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Answer :

Let A = (1, 5), B = (2, 3) and C = (–2, –11)

Using Distance Formula to find distance AB, BC and CA.

AB = \(\sqrt{(2-1)^2 + (3 - 5)^2} \)

\( = \sqrt{1^2 - (-2)^2} \)

\( = \sqrt{1 + 5} \)

\( = \sqrt{5} \)

BC = \(\sqrt{(-2-2)^2 + (-11 - 3)^2} \)

\( = \sqrt{(-4)^2 - (-14)^2} \)

\( = \sqrt{16 + 196} \)

\( = \sqrt{212}\)

CA = \(\sqrt{(-2-1)^2 + (-11 - 5)^2} \)

\( = \sqrt{(-3)^2 - (-15)^2} \)

\( = \sqrt{9 + 256} \)

\( = \sqrt{265}\)

Since AB + AC \(\ne\) BC, BC + AC \(\ne\) AB and AC \(\ne\) BC.

Therefore, the points A, B and C are not collinear.

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