Consider, A = (1, 5) B = (2, 3) and C = (-2, -11)

Find the distance between points; say AB, BC and CA

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

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

= \(\sqrt{1+4}\)

= \(\sqrt{5}\)

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

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

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

= \(\sqrt{212}\)

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

= \(\sqrt{(-3)^2 + (-16)^2}\)

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

= \(\sqrt{265}\)

Since AB + BC ≠ CA

Therefore, the points (1, 5), (2, 3), and (-2, -11) are not collinear.