To prove that the points A(1, -3), B(-3, 0), and C(4, 1) are the vertices of an isosceles right triangle and to find the area of the triangle, we will follow these steps:
### Step 1: Calculate the lengths of the sides of the triangle using the distance formula.
The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
#### Length of AB:
Let’s calculate the length of side AB:
- A(1, -3) and B(-3, 0)
\[
AB = \sqrt{((-3) - 1)^2 + (0 - (-3))^2} = \sqrt{(-4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]
#### Length of BC:
Now, let’s calculate the length of side BC:
- B(-3, 0) and C(4, 1)
\[
BC = \sqrt{(4 - (-3))^2 + (1 - 0)^2} = \sqrt{(4 + 3)^2 + (1)^2} = \sqrt{(7)^2 + (1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}
\]
#### Length of AC:
Finally, let’s calculate the length of side AC:
- A(1, -3) and C(4, 1)
\[
AC = \sqrt{(4 - 1)^2 + (1 - (-3))^2} = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
### Step 2: Verify 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: Verify if the triangle is a right triangle using the Pythagorean theorem.
For triangle ABC to be a right triangle, it must satisfy the Pythagorean theorem:
\[
BC^2 = AB^2 + AC^2
\]
Calculating:
\[
(5\sqrt{2})^2 = 5^2 + 5^2
\]
\[
50 = 25 + 25
\]
\[
50 = 50
\]
Since this holds true, triangle ABC is a right triangle.
### Step 4: Calculate the area of triangle ABC.
The area \(A\) of a triangle can be calculated using the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
We can take AB as the base and AC as the height:
\[
A = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 5 \times 5 = \frac{25}{2} = 12.5 \text{ square units}
\]
### Conclusion:
Thus, we have proved that the points A(1, -3), B(-3, 0), and C(4, 1) form an isosceles right triangle, and the area of the triangle is \(12.5\) square units.
