91Ó°ÊÓ

To determine if a triangle is a right triangle, it's crucial to find the lengths of its sides. We accomplish this using the distance formula. The distance formula, derived from the Pythagorean theorem, is used to calculate the distance between two points in a plane.
The formula is:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Let's use it to find the lengths of the sides in our triangle with vertices at points \(A(3,-6)\), \(B(8,-2)\), and \(C(-1,-1)\).
First, find the length of side AB:
\[ d_{AB} = \sqrt{(8 - 3)^2 + (-2 + 6)^2} = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \]
Second, find the length of side BC:
\[ d_{BC} = \sqrt{(-1 - 8)^2 + (-1 + 2)^2} = \sqrt{(-9)^2 + 1} = \sqrt{81 + 1} = \sqrt{82} \]
Finally, find the length of side AC:
\[ d_{AC} = \sqrt{(-1 - 3)^2 + (-1 + 6)^2} = \sqrt{(-4)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \]
Now we have the lengths of all sides:
AB = \( \sqrt{41} \)
BC = \( \sqrt{82} \)
AC = \( \sqrt{41} \).

The next step is to determine if our triangle is a right triangle. For this, we'll use the Pythagorean theorem, which states:
\[ a^2 + b^2 = c^2 \]
In a right triangle, the sum of the squares of the lengths of the two shorter sides should equal the square of the length of the longest side. This is called the converse of the Pythagorean theorem.
For triangle ABC, we have:
\[ AB^2 = 41, \ AC^2 = 41, \ BC^2 = 82 \]
Now, if \( AB^2 + AC^2 \) equals \( BC^2 \), then triangle ABC is a right triangle:
\[ AB^2 + AC^2 = 41 + 41 = 82 \]
\[ AB^2 + AC^2 = BC^2 \], thus the triangle is right.
So, triangle ABC is indeed a right triangle.

Once we've established that our triangle is a right triangle, calculating its area becomes straightforward. The formula for the area of a triangle is:
\[ Area = \frac{1}{2} \times base \times height \]
In a right triangle, any two sides can be treated as the base and the height because they form the right angle. For our triangle ABC, let's treat AB as the base and AC as the height, both with a length of \( \sqrt{41} \).
Now plug in the values:
\[ Area = \frac{1}{2} \times \sqrt{41} \times \sqrt{41} = \frac{1}{2} \times 41 = 20.5 \]
Thus, the area of the right triangle ABC is 20.5 square units.