If A(-1, 4), B(2, 3) and C(8, 1),then which o | vedclass

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(A) To check if the points $A(-1, 4), B(2, 3),$ and $C(8, 1)$ are collinear,we calculate the slopes of the segments $AB$ and $BC$.Slope of $AB = \frac

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$A-B-C$

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(A) To check if the points $A(-1, 4), B(2, 3),$ and $C(8, 1)$ are collinear,we calculate the slopes of the segments $AB$ and $BC$.Slope of $AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 4}{2 - (-1)} = \frac{-1}{3}$.Slope of $BC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 3}{8 - 2} = \frac{-2}{6} = \frac{-1}{3}$.Since the slope of $AB$ is equal to the slope of $BC$,the points $A, B,$ and $C$ are collinear.Now,we check the distances between the points:$AB = \sqrt{(2 - (-1))^2 + (3 - 4)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.$BC = \sqrt{(8 - 2)^2 + (1 - 3)^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$.$AC = \sqrt{(8 - (-1))^2 + (1 - 4)^2} = \sqrt{9^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}$.Since $AB + BC = \sqrt{10} + 2\sqrt{10} = 3\sqrt{10} = AC$,the point $B$ lies between $A$ and $C$.Therefore,the order of the points is $A-B-C$.

If $A(-1, 4), B(2, 3)$ and $C(8, 1)$,then which of the following is true regarding the collinearity of points $A, B,$ and $C$?

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