To show that the given points are the vertices of a rectangle, we need to verify two main properties of a rectangle:
1. Opposite sides are equal in length.
2. The diagonals are equal in length.
We will use the distance formula to calculate the lengths of the sides and diagonals. The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
### Part (i): Points A(-4, -1), B(-2, -4), C(4, 0), D(2, 3)
1. **Calculate the distances:**
- **AB:**
\[
AB = \sqrt{(-2 - (-4))^2 + (-4 - (-1))^2} = \sqrt{(2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}
\]
- **BC:**
\[
BC = \sqrt{(4 - (-2))^2 + (0 - (-4))^2} = \sqrt{(6)^2 + (4)^2} = \sqrt{36 + 16} = \sqrt{52}
\]
- **CD:**
\[
CD = \sqrt{(2 - 4)^2 + (3 - 0)^2} = \sqrt{(-2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13}
\]
- **DA:**
\[
DA = \sqrt{(-4 - 2)^2 + (-1 - 3)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}
\]
2. **Check the diagonals:**
- **AC:**
\[
AC = \sqrt{(4 - (-4))^2 + (0 - (-1))^2} = \sqrt{(8)^2 + (1)^2} = \sqrt{64 + 1} = \sqrt{65}
\]
- **BD:**
\[
BD = \sqrt{(2 - (-2))^2 + (3 - (-4))^2} = \sqrt{(4)^2 + (7)^2} = \sqrt{16 + 49} = \sqrt{65}
\]
3. **Conclusion:**
- Opposite sides are equal: \(AB = CD\) and \(BC = DA\).
- Diagonals are equal: \(AC = BD\).
- Therefore, points A, B, C, and D form a rectangle.
### Part (ii): Points A(2, -2), B(14, 10), C(11, 13), D(-1, 1)
1. **Calculate the distances:**
- **AB:**
\[
AB = \sqrt{(14 - 2)^2 + (10 - (-2))^2} = \sqrt{(12)^2 + (12)^2} = \sqrt{144 + 144} = 12\sqrt{2}
\]
- **BC:**
\[
BC = \sqrt{(11 - 14)^2 + (13 - 10)^2} = \sqrt{(-3)^2 + (3)^2} = \sqrt{9 + 9} = 3\sqrt{2}
\]
- **CD:**
\[
CD = \sqrt{(-1 - 11)^2 + (1 - 13)^2} = \sqrt{(-12)^2 + (-12)^2} = \sqrt{144 + 144} = 12\sqrt{2}
\]
- **DA:**
\[
DA = \sqrt{(2 - (-1))^2 + (-2 - 1)^2} = \sqrt{(3)^2 + (-3)^2} = \sqrt{9 + 9} = 3\sqrt{2}
\]
2. **Check the diagonals:**
- **AC:**
\[
AC = \sqrt{(11 - 2)^2 + (13 - (-2))^2} = \sqrt{(9)^2 + (15)^2} = \sqrt{81 + 225} = \sqrt{306}
\]
- **BD:**
\[
BD = \sqrt{(-1 - 14)^2 + (1 - 10)^2} = \sqrt{(-15)^2 + (-9)^2} = \sqrt{225 + 81} = \sqrt{306}
\]
3. **Conclusion:**
- Opposite sides are equal: \(AB = CD\) and \(BC = DA\).
- Diagonals are equal: \(AC = BD\).
- Therefore, points A, B, C, and D form a rectangle.
### Part (iii): Points A(0, -4), B(6, 2), C(3, 5), D(-3, -1)
1. **Calculate the distances:**
- **AB:**
\[
AB = \sqrt{(6 - 0)^2 + (2 - (-4))^2} = \sqrt{(6)^2 + (6)^2} = \sqrt{36 + 36} = 6\sqrt{2}
\]
- **BC:**
\[
BC = \sqrt{(3 - 6)^2 + (5 - 2)^2} = \sqrt{(-3)^2 + (3)^2} = \sqrt{9 + 9} = 3\sqrt{2}
\]
- **CD:**
\[
CD = \sqrt{(-3 - 3)^2 + (-1 - 5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = 6\sqrt{2}
\]
- **DA:**
\[
DA = \sqrt{(0 - (-3))^2 + (-4 - (-1))^2} = \sqrt{(3)^2 + (-3)^2} = \sqrt{9 + 9} = 3\sqrt{2}
\]
2. **Check the diagonals:**
- **AC:**
\[
AC = \sqrt{(3 - 0)^2 + (5 - (-4))^2} = \sqrt{(3)^2 + (9)^2} = \sqrt{9 + 81} = \sqrt{90}
\]
- **BD:**
\[
BD = \sqrt{(-3 - 6)^2 + (-1 - 2)^2} = \sqrt{(-9)^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90}
\]
3. **Conclusion:**
- Opposite sides are equal: \(AB = CD\) and \(BC = DA\).
- Diagonals are equal: \(AC = BD\).
- Therefore, points A, B, C, and D form a rectangle.
