Using distance formula, show that the points A(3, 1), B(6, 4) and C(8, 6) are collinear.

By distance formula,

d = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

$AB = \sqrt{(6 - 3)^2 + (4 - 1)^2} \\[1em] = \sqrt{3^2 + 3^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2}. \\[1em] BC = \sqrt{(8 - 6)^2 + (6 - 4)^2} \\[1em] = \sqrt{2^2 + 2^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \\[1em] = 2\sqrt{2}. \\[1em] AC = \sqrt{(8 - 3)^2 + (6 - 1)^2} \\[1em] = \sqrt{5^2 + 5^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}. \\[1em] \Rightarrow AB + BC = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2} = AC.$

Thus, AC = AB + BC.

Hence, proved that A, B and C are collinear.