Show that the points (-2,3), (8, 3) and (6, 7) are the vertices of a right angle triangle .

Answer

To show that the points (-2, 3), (8, 3), and (6, 7) are the vertices of a right-angled triangle, we will use the distance formula to calculate the lengths of the sides of the triangle formed by these points and then apply the Pythagorean theorem.

### Step-by-Step Solution:

1. **Identify the Points**:
Let the points be:
- A = (-2, 3)
- B = (8, 3)
- C = (6, 7)

2. **Calculate the Length of Side AB**:
We use the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For points A and B:
\[ AB = \sqrt{(8 - (-2))^2 + (3 - 3)^2} = \sqrt{(8 + 2)^2 + 0^2} = \sqrt{10^2} = 10 \]

3. **Calculate the Length of Side BC**:
For points B and C:
\[ BC = \sqrt{(6 - 8)^2 + (7 - 3)^2} = \sqrt{(-2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} \]

4. **Calculate the Length of Side AC**:
For points A and C:
\[ AC = \sqrt{(6 - (-2))^2 + (7 - 3)^2} = \sqrt{(6 + 2)^2 + (4)^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} \]

5. **Check Pythagorean Theorem**:
According to the Pythagorean theorem, for a right-angled triangle:
\[ c^2 = a^2 + b^2 \]
where c is the longest side. Here, we have:
- \( AB = 10 \)
- \( BC = \sqrt{20} \)
- \( AC = \sqrt{80} \)

We need to check if:
\[ AB^2 = BC^2 + AC^2 \]
Calculating:
\[ AB^2 = 10^2 = 100 \]
\[ BC^2 = (\sqrt{20})^2 = 20 \]
\[ AC^2 = (\sqrt{80})^2 = 80 \]
Now, adding \( BC^2 \) and \( AC^2 \):
\[ BC^2 + AC^2 = 20 + 80 = 100 \]

6. **Conclusion**:
Since \( AB^2 = BC^2 + AC^2 \) holds true, the triangle formed by the points A, B, and C is a right-angled triangle.