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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 'β'.
Step II: Mark points A, B, C, D, E on the straight line 'β'.
Thus, we have drawn the collinear points A, B, C, D and E on the line 'β'.
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 = β(β2β1)2+(7β1)2 = β(β3)2+62 = β9+36 = β45 = 3β5 units.
BC = β(3+2)2+(β3β7)2 = β52+(β10)2 = β25+100 = β125 = 5β5 units.
AC = β(3β1)2+(β3β1)2 = β22+(β4)2 = β4+16 = β20 = 2β5 units.
Therefore, AB + AC = 3β5 + 2β5 units = 5β5 = 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 = β(6β1)2+(4+1)2 = β52+52 = β25+25 = β50 = 5β2
BC = β(4β6)2+(2β4)2 = β(β2)2+(β2)2 = β4+4 = β8 = 2β2
and
AC = β(4β1)2+(2+1)2 = β32+32 = β9+9 = β18 = 3β2
βΉ BC + AC = 2β2 + 3β2 = 5β2 = 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 = β(2β8)2+(3β11)2 = β62+(β8)2 = β36+64 = β100 = 10
BC = β(8β(β1))2+(11β(β1))2 = β92+122 = β81+144 = β225 = 15
and
CA = β((β1)β2)2+((β1)+3)2 = β(β3)2+(β4)2 = β9+16 = β25 = 5
βΉ AB + CA = 10 + 5 = 15 = BC
Hence, the given points A, B, C are collinear.
β Distance and Section Formulae
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