Collinearity of Three Points

What is the condition of collinearity of three points?

We will find the condition of collinearity of three given points by using the concept of slope.

Let P(x\(_{1}\), y\(_{1}\)) , Q (x\(_{2}\), y\(_{2}\)) and R (x\(_{3}\), y\(_{3}\)) are three given points. If the points P, Q and R are collinearity then we must have,

Slop of the line PQ = slop of the line PR

Therefore, \(\frac{y_{1} - y_{2}}{x_{1} - x_{2}}\) = \(\frac{y_{1} - y_{3}}{x_{1} - x_{3}}\)

⇒ (y\(_{1}\) - y\(_{2}\)) (x\(_{1}\) - x\(_{3}\)) = (y\(_{1}\) - y\(_{3}\)) (x\(_{1}\) - x\(_{3}\))

⇒ x\(_{1}\) (y\(_{2}\)  - y\(_{3}\)) + x\(_{2}\) (y\(_{3}\) - y\(_{1}\)) + x\(_{3}\) (y\(_{1}\) - y\(_{2}\)) = 0

Which is the required condition of collinearity of the points P, Q and R.

Solved examples using the concept of slope to find the condition of collinearity of three given points:

1. Using the method of slope, show that the points P(4, 8), Q (5, 12) and R (9, 28) are collinear.

Solution:

The given three points are P(4, 8), Q (5, 12) and R (9, 28).

If the points P, Q and R are collinear then we must have,

x\(_{1}\) (y\(_{2}\)  - y\(_{3}\)) + x\(_{2}\) (y\(_{3}\) - y\(_{1}\)) + x\(_{3}\) (y\(_{1}\) - y\(_{2}\)) = 0, where x\(_{1}\) = 4, y\(_{1}\) = 8, x\(_{2}\) = 5, y\(_{2}\) = 12, x\(_{3}\) = 9 and y\(_{3}\) = 28

Now, x\(_{1}\) (y\(_{2}\)  - y\(_{3}\)) + x\(_{2}\) (y\(_{3}\) - y\(_{1}\)) + x\(_{3}\) (y\(_{1}\) - y\(_{2}\))

= 4(12 - 28) + 5(28 - 8) + 9(8 - 12)

= 4(-16) + 5(20) + 9(-4)

= -64 + 100 - 36

= 0

Therefore, the given three points P(4, 8), Q (5, 12) and R (9, 28) are collinear.

2. Using the method of slope, show that the points A (1, -1), B (5, 5) and C (-3, -7) are collinear.

Solution:

The given three points are A (1, -1), B (5, 5) and C (-3, -7).

If the points A, B and C are collinear then we must have,

x\(_{1}\) (y\(_{2}\)  - y\(_{3}\)) + x\(_{2}\) (y\(_{3}\) - y\(_{1}\)) + x\(_{3}\) (y\(_{1}\) - y\(_{2}\)) = 0, where x\(_{1}\) = 1, y\(_{1}\) = -1, x\(_{2}\) = 5, y\(_{2}\) = 5, x\(_{3}\) = -3 and y\(_{3}\) = -7

Now, x\(_{1}\) (y\(_{2}\)  - y\(_{3}\)) + x\(_{2}\) (y\(_{3}\) - y\(_{1}\)) + x\(_{3}\) (y\(_{1}\) - y\(_{2}\))

= 1{5 - (-7)} + 5{(-7) - (-1)} + (-3){(-1) - 5)}

= 1(5 + 7) + 5(-7 + 1) - 3(-1 - 5)

= 1(12) + 5(-6) - 3(-6)

= 12 - 30 + 18

= 0

Therefore, the given three points A (1, -1), B (5, 5) and C (-3, -7) are collinear.

● The Straight Line

• Straight Line
• Slope of a Straight Line
• Slope of a Line through Two Given Points
• Collinearity of Three Points
• Equation of a Line Parallel to x-axis
• Equation of a Line Parallel to y-axis
• Slope-intercept Form
• Point-slope Form
• Straight line in Two-point Form
• Straight Line in Intercept Form
• Straight Line in Normal Form
• General Form into Slope-intercept Form
• General Form into Intercept Form
• General Form into Normal Form
  
• Point of Intersection of Two Lines
  
• Concurrency of Three Lines
• Angle between Two Straight Lines
• Condition of Parallelism of Lines
• Equation of a Line Parallel to a Line
• Condition of Perpendicularity of Two Lines
• Equation of a Line Perpendicular to a Line
• Identical Straight Lines
• Position of a Point Relative to a Line
• Distance of a Point from a Straight Line
• Equations of the Bisectors of the Angles between Two Straight Lines
• Bisector of the Angle which Contains the Origin
• Straight Line Formulae
• Problems on Straight Lines
• Word Problems on Straight Lines
• Problems on Slope and Intercept

11 and 12 Grade Math

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