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.
Working Rules to Draw Collinear Points:
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 adjoining 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\(\sqrt{5}\) = BCThus, AB + AC = BCHence, 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\(\sqrt{2}\)andAC = \(\sqrt{(4 – 1)^{2} + (2 + 1)^{2}}\) = \(\sqrt{3^{2} +
3^{2}}\) = \(\sqrt{9 + 9}\) = \(\sqrt{18}\) = 3\(\sqrt{2}\)⟹ BC + AC = 2\(\sqrt{2}\) + 3\(\sqrt{2}\) =
5\(\sqrt{2}\) = ABSo, 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 = \(\sqrt{(2 – 8)^{2} + (3 – 11)^{2}}\) = \(\sqrt{6^{2} +
(-8)^{2}}\) = \(\sqrt{36 + 64}\) = \(\sqrt{100}\) = 10BC = \(\sqrt{(8 – (-1))^{2} + (11 – (-1))^{2}}\) =
\(\sqrt{9^{2} + 12^{2}}\) = \(\sqrt{81 + 144}\) = \(\sqrt{225}\) = 15andCA = \(\sqrt{((-1) – 2)^{2} + ((-1) + 3)^{2}}\) = \(\sqrt{(-3)^{2}
+ (-4)^{2}}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5⟹ AB + CA = 10 + 5 = 15 = BCHence, the given points A, B, C are collinear.
● Distance and Section Formulae
10th Grade MathFrom Conditions of Collinearity of Three Points to HOME PAGE
Didn’t find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.
