We will discuss here how to prove the conditions of collinearity of three points.
Definition of Collinear Points:
Three or more points in a plane are said to be collinear if they all he on the same line.
Step I: Draw a straight line 'ℓ'.
Step II: Mark points A, B, C, D, E on the straight line 'ℓ'.
Thus, we have drawn the collinear points A, B, C, D and E on the line 'ℓ'.
NOTE: If the points do not lie on the line, they are called non-collinear points.
Three points A, B and C are said to be collinear if they lie on the same straight line.
There points A, B and C will be collinear if AB + BC = AC as is clear from the above figure.
In general, three points A, B and C are collinear if the sum of the lengths of any two line segments among AB, BC and CA is equal to the length of the remaining line segment, that is,
either AB + BC = AC or AC + CB = AB or BA + AC = BC.
In other words,
There points A, B and C are collinear iff:
(i) AB + BC = AC i.e.,
Or, (ii) AB + AC = BC i.e. ,
Or, AC + BC = AB i.e.,
Solved examples to prove the collinearity of three points:
1. Prove that the points A (1, 1), B (-2, 7) and (3, -3) are collinear.
Solution:
Let A (1, 1), B (-2, 7) and C (3, -3) be the given points. Then,
AB = \(\sqrt{(-2 - 1)^{2} + (7 - 1)^{2}}\) = \(\sqrt{(-3)^{2} + 6^{2}}\) = \(\sqrt{9 + 36}\) = \(\sqrt{45}\) = 3\(\sqrt{5}\) units.
BC = \(\sqrt{(3 + 2)^{2} + (-3 - 7)^{2}}\) = \(\sqrt{5^{2} + (-10)^{2}}\) = \(\sqrt{25 + 100}\) = \(\sqrt{125}\) = 5\(\sqrt{5}\) units.
AC = \(\sqrt{(3 - 1)^{2} + (-3 - 1)^{2}}\) = \(\sqrt{2^{2} + (-4)^{2}}\) = \(\sqrt{4 + 16}\) = \(\sqrt{20}\) = 2\(\sqrt{5}\) units.
Therefore, AB + AC = 3\(\sqrt{5}\) + 2\(\sqrt{5}\) units = 5\(\sqrt{5}\) = BC
Thus, AB + AC = BC
Hence, the given points A, B, C are collinear.
2. Use the distance formula to show the points (1, -1), (6, 4) and (4, 2) are collinear.
Solution:
Let the points be A (1, -1), B (6, 4) and C (4, 2). Then,
AB = \(\sqrt{(6 - 1)^{2} + (4 + 1)^{2}}\) = \(\sqrt{5^{2} + 5^{2}}\) = \(\sqrt{25 + 25}\) = \(\sqrt{50}\) = 5\(\sqrt{2}\)
BC = \(\sqrt{(4 - 6)^{2} + (2 - 4)^{2}}\) = \(\sqrt{(-2)^{2} + (-2)^{2}}\) = \(\sqrt{4 + 4}\) = \(\sqrt{8}\) = 2\(\sqrt{2}\)
and
AC = \(\sqrt{(4 - 1)^{2} + (2 + 1)^{2}}\) = \(\sqrt{3^{2} + 3^{2}}\) = \(\sqrt{9 + 9}\) = \(\sqrt{18}\) = 3\(\sqrt{2}\)
⟹ BC + AC = 2\(\sqrt{2}\) + 3\(\sqrt{2}\) = 5\(\sqrt{2}\) = AB
So, the points A, B and C are collinear with C lying between A and B.
3. Use the distance formula to show the points (2, 3), (8, 11) and (-1, -1) are collinear.
Solution:
Let the points be A (2, 3), B (8, 11) and C (-1, -1). Then,
AB = \(\sqrt{(2 - 8)^{2} + (3 - 11)^{2}}\) = \(\sqrt{6^{2} + (-8)^{2}}\) = \(\sqrt{36 + 64}\) = \(\sqrt{100}\) = 10
BC = \(\sqrt{(8 - (-1))^{2} + (11 - (-1))^{2}}\) = \(\sqrt{9^{2} + 12^{2}}\) = \(\sqrt{81 + 144}\) = \(\sqrt{225}\) = 15
and
CA = \(\sqrt{((-1) - 2)^{2} + ((-1) + 3)^{2}}\) = \(\sqrt{(-3)^{2} + (-4)^{2}}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5
⟹ AB + CA = 10 + 5 = 15 = BC
Hence, the given points A, B, C are collinear.
● Distance and Section Formulae
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