find whether if it is a square, whether the distance is the same, whether if it a 90 degree, and whther if it is parallel with it's opposite line

To determine if the given points \((-1,5)\), \((3,2)\), \((4,1)\), and \((0,-2)\) form a square, we need to check the following:

1. **Distance Between Adjacent Points:** All four sides of a square should have equal length.
2. **90-Degree Angles:** The angles between the adjacent sides should be 90 degrees.
3. **Parallel Opposite Lines:** Opposite sides should be parallel.

Let's go through each step systematically.

### Step 1: Calculate the Distances Between Adjacent Points

The distance \( d \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

#### Distances:
1. Between \((-1,5)\) and \((3,2)\):

\[
d_1 = \sqrt{(3 - (-1))^2 + (2 - 5)^2} = \sqrt{(3 + 1)^2 + (-3)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]

2. Between \((3,2)\) and \((4,1)\):

\[
d_2 = \sqrt{(4 - 3)^2 + (1 - 2)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \approx 1.414
\]

3. Between \((4,1)\) and \((0,-2)\):

\[
d_3 = \sqrt{(0 - 4)^2 + (-2 - 1)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]

4. Between \((0,-2)\) and \((-1,5)\):

\[
d_4 = \sqrt{(-1 - 0)^2 + (5 - (-2))^2} = \sqrt{(-1)^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50} \approx 7.071
\]

### Step 2: Check if Adjacent Sides are Equal
- \( d_1 = 5 \)
- \( d_2 \approx 1.414 \)
- \( d_3 = 5 \)
- \( d_4 \approx 7.071 \)

Since all sides are not equal, the figure cannot be a square.

### Step 3: Check if Angles Between Adjacent Sides are 90 Degrees
To verify if the angles are 90 degrees, we can calculate the slopes of the sides and check if the product of the slopes of two adjacent sides is \(-1\) (which indicates perpendicularity).

The slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

#### Slopes:
1. Between \((-1,5)\) and \((3,2)\):

\[
m_1 = \frac{2 - 5}{3 - (-1)} = \frac{-3}{4}
\]

2. Between \((3,2)\) and \((4,1)\):

\[
m_2 = \frac{1 - 2}{4 - 3} = \frac{-1}{1} = -1
\]

3. Between \((4,1)\) and \((0,-2)\):

\[
m_3 = \frac{-2 - 1}{0 - 4} = \frac{-3}{-4} = \frac{3}{4}
\]

4. Between \((0,-2)\) and \((-1,5)\):

\[
m_4 = \frac{5 - (-2)}{-1 - 0} = \frac{7}{-1} = -7
\]

### Perpendicular Check:
- \( m_1 \times m_2 = \frac{-3}{4} \times (-1) = \frac{3}{4} \) (Not equal to \(-1\))
- \( m_2 \times m_3 = -1 \times \frac{3}{4} = \frac{-3}{4} \) (Not equal to \(-1\))
- \( m_3 \times m_4 = \frac{3}{4} \times -7 = \frac{-21}{4} \) (Not equal to \(-1\))

None of the adjacent sides are perpendicular, so the angles are not 90 degrees.

### Step 4: Check if Opposite Sides are Parallel
Two lines are parallel if their slopes are equal.

- \( m_1 \) and \( m_3 \) (opposite sides): \(\frac{-3}{4}\) and \(\frac{3}{4}\) (Not equal, so not parallel).
- \( m_2 \) and \( m_4 \) (opposite sides): \(-1\) and \(-7\) (Not equal, so not parallel).

### Conclusion:
The given points do not form a square. The sides are not of equal length, the angles are not 90 degrees, and the opposite sides are not parallel.

Would you like more details or have any questions?

---

### Follow-Up Questions:
1. What conditions must be met for four points to form a rectangle?
2. How do you determine if a quadrilateral is a parallelogram using its vertices?
3. Can the given points form any other special quadrilateral, such as a rhombus or trapezoid?
4. How can we verify if the given points are collinear?
5. What is the role of the distance formula in coordinate geometry?

### Tip:
In coordinate geometry, always verify the distances and angles when identifying shapes based on their vertices. This helps in accurately classifying the shapes.

Explore how to determine if four given points \((-1,5)\), \((3,2)\), \((4,1)\), and \((0,-2)\) form a square. Check distances between points, angles between sides, and parallelism of opposite lines using coordinate geometry concepts. Suitable for high school students learning about geometric shapes and coordinate geometry.

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