Using the distance formula,show that $(4, 3)$,$(5, 1)$,and $(1, 9)$ are collinear.

(N/A) Let the points be $A(4, 3)$,$B(5, 1)$,and $C(1, 9)$.
Three points are collinear if the sum of the lengths of any two segments is equal to the length of the third segment.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$1$. Distance $AB = \sqrt{(5 - 4)^2 + (1 - 3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.
$2$. Distance $BC = \sqrt{(1 - 5)^2 + (9 - 1)^2} = \sqrt{(-4)^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$.
$3$. Distance $AC = \sqrt{(1 - 4)^2 + (9 - 3)^2} = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$.
Since $AB + AC = \sqrt{5} + 3\sqrt{5} = 4\sqrt{5} = BC$,the points $A$,$B$,and $C$ are collinear.