Points Collinear Dc2313

1. **State the problem:** We need to prove that points A(1, 1), B(-2, 7), and C(3, -3) lie on the same straight line, i.e., they are collinear. 2. **Formula used:** The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Calculate distances:** - Distance $AB$: $$AB = \sqrt{(-2 - 1)^2 + (7 - 1)^2} = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$$ - Distance $BC$: $$BC = \sqrt{(3 + 2)^2 + (-3 - 7)^2} = \sqrt{5^2 + (-10)^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5}$$ - Distance $AC$: $$AC = \sqrt{(3 - 1)^2 + (-3 - 1)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$$ 4. **Check collinearity condition:** If points are collinear, the sum of the lengths of two segments equals the third: $$AB + AC = 3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5} = BC$$ 5. **Conclusion:** Since $AB + AC = BC$, points A, B, and C are collinear. **Final answer:** The points A(1, 1), B(-2, 7), and C(3, -3) are collinear because $AB + AC = BC$.