is at \((-2, -1)\) and the magnitude \(k\) is 2.

We can rewrite the formula with specific coordinates:

\[
P' = (-2, -1) + 2 \left((x_A, y_A) - (-2, -1)\right)
\]

Calculating \(P - C\):

\[
P - C = (x_A + 2, y_A + 1)
\]

Now plugging that into the dilation formula:

\[
P' = (-2, -1) + 2 \cdot (x_A + 2, y_A + 1)
\]

This becomes:

\[
P' = (-2, -1) + (2(x_A + 2), 2(y_A + 1))
\]
\[
= (-2 + 2(x_A + 2), -1 + 2(y_A + 1))
\]
\[
= (2x_A + 2, 2y_A + 1)
\]

For the resulting coordinates, let's consider the possible responses provided and pick the one that makes sense when substituting point A's original coordinates.

Since the coordinates of point A are not given in the problem statement, let's test each option provided to check which might correspond to point A before dilation:

1. If \(P' = (-4, 3)\):
\[
2x_A + 2 = -4 \rightarrow 2x_A = -6 \rightarrow x_A = -3
\]
\[
2y_A + 1 = 3 \rightarrow 2y_A = 2 \rightarrow y_A = 1
\]
Then \(A = (-3, 1)\).

2. If \(P' = (-5, 0)\):
\[
2x_A + 2 = -5 \rightarrow 2x_A = -7 \rightarrow x_A = -3.5 \quad (\text{Not an integer})
\]
\[
2y_A + 1 = 0 \rightarrow 2y_A = -1 \rightarrow y_A = -0.5 \quad (\text{Not an integer})
\]

3. If \(P' = (-3, 0)\):
\[
2x_A + 2 = -3 \rightarrow 2x_A = -5 \rightarrow x_A = -2.5 \quad (\text{Not an integer})
\]
\[
2y_A + 1 = 0 \rightarrow 2y_A = -1 \rightarrow y_A = -0.5 \quad (\text{Not an integer})
\]

4. If \(P' = (-1, 2)\):
\[
2x_A + 2 = -1 \rightarrow 2x_A = -3 \rightarrow x_A = -1.5 \quad (\text{Not an integer})
\]
\[
2y_A + 1 = 2 \rightarrow 2y_A = 1 \rightarrow y_A = 0.5 \quad (\text{Not an integer})
\]

After testing, only point A corresponding to (-4, 3) has integer values for point A as well.

Thus, the answer is:

**(−4,3)**