To find a possible coordinate for Mia's house, we need to determine a point that is 7 units away from the park located at (3, -5). This can be done using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
In this case, the distance \( d = 7 \), and the park coordinates are \( (x_2, y_2) = (3, -5) \). Let \( (x_1, y_1) \) represent the coordinates of Mia's house.
Applying the distance formula, we rearrange it for Mia's house:
\[ 7 = \sqrt{(x_1 - 3)^2 + (y_1 + 5)^2} \]
Squaring both sides, we get:
\[ 49 = (x_1 - 3)^2 + (y_1 + 5)^2 \]
This equation describes a circle centered at the point (3, -5) with a radius of 7. We can find potential coordinates by choosing values for \( x_1 \) and calculating \( y_1 \).
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Let's say we choose \( x_1 = 3 \) (the same x-coordinate as the park):
\[ 49 = (3 - 3)^2 + (y_1 + 5)^2 \] \[ 49 = 0 + (y_1 + 5)^2 \] \[ (y_1 + 5)^2 = 49 \] \[ y_1 + 5 = 7 \quad \text{or} \quad y_1 + 5 = -7 \] This gives:
- \( y_1 = 2 \) (one possible coordinate is \( (3, 2) \))
- \( y_1 = -12 \) (another possible coordinate is \( (3, -12) \))
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Alternatively, let’s choose \( x_1 = 0 \):
\[ 49 = (0 - 3)^2 + (y_1 + 5)^2 \] \[ 49 = 9 + (y_1 + 5)^2 \] \[ 40 = (y_1 + 5)^2 \] \[ y_1 + 5 = \sqrt{40} \quad \text{or} \quad y_1 + 5 = -\sqrt{40} \] This gives: \[ y_1 \approx 2.32 \quad \text{or} \quad y_1 \approx -12.32 \] So we could also have \( (0, 2.32) \) or \( (0, -12.32) \).
Thus, some possible coordinates for Mia's house include \( (3, 2) \), \( (3, -12) \), \( (0, 2.32) \), and \( (0, -12.32) \).