Are the three points A(2, 3), B(5, 6) and C(0, 2) collinear?

Engineering

Mathematics

Introduction to Determinants

Question

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Three points A, B and C are collinear if AB + BC = AC.

Therefore, we calculate:

AB = \sqrt{(5 - 2)^{2} + (6 - 3)^{2}} = \sqrt{3^{2} + 3^{2}} = \sqrt{9 + 9} = \sqrt{18}

BC = \sqrt{(0 - 5)^{2} + (2 - 6)^{2}} = \sqrt{(- 5)^{2} + (- 4)^{2}} = \sqrt{25 + 16} = \sqrt{41}

AC = \sqrt{(0 - 2)^{2} + (2 - 3)^{2}} = \sqrt{(- 2)^{2} + (- 1)^{2}} = \sqrt{4 + 1} = \sqrt{5}

Since,

\sqrt{18} + \sqrt{41} \neq \sqrt{5}

that is AB + BC ≠ AC.

Hence, the given points are not collinear.

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