Show in two ways that the points (6,1), (2,-3) and (4,-5) are vertices of a right triangle 

To demonstrate that the points \( (6, 1) \), \( (2, -3) \), and \( (4, -5) \) are vertices of a right triangle, we'll use two methods: **distance formula** and **slope analysis**.

### Method 1: Distance Formula (Pythagorean Theorem)
We will calculate the distances between each pair of points and check if they satisfy the Pythagorean theorem, i.e., if \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse.

The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

1. **Distance between \( (6, 1) \) and \( (2, -3) \):**
\[
d_1 = \sqrt{(2 - 6)^2 + (-3 - 1)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
\]

2. **Distance between \( (2, -3) \) and \( (4, -5) \):**
\[
d_2 = \sqrt{(4 - 2)^2 + (-5 - (-3))^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]

3. **Distance between \( (6, 1) \) and \( (4, -5) \):**
\[
d_3 = \sqrt{(4 - 6)^2 + (-5 - 1)^2} = \sqrt{(-2)^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}
\]

Now check if the Pythagorean theorem holds:
\[
(2\sqrt{2})^2 + (4\sqrt{2})^2 = (2\sqrt{10})^2
\]
\[
8 + 32 = 40 \quad \text{(True)}
\]

Since the distances satisfy the Pythagorean theorem, the points form a right triangle, with \( 2\sqrt{10} \) as the hypotenuse.

### Method 2: Slope Analysis (Perpendicular Slopes)
For a right triangle, two sides must be perpendicular to each other, meaning the product of their slopes should be \( -1 \).

The slope between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

1. **Slope between \( (6, 1) \) and \( (2, -3) \):**
\[
m_1 = \frac{-3 - 1}{2 - 6} = \frac{-4}{-4} = 1
\]

2. **Slope between \( (2, -3) \) and \( (4, -5) \):**
\[
m_2 = \frac{-5 - (-3)}{4 - 2} = \frac{-2}{2} = -1
\]

3. **Slope between \( (6, 1) \) and \( (4, -5) \):**
\[
m_3 = \frac{-5 - 1}{4 - 6} = \frac{-6}{-2} = 3
\]

Now check if any two slopes are perpendicular:
\[
m_1 \times m_2 = 1 \times (-1) = -1
\]
Since \( m_1 \times m_2 = -1 \), the sides corresponding to these slopes are perpendicular, confirming that the triangle is a right triangle.

### Conclusion:
The points \( (6, 1) \), \( (2, -3) \), and \( (4, -5) \) form a right triangle. This is confirmed through both the distance formula (Pythagorean theorem) and slope analysis (perpendicular slopes).

---

Would you like more details on any step or have any other questions?

Here are 5 related questions:
1. How do we find the area of this right triangle using its vertices?
2. What are the equations of the lines passing through each side of the triangle?
3. How can we find the coordinates of the circumcenter of this triangle?
4. What is the radius of the circumcircle of this triangle?
5. Can we verify the angles of this triangle using the cosine rule?

**Tip:** When two lines are perpendicular, their slopes multiply to give \(-1\), which is a quick check for right angles in coordinate geometry.

Show in two ways that the points (6,1), (2,-3) and (4,-5) are vertices of a right triangle

Learn how to prove that the points (6,1), (2,-3), and (4,-5) form a right triangle using the distance formula and slope analysis. This step-by-step solution covers the Pythagorean theorem and perpendicular slopes, perfect for students in Grades 9-11.

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