Prove that the points A(1, -3), B (-3, 0) and C(4, 1) are the vertices of an isosceles right- angled triangle. Find the area of the triangle.

Answer

To prove that the points A(1, -3), B(-3, 0), and C(4, 1) are the vertices of an isosceles right-angled triangle and to find the area of the triangle, we will follow these steps:

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

**Distance Formula:**
The distance \( d \) between two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

#### 1.1: Calculate AB
Coordinates of A = (1, -3) and B = (-3, 0)

Using the distance formula:
\[ AB = \sqrt{((-3) - 1)^2 + (0 - (-3))^2} \]
\[ = \sqrt{(-4)^2 + (3)^2} \]
\[ = \sqrt{16 + 9} \]
\[ = \sqrt{25} = 5 \]

#### 1.2: Calculate BC
Coordinates of B = (-3, 0) and C = (4, 1)

Using the distance formula:
\[ BC = \sqrt{(4 - (-3))^2 + (1 - 0)^2} \]
\[ = \sqrt{(4 + 3)^2 + (1)^2} \]
\[ = \sqrt{(7)^2 + (1)^2} \]
\[ = \sqrt{49 + 1} \]
\[ = \sqrt{50} = 5\sqrt{2} \]

#### 1.3: Calculate AC
Coordinates of A = (1, -3) and C = (4, 1)

Using the distance formula:
\[ AC = \sqrt{(4 - 1)^2 + (1 - (-3))^2} \]
\[ = \sqrt{(3)^2 + (4)^2} \]
\[ = \sqrt{9 + 16} \]
\[ = \sqrt{25} = 5 \]

### Step 2: Determine if the triangle is isosceles
From the calculations:
- \( AB = 5 \)
- \( BC = 5\sqrt{2} \)
- \( AC = 5 \)

Since \( AB = AC \), triangle ABC is isosceles.

### Step 3: Determine if the triangle is right-angled
To check if triangle ABC is right-angled, we can use the Pythagorean theorem:
\[ AB^2 + AC^2 = BC^2 \]

Calculating:
- \( AB^2 = 5^2 = 25 \)
- \( AC^2 = 5^2 = 25 \)
- \( BC^2 = (5\sqrt{2})^2 = 50 \)

Now, check:
\[ AB^2 + AC^2 = 25 + 25 = 50 = BC^2 \]

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

### Step 4: Calculate the area of triangle ABC
The area \( A \) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, we can take \( AC \) as the base and \( AB \) as the height:
\[ A = \frac{1}{2} \times AC \times AB \]
\[ = \frac{1}{2} \times 5 \times 5 \]
\[ = \frac{25}{2} = 12.5 \]

### Final Result
The points A(1, -3), B(-3, 0), and C(4, 1) are the vertices of an isosceles right-angled triangle, and the area of the triangle is \( 12.5 \) square units.

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