To determine where a third alien could have been located in order to be rescued on the same trip, we need to first find the positions of the two stranded aliens you want to pick up in a single trip.
Let's assume you pick up the two aliens located at (2, -5) and (-2, -3). The distance from the origin (the starting point) to these locations can be calculated using the distance formula:
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Distance to (2, -5): \[ d_1 = \sqrt{(2 - 0)^2 + (-5 - 0)^2} = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \]
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Distance to (-2, -3): \[ d_2 = \sqrt{(-2 - 0)^2 + (-3 - 0)^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \]
The total distance traveled to pick both aliens from the origin can be represented as going from the origin to (2, -5) and then to (-2, -3), or vice versa.
Next, to find a possible location for the third alien that could be picked up on the same trip, we look for a point that falls within the triangle or path formed by the two chosen points. A straightforward way is to consider the midpoint of the segment between the two chosen positions, which could be an ideal location for a third alien.
Let's calculate the midpoint of (2, -5) and (-2, -3):
\[ \text{Midpoint} = \left(\frac{2 + (-2)}{2}, \frac{-5 + (-3)}{2}\right) = \left(0, -4\right) \]
This point (0, -4) is equidistant from both aliens and is in line between them. Thus:
To answer your question, the third alien could have been located at (0, -4) so that it could have been picked up on the same trip as the other two aliens. However, other locations could exist depending on the allowed route taken. If there's specific criteria for selecting the positions, those should be considered as well.
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