Show that the points A(-5,6) ,B(3,0) and C(9,8) are vertices of an isosceles right angled triangles. Find the area of this triangles.

Answer

To show that the points A(-5, 6), B(3, 0), and C(9, 8) are the vertices of an isosceles right-angled triangle and to find the area of this triangle, we will follow these steps:

### Step 1: Calculate the distances between the points A, B, and C.

1. **Distance AB**:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3 - (-5))^2 + (0 - 6)^2} \]
\[ = \sqrt{(3 + 5)^2 + (-6)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \]

2. **Distance BC**:
\[ BC = \sqrt{(9 - 3)^2 + (8 - 0)^2} = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]

3. **Distance AC**:
\[ AC = \sqrt{(9 - (-5))^2 + (8 - 6)^2} = \sqrt{(9 + 5)^2 + (2)^2} = \sqrt{14^2 + 2^2} = \sqrt{196 + 4} = \sqrt{200} = 10\sqrt{2} \]

### Step 2: Determine if the triangle is isosceles.

From the distances calculated:
- \( AB = 10 \)
- \( BC = 10 \)
- \( AC = 10\sqrt{2} \)

Since \( AB = BC \), the triangle is isosceles.

### Step 3: Check if the triangle is a right-angled triangle using the Pythagorean theorem.

We need to check if:
\[ AC^2 = AB^2 + BC^2 \]
Calculating:
\[ AC^2 = (10\sqrt{2})^2 = 200 \]
\[ AB^2 + BC^2 = 10^2 + 10^2 = 100 + 100 = 200 \]

Since \( AC^2 = AB^2 + BC^2 \), the triangle is also a right-angled triangle.

### Step 4: Calculate the area of the triangle.

For an isosceles right triangle, the area can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can take \( AB \) and \( BC \) as the base and height:
\[ \text{Area} = \frac{1}{2} \times 10 \times 10 = \frac{100}{2} = 50 \]

### Final Result:
The points A(-5, 6), B(3, 0), and C(9, 8) are the vertices of an isosceles right-angled triangle, and the area of this triangle is \( 50 \) square units.

---