The problem asks us to prove that the points $(-2, 3)$, $(8, 3)$, and $(6, 7)$ are the vertices of a right-angled triangle. To do this, we can use the distance formula to calculate the lengths of the sides of the triangle and then apply the Pythagorean theorem.

Step 1: Calculate the distances between each pair of points (the sides of the triangle)

The distance formula is:

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

Distance between $(-2, 3)$ and $(8, 3)$:

$d_1 = \sqrt{(8 - (-2))^2 + (3 - 3)^2} = \sqrt{(8 + 2)^2} = \sqrt{10^2} = 10$

Distance between $(8, 3)$ and $(6, 7)$:

$d_2 = \sqrt{(6 - 8)^2 + (7 - 3)^2} = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \approx 4.47$

Distance between $(6, 7)$ and $(-2, 3)$:

$d_3 = \sqrt{(6 - (-2))^2 + (7 - 3)^2} = \sqrt{(6 + 2)^2 + 4^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} \approx 8.94$

Step 2: Check if the Pythagorean theorem holds

The Pythagorean theorem states that for a right-angled triangle, the sum of the squares of the two shorter sides must equal the square of the longest side. Let's check this condition.

• The longest side is $d_1 = 10$.
• Check if $d_1^2 = d_2^2 + d_3^2$: $10^2 = 4.47^2 + 8.94^2$ $100 = 20 + 80$ $100 = 100$

Since the Pythagorean theorem holds, the points indeed form a right-angled triangle.

Would you like a further explanation or clarification on this solution?

Here are some related questions to explore:

1. How can the distance formula be derived from the Pythagorean theorem?
2. Can you prove that another set of points forms a right-angled triangle using the same method?
3. How do you find the area of a triangle when you know the coordinates of its vertices?
4. What other methods can be used to prove that a triangle is right-angled?
5. How does the Pythagorean theorem generalize to 3-dimensional space?

Tip: Always double-check the distance calculations when working with coordinate geometry problems to avoid small errors.