Three points are said to be collinear, if they are on the same straight line.
We can prove the collinearity of three points using one of the four concepts given below.
1. Slope of a line
2. Distance between the two points
3. Area of triangle
4. Equation of a line
1. Using slope of a line :
Let A, B and C be the given three points.
Find the slope of AB and BC.
(i) Slope of AB and slope of BC must be equal.
(ii) There must be a common point in AB and BC.
If the above two conditions are met, then the given three points are collinear.
2. Using distance between two points :
Let A, B and C be the given three points.
We have to find the three lengths AB, BC and AC among the given three points A, B and C.
The three points A, B and C are collinear, if the sum of the lengths of any two line segments among AB, BC and AC is equal to the length of the remaining line segment.
AB + BC = AC
(or)
AB + AC = BC
(or)
AC + BC = AB
3. Using area of triangle :
A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) be the given three points.
If the three points, A, B and C are collinear, they will lie on the same and they cannot form a triangle. then the area of triangle is equal to zero.
(½){(x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1}) - (x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3})} = 0
Multiplying both sides by 2,
(x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1}) - (x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3}) = 0
x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1} = x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3}
4. Using equation of a line :
A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) be the given three points.
Find the equation of the line joining the points A and B. Substitute the third point C into the equation of the line AB. If the point C satisfies the equation of line AB, then the point C lies on the line AB. Then, the three points A, B and C are collinear.
Example 1 :
Using the concept of area of triangle, show that the points (5, -2), (4, -1) and (1, 2) are collinear.
Solution :
Using the concept of area of triangle, if the three points (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are collinear, then
x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1} = x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3}
Here,
(x_{1}, y_{1}) = (5, -2)
(x_{2}, y_{2}) = (4, -1)
(x_{3}, y_{3}) = (1, 2)
x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1 }= 5(-1) + 4(2) + 1(-2)
= -5 + 8 -2
= 1 ----(1)
x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3} = 4(-2) + 1(-1) + 5(2)
= -8 - 1 + 10
= 1 ----(2)
From (1) and (2),
x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1} = x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3}
Hence, the three points A, B and C are collinear.
Example 2 :
Using the concept of slope of a line, show that the following points are collinear.
(5, -2), (4, -1), (1, 2)
Solution :
Let A(5, -2), B(4, -1) and C(1, 2).
Slope of the line joining (x_{1}, y_{1}) and (x_{2}, y_{2}) is,
Slope of the line joining the points A(5, - 2) and B(4, -1) substitute (x_{1}, y_{1}) = (5, -2) and (x_{2}, y_{2}) = (4, -1) into the above formula.
= ⁽⁻¹ ⁺ ²⁾⁄₍₄ ₋ ₅₎
= -1
Slope of the line joining the points B(4, -1) and C(1, 2) substitute (x_{1}, y_{1}) = (4, -1) and (x_{2}, y_{2}) = (1, 2) into the above formula.
= ⁽² ⁺ ¹⁾⁄₍₁ ₋ ₄₎
= -1
slope of AB = slope of BC
And also, B is the common point.
Hence, the points A, B and C are collinear.
Example 3 :
Using the concept of distance between two points, show that the points (5, -2), (4, -1) and (1, 2) are collinear.
Solution :
Let A(5, -2), B(4, -1) and C(1, 2).
Distance between two points (x_{1}, y_{1}) and (x_{2}, y_{2}) :
Distance between the two points A(5, - 2) and B(4, -1) :
AB = √[(4 - 5)^{2} + (-1 + 2)^{2}]
= √[(-1)^{2} + (1)^{2}]
= √[1 + 1]
= √2
Distance between B and C :
BC = √[(1 - 4)^{2} + (2 + 1)^{2}]
= √[(-3)^{2} + (3)^{2}]
= √[9 + 9]
= √18
= 3√2
Distance between A and C :
AC = √[(1 - 5)^{2} + (2 + 2)^{2}]
= √[(-4)^{2} + (4)^{2}]
= √[16 + 16]
= √[32]
= 4√2
AB + BC = √2 + 3√2
= 4√2
= AC
AB + BC = √2 + 3√2
AB + BC = 4√2
AB + BC = AC
Hence, the three points A, B and C are collinear.
Example 4 :
Using the concept of equation of line, show that the points (5, -2), (4, -1) and (1, 2) are collinear.
Solution :
Equation of the straight line in two-points form :
Find the equation of the line through the points
(5, -2) and (4, -1)
Substitute (x_{1}, y_{1}) = (5, -2) and (x_{2}, y_{2}) = (4, -1).
⁽ʸ ⁺ ²⁾⁄₍₋₁ ₊ ₂₎ = ⁽ˣ ⁻ ⁵⁾⁄₍₄ ₋ ₅₎
⁽ʸ ⁺ ²⁾⁄₁ = ⁽ˣ ⁻ ⁵⁾⁄₍₋₁₎
y + 2 = -x + 5
y = -x + 3
Substitute the third point (1, 2).
2 = -1 + 3
2 = 2
1 + 2 - 3 = 0
0 = 0
Therefore, the third point (1, 2) satisfies the equation.
Hence, the given three points are collinear.
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