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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