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To determine which points are exactly 5 units away from the point [tex]\( P(-1, 2) \)[/tex], we need to use the distance formula. For two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex], the distance between them is given by:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Here, [tex]\( P = (-1, 2) \)[/tex] is our reference point. We will calculate the distance from [tex]\( P \)[/tex] to each of the given points and see if the distance equals 5 units.
1. For the point (4, 2):
[tex]\[ d = \sqrt{(4 - (-1))^2 + (2 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{(4 + 1)^2 + 0^2} \][/tex]
[tex]\[ d = \sqrt{5^2} \][/tex]
[tex]\[ d = 5 \][/tex]
So, [tex]\( (4, 2) \)[/tex] is 5 units away from point [tex]\( P \)[/tex].
2. For the point (1, 5):
[tex]\[ d = \sqrt{(1 - (-1))^2 + (5 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{(1 + 1)^2 + 3^2} \][/tex]
[tex]\[ d = \sqrt{2^2 + 3^2} \][/tex]
[tex]\[ d = \sqrt{4 + 9} \][/tex]
[tex]\[ d = \sqrt{13} \][/tex]
Since [tex]\(\sqrt{13} \neq 5\)[/tex], [tex]\( (1, 5) \)[/tex] is not 5 units away from point [tex]\( P \)[/tex].
3. For the point (2, -2):
[tex]\[ d = \sqrt{(2 - (-1))^2 + (-2 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{(2 + 1)^2 + (-2 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{3^2 + (-4)^2} \][/tex]
[tex]\[ d = \sqrt{9 + 16} \][/tex]
[tex]\[ d = \sqrt{25} \][/tex]
[tex]\[ d = 5 \][/tex]
So, [tex]\( (2, -2) \)[/tex] is 5 units away from point [tex]\( P \)[/tex].
4. For the point (-6, 7):
[tex]\[ d = \sqrt{(-6 - (-1))^2 + (7 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{(-6 + 1)^2 + 5^2} \][/tex]
[tex]\[ d = \sqrt{(-5)^2 + 5^2} \][/tex]
[tex]\[ d = \sqrt{25 + 25} \][/tex]
[tex]\[ d = \sqrt{50} \][/tex]
Since [tex]\(\sqrt{50} \neq 5\)[/tex], [tex]\( (-6, 7) \)[/tex] is not 5 units away from point [tex]\( P \)[/tex].
5. For the point (-1, -3):
[tex]\[ d = \sqrt{(-1 - (-1))^2 + (-3 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{0^2 + (-3 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{(-5)^2} \][/tex]
[tex]\[ d = \sqrt{25} \][/tex]
[tex]\[ d = 5 \][/tex]
So, [tex]\( (-1, -3) \)[/tex] is 5 units away from point [tex]\( P \)[/tex].
Therefore, the three ordered pairs that describe points exactly 5 units away from [tex]\( P(-1, 2) \)[/tex] are:
[tex]\[ (4, 2), (2, -2), \text{ and } (-1, -3). \][/tex]
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Here, [tex]\( P = (-1, 2) \)[/tex] is our reference point. We will calculate the distance from [tex]\( P \)[/tex] to each of the given points and see if the distance equals 5 units.
1. For the point (4, 2):
[tex]\[ d = \sqrt{(4 - (-1))^2 + (2 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{(4 + 1)^2 + 0^2} \][/tex]
[tex]\[ d = \sqrt{5^2} \][/tex]
[tex]\[ d = 5 \][/tex]
So, [tex]\( (4, 2) \)[/tex] is 5 units away from point [tex]\( P \)[/tex].
2. For the point (1, 5):
[tex]\[ d = \sqrt{(1 - (-1))^2 + (5 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{(1 + 1)^2 + 3^2} \][/tex]
[tex]\[ d = \sqrt{2^2 + 3^2} \][/tex]
[tex]\[ d = \sqrt{4 + 9} \][/tex]
[tex]\[ d = \sqrt{13} \][/tex]
Since [tex]\(\sqrt{13} \neq 5\)[/tex], [tex]\( (1, 5) \)[/tex] is not 5 units away from point [tex]\( P \)[/tex].
3. For the point (2, -2):
[tex]\[ d = \sqrt{(2 - (-1))^2 + (-2 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{(2 + 1)^2 + (-2 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{3^2 + (-4)^2} \][/tex]
[tex]\[ d = \sqrt{9 + 16} \][/tex]
[tex]\[ d = \sqrt{25} \][/tex]
[tex]\[ d = 5 \][/tex]
So, [tex]\( (2, -2) \)[/tex] is 5 units away from point [tex]\( P \)[/tex].
4. For the point (-6, 7):
[tex]\[ d = \sqrt{(-6 - (-1))^2 + (7 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{(-6 + 1)^2 + 5^2} \][/tex]
[tex]\[ d = \sqrt{(-5)^2 + 5^2} \][/tex]
[tex]\[ d = \sqrt{25 + 25} \][/tex]
[tex]\[ d = \sqrt{50} \][/tex]
Since [tex]\(\sqrt{50} \neq 5\)[/tex], [tex]\( (-6, 7) \)[/tex] is not 5 units away from point [tex]\( P \)[/tex].
5. For the point (-1, -3):
[tex]\[ d = \sqrt{(-1 - (-1))^2 + (-3 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{0^2 + (-3 - 2)^2} \][/tex]
[tex]\[ d = \sqrt{(-5)^2} \][/tex]
[tex]\[ d = \sqrt{25} \][/tex]
[tex]\[ d = 5 \][/tex]
So, [tex]\( (-1, -3) \)[/tex] is 5 units away from point [tex]\( P \)[/tex].
Therefore, the three ordered pairs that describe points exactly 5 units away from [tex]\( P(-1, 2) \)[/tex] are:
[tex]\[ (4, 2), (2, -2), \text{ and } (-1, -3). \][/tex]
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