Using the distance formula,show that the points $A(-1, 4)$,$B(2, 3)$,and $C(5, 2)$ are collinear.

(N/A) To show that points $A(-1, 4)$,$B(2, 3)$,and $C(5, 2)$ are collinear,we must show that the sum of the lengths of two segments equals the length of the third segment.
$1$. Calculate the distance $AB$ using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(2 - (-1))^2 + (3 - 4)^2} = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
$2$. Calculate the distance $BC$:
$BC = \sqrt{(5 - 2)^2 + (2 - 3)^2} = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
$3$. Calculate the distance $AC$:
$AC = \sqrt{(5 - (-1))^2 + (2 - 4)^2} = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$.
$4$. Since $AB + BC = \sqrt{10} + \sqrt{10} = 2\sqrt{10} = AC$,the points $A$,$B$,and $C$ lie on the same line and are therefore collinear.