Show that (-1,-1),(9,4),(20,6), and (10,1) are the vertices of a rhombus, and then find the area of this rhombus.

To prove that the points \((-1, -1)\), \((9, 4)\), \((20, 6)\), and \((10, 1)\) are the vertices of a rhombus and to find the area of the rhombus, we will follow these steps:

### 1. Show that the four sides are equal.
The first step is to calculate the lengths of the sides between each pair of adjacent vertices.

#### Distance between \((-1, -1)\) and \((9, 4)\):
\[
\text{Length}_1 = \sqrt{(9 - (-1))^2 + (4 - (-1))^2} = \sqrt{(9 + 1)^2 + (4 + 1)^2} = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}
\]

#### Distance between \((9, 4)\) and \((20, 6)\):
\[
\text{Length}_2 = \sqrt{(20 - 9)^2 + (6 - 4)^2} = \sqrt{11^2 + 2^2} = \sqrt{121 + 4} = \sqrt{125} = 5\sqrt{5}
\]

#### Distance between \((20, 6)\) and \((10, 1)\):
\[
\text{Length}_3 = \sqrt{(10 - 20)^2 + (1 - 6)^2} = \sqrt{(-10)^2 + (-5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}
\]

#### Distance between \((10, 1)\) and \((-1, -1)\):
\[
\text{Length}_4 = \sqrt{(10 - (-1))^2 + (1 - (-1))^2} = \sqrt{(10 + 1)^2 + (1 + 1)^2} = \sqrt{11^2 + 2^2} = \sqrt{125} = 5\sqrt{5}
\]

Since all four sides are equal (\(5\sqrt{5}\)), the quadrilateral is a rhombus.

### 2. Show that the diagonals are perpendicular.
Next, let's find the lengths of the diagonals to verify that they are perpendicular.

#### Diagonal between \((-1, -1)\) and \((20, 6)\):
\[
\text{Diagonal}_1 = \sqrt{(20 - (-1))^2 + (6 - (-1))^2} = \sqrt{(20 + 1)^2 + (6 + 1)^2} = \sqrt{21^2 + 7^2} = \sqrt{441 + 49} = \sqrt{490} = 7\sqrt{10}
\]

#### Diagonal between \((9, 4)\) and \((10, 1)\):
\[
\text{Diagonal}_2 = \sqrt{(10 - 9)^2 + (1 - 4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]

The diagonals of a rhombus are perpendicular to each other. This can be shown by verifying that the dot product of their direction vectors equals zero:

- Direction vector for \((-1, -1)\) to \((20, 6)\) is \((21, 7)\).
- Direction vector for \((9, 4)\) to \((10, 1)\) is \((1, -3)\).

Calculate the dot product:
\[
\text{Dot product} = (21 \times 1) + (7 \times -3) = 21 - 21 = 0
\]
Since the dot product is zero, the diagonals are perpendicular, confirming that the quadrilateral is indeed a rhombus.

### 3. Calculate the area of the rhombus.
The area \(A\) of a rhombus is given by:
\[
A = \frac{1}{2} \times \text{Diagonal}_1 \times \text{Diagonal}_2
\]
Substituting the lengths of the diagonals:
\[
A = \frac{1}{2} \times 7\sqrt{10} \times \sqrt{10} = \frac{1}{2} \times 7 \times 10 = \frac{70}{2} = 35 \text{ square units}
\]

### Conclusion
The given points are the vertices of a rhombus, and the area of this rhombus is \(35\) square units.

Would you like any further details on this problem or have any questions? Here are 5 related questions:

1. How do you find the area of a rhombus if only the side length and one angle are given?
2. What are the properties of a rhombus that distinguish it from other quadrilaterals?
3. How can you prove that the diagonals of a rhombus bisect each other at right angles?
4. How would the area change if the coordinates of one vertex were different?
5. Can the area of a rhombus be found using trigonometry if angles and side lengths are known?

**Tip:** The diagonals of a rhombus not only bisect each other at right angles, but they also bisect the angles of the rhombus.

Learn how to prove that the points (-1, -1), (9, 4), (20, 6), and (10, 1) form a rhombus using coordinate geometry and find its area. This problem covers the distance formula, properties of rhombus, and calculation of area. Suitable for students in Grades 9-12.

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