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.
11 and 12 Grade Math
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