Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
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]
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
