Answer :
To determine which set of points forms the vertices of a right triangle, we need to verify if these points satisfy the Pythagorean theorem: \(a^2 + b^2 = c^2\), for some permutation of sides \(a\), \(b\), and \(c\), where \(c\) is the hypotenuse.
### Set 1: \(A(-1,-3), B(4,-3), C(2,-1)\)
1. Calculate the distances between the points:
- \(AB\): \(\sqrt{(4 - (-1))^2 + ((-3) - (-3))^2} = \sqrt{5^2 + 0} = 5\)
- \(BC\): \(\sqrt{(2 - 4)^2 + ((-1) - (-3))^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \approx 2\sqrt{2}\)
- \(CA\): \(\sqrt{(2 - (-1))^2 + ((-1) - (-3))^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}\)
2. Check for the Pythagorean theorem:
- \(5^2 + (\sqrt{8})^2 = 25 + 8 \neq 13\)
- \(5^2 + (\sqrt{13})^2 = 25 + 13 \neq 8\)
- \((\sqrt{8})^2 + (\sqrt{13})^2 = 8 + 13 \neq 25\)
This set does not form a right triangle.
### Set 2: \(A(-1,1), B(3,5), C(4,-4)\)
1. Calculate the distances between the points:
- \(AB\): \(\sqrt{(3 - (-1))^2 + (5 - 1)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = 4\sqrt{2}\)
- \(BC\): \(\sqrt{(4 - 3)^2 + (-4 - 5)^2} = \sqrt{1^2 + (-9)^2} = \sqrt{1 + 81} = \sqrt{82}\)
- \(CA\): \(\sqrt{(4 - (-1))^2 + (-4 - 1)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = 5\sqrt{2}\)
2. Check for the Pythagorean theorem:
- \((4\sqrt{2})^2 + (5\sqrt{2})^2 = 32 + 50 = 82\)
- \(\sqrt{82}^2 = 82\)
This set satisfies the Pythagorean theorem, implying it forms a right triangle.
### Set 3: \(A(-1,3), B(-4,-3), C(-4,1)\)
1. Calculate the distances between the points:
- \(AB\): \(\sqrt{(-4 - (-1))^2 + (-3 - 3)^2} = \sqrt{(-3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\)
- \(BC\): \(\sqrt{(-4 - (-4))^2 + (1 - (-3))^2} = \sqrt{0 + 4^2} = 4\)
- \(CA\): \(\sqrt{(-4 - (-1))^2 + (1 - 3)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}\)
2. Check for the Pythagorean theorem:
- \((3\sqrt{5})^2 + 4^2 = 45 + 16 \neq 13\)
- \((4)^2 + (\sqrt{13})^2 = 16 + 13 \neq 45\)
- \((3\sqrt{5})^2 + (\sqrt{13})^2 = 45 + 13 \neq 16\)
This set does not form a right triangle.
### Set 4: \(A(-2,2), B(1,-4), C(4,2)\)
1. Calculate the distances between the points:
- \(AB\): \(\sqrt{(1 - (-2))^2 + (-4 - 2)^2} = \sqrt{3^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\)
- \(BC\): \(\sqrt{(4 - 1)^2 + (2 - (-4))^2} = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\)
- \(CA\): \(\sqrt{(4 - (-2))^2 + (2 - 2)^2} = \sqrt{6^2 + 0} = 6\)
2. Check for the Pythagorean theorem:
- \((3\sqrt{5})^2 + 6^2 = 45 + 36 \neq 45\)
- \((3\sqrt{5})^2 + (3\sqrt{5})^2 = 45 + 45 \neq 36\)
- \(6^2 + (3\sqrt{5})^2 = 36 + 45 \neq 45\)
This set does not form a right triangle.
### Conclusion:
The second set of points, [tex]\(A(-1,1), B(3,5), C(4,-4)\)[/tex], is the only set that forms the vertices of a right triangle.
### Set 1: \(A(-1,-3), B(4,-3), C(2,-1)\)
1. Calculate the distances between the points:
- \(AB\): \(\sqrt{(4 - (-1))^2 + ((-3) - (-3))^2} = \sqrt{5^2 + 0} = 5\)
- \(BC\): \(\sqrt{(2 - 4)^2 + ((-1) - (-3))^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \approx 2\sqrt{2}\)
- \(CA\): \(\sqrt{(2 - (-1))^2 + ((-1) - (-3))^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}\)
2. Check for the Pythagorean theorem:
- \(5^2 + (\sqrt{8})^2 = 25 + 8 \neq 13\)
- \(5^2 + (\sqrt{13})^2 = 25 + 13 \neq 8\)
- \((\sqrt{8})^2 + (\sqrt{13})^2 = 8 + 13 \neq 25\)
This set does not form a right triangle.
### Set 2: \(A(-1,1), B(3,5), C(4,-4)\)
1. Calculate the distances between the points:
- \(AB\): \(\sqrt{(3 - (-1))^2 + (5 - 1)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = 4\sqrt{2}\)
- \(BC\): \(\sqrt{(4 - 3)^2 + (-4 - 5)^2} = \sqrt{1^2 + (-9)^2} = \sqrt{1 + 81} = \sqrt{82}\)
- \(CA\): \(\sqrt{(4 - (-1))^2 + (-4 - 1)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = 5\sqrt{2}\)
2. Check for the Pythagorean theorem:
- \((4\sqrt{2})^2 + (5\sqrt{2})^2 = 32 + 50 = 82\)
- \(\sqrt{82}^2 = 82\)
This set satisfies the Pythagorean theorem, implying it forms a right triangle.
### Set 3: \(A(-1,3), B(-4,-3), C(-4,1)\)
1. Calculate the distances between the points:
- \(AB\): \(\sqrt{(-4 - (-1))^2 + (-3 - 3)^2} = \sqrt{(-3)^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\)
- \(BC\): \(\sqrt{(-4 - (-4))^2 + (1 - (-3))^2} = \sqrt{0 + 4^2} = 4\)
- \(CA\): \(\sqrt{(-4 - (-1))^2 + (1 - 3)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}\)
2. Check for the Pythagorean theorem:
- \((3\sqrt{5})^2 + 4^2 = 45 + 16 \neq 13\)
- \((4)^2 + (\sqrt{13})^2 = 16 + 13 \neq 45\)
- \((3\sqrt{5})^2 + (\sqrt{13})^2 = 45 + 13 \neq 16\)
This set does not form a right triangle.
### Set 4: \(A(-2,2), B(1,-4), C(4,2)\)
1. Calculate the distances between the points:
- \(AB\): \(\sqrt{(1 - (-2))^2 + (-4 - 2)^2} = \sqrt{3^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\)
- \(BC\): \(\sqrt{(4 - 1)^2 + (2 - (-4))^2} = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\)
- \(CA\): \(\sqrt{(4 - (-2))^2 + (2 - 2)^2} = \sqrt{6^2 + 0} = 6\)
2. Check for the Pythagorean theorem:
- \((3\sqrt{5})^2 + 6^2 = 45 + 36 \neq 45\)
- \((3\sqrt{5})^2 + (3\sqrt{5})^2 = 45 + 45 \neq 36\)
- \(6^2 + (3\sqrt{5})^2 = 36 + 45 \neq 45\)
This set does not form a right triangle.
### Conclusion:
The second set of points, [tex]\(A(-1,1), B(3,5), C(4,-4)\)[/tex], is the only set that forms the vertices of a right triangle.
