Here we will learn about condition of collinearity of three points.
First Method:
Let us assume that the three noncoincident points A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are collinear. Then, one of these three points will divide the line segment joining the other two internally in a definite ratio. Suppose, the point B divides the line segment AC internally in the ratio λ : 1.
Hence, we have,
(λx₃ + 1 ∙ x₁)/(λ + 1) = x₂ …..(1)
and (λy₃ + 1 ∙ y₁)/(λ+1) = y₂ ..…(2)
From (1) we get,
λx₂ + x₂ = λx₃ + x₁
or, λ (x₂  x₃) = x₁  x₂
or, λ = (x₁  x₂)/(x₂  x₃)
Similarly, from (2) we get, λ = (y₁  y₂)/(y₂  y₃)
Therefore, (x₁  x₂)/(x₂  x₃) = (y₁ y₂)/(y₂  y₃)
or, (x₁  x ₂)(y₂  y₃) = (y₁  y₂) (x₂  x₃ )
or, x₁ (y₂  y₃) + x₂ y₃  y₁) + x₃ (y₁  y₂) = 0
which is the required condition of collinearity ofthe three given points.
Second Method:
Let A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃)be three noncoincident points and they are collinear. Since area of a triangle = ½ ∙ base × altitude, hence it is evident that the altitude of the triangle ABC is zero, when the points A, B, and C are collinear. Thus, the area of the triangle is zero if the points A, B and Care collinear. Therefore, the required condition of collinearity is
1/2 [x₁ (y₂  y₃) + x₂(y₃  y₁) + x₃ (y₁  y₂)] = 0
or, x₁ (y₂  y₃) + x₂ (y₃  y₁) + x₃ (y₁  y₂) = 0.
Examples on Condition of Collinearity of Three Points:
1. Show that the points (0, 2) , (2, 4) and (1, 5) are collinear.
Solution:
The area of the triangle formed by joining the given points
= 1/2 [(0  10 + 2)  (4 4 + 0)] = 1/2 (8 + 8) = 0.
Since the area of the triangle formed by joining the given points is zero, hence the given points are collinear. Proved
2. Show that the straight line joining the points (4, 3) and (8, 6) passes through the origin.
Solution:
The area of the triangle formed by joining the points (4, 3), (8, 6) and (0, 0) is 1/2 [24  24] = 0.
Since the area of the triangle formed by joining the points (4, 3), (8, 6) and (0, 0) is zero, hence the three points are collinear : therefore, the straight line joining the points (4, 3) and (8, 6)passes through the origin.
3. Find the condition that the points (a, b), (b, a) and (a², – b²) are in a straight line.
Solution:
Since the three given points are in a straight line, hence the area of the triangle formed by the points must be zero.
Therefore, 1/2  (a²  b³ + a²b) – (b² + a³  ab²)  = 0
or, a²  b³ + a²b – b² – a³ + ab² = 0
or, a² – b² – (a³ + b³) + ab (a + b) = 0
or, (a + b) [a  b  (a²  ab + b²) + ab] = 0
or, (a + b) [(a  b) (a²  ab + b²  ab)] = 0
or, (a + b) [(a  b)  (a  b)²] = 0
or, (a + b) (a  b) (1  a + b) = 0
Therefore, either a + b = 0 or, a – b = 0 or, 1  a + b = 0.
● Coordinate Geometry
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