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Using distance formula, prove that the points A(2, 3), B(-7, 0) and C(-1, 2) are collinear.
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Sol. $AB = \sqrt{(-7 -2)^2 + (0 - 3)^2} = \sqrt{90} = 3\sqrt{10}$ (I) (1/2)
$BC = \sqrt{(-1 + 7)^2 + (2 - 0)^2} = \sqrt{40} = 2\sqrt{10}$ (II) (1/2)
$AC = \sqrt{(-1-2)^2 + (2 - 3)^2} = \sqrt{10} = \sqrt{10}$ (III) (1/2)
AC + BC = AB (IV) (1/2)
$\therefore$ A, B, C are collinear.
$BC = \sqrt{(-1 + 7)^2 + (2 - 0)^2} = \sqrt{40} = 2\sqrt{10}$ (II) (1/2)
$AC = \sqrt{(-1-2)^2 + (2 - 3)^2} = \sqrt{10} = \sqrt{10}$ (III) (1/2)
AC + BC = AB (IV) (1/2)
$\therefore$ A, B, C are collinear.