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





11 and 12 Grade Math

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