Let the points (1, 5), (2, 3) and (- 2,-11) be representing the vertices A, B and C of a triangle respectively.
Let A = (1, 5), B = (2, 3) and C = (- 2,-11)
Distance between the points is given by
√(x1−x2)2+(y1−y2)2
∴AB=√(1−2)2+(5−3)2=√5
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BC=√(2−(−2))2+(3−(−11))2
=√42+142=√16+196=√212
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AC=√(1−(−2))2+(5−(−11))2
=√32+162=√9+256=√265
Since AB+BC≠AC
Therefore, the points (1, 5), (2, 3), and ( - 2, - 11) are not collinear.
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