Using the distance formula,show that the points $(0, -3), (1, -1),$ and $(2, 1)$ are collinear.

(N/A) Let $A(0, -3), B(1, -1),$ and $C(2, 1)$ be the given points.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(1 - 0)^2 + (-1 - (-3))^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$
$BC = \sqrt{(2 - 1)^2 + (1 - (-1))^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$
$AC = \sqrt{(2 - 0)^2 + (1 - (-3))^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$
Since $AB + BC = \sqrt{5} + \sqrt{5} = 2\sqrt{5} = AC$,the sum of the lengths of two segments is equal to the length of the third segment.
Therefore,the points $A, B,$ and $C$ are collinear.