1. **State the problem:** We need to prove that points A(1,-3), B(-3,0), and C(4,1) form an isosceles right-angled triangle and then find its area. 2. **Formula and rules:** - Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ - A triangle is right-angled if the square of the longest side equals the sum of the squares of the other two sides (Pythagoras theorem). - A triangle is isosceles if at least two sides are equal. - Area of a triangle with base $b$ and height $h$ is $$\frac{1}{2}bh$$. 3. **Calculate side lengths:** - $AB = \sqrt{(-3 - 1)^2 + (0 + 3)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ - $BC = \sqrt{(4 + 3)^2 + (1 - 0)^2} = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$ - $AC = \sqrt{(4 - 1)^2 + (1 + 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ 4. **Check for isosceles:** - Sides $AB = 5$, $AC = 5$, so two sides are equal. The triangle is isosceles. 5. **Check for right angle:** - Longest side is $BC = 5\sqrt{2}$. - Check Pythagoras: $AB^2 + AC^2 = 5^2 + 5^2 = 25 + 25 = 50$ - $BC^2 = (5\sqrt{2})^2 = 25 \times 2 = 50$ - Since $AB^2 + AC^2 = BC^2$, the triangle is right-angled at point A. 6. **Find area:** - The legs forming the right angle are $AB$ and $AC$, both length 5. - Area = $\frac{1}{2} \times 5 \times 5 = \frac{25}{2} = 12.5$ **Final answer:** The points form an isosceles right-angled triangle with area $12.5$ square units.