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 are equal to the length of the remaining line segment. We hope the information provided in this article will be helpful for all the students to understand the conditions of collinearity of three points. To know more about the concept of Distance and Section Formulae click on the given links and start preparing for the exams.
Also Read:
- Worksheet on Collinearity of Three Points
- Points and Line Segment
Conditions of Collinearity of Three Points
We are going to discuss the conditions of collinearity of three points in this section
1. If all the three points are collinear, then they must lie on the same line.
2. We can prove the collinearity of three points using one of the below concepts.
i. Concept of the distance between the two points.
ii. Concept of slope
iii. Concept of the equation of the line
iv. Concept of the area of a triangle.
Concept of Slope
Let A, B, and C be the three points.
If we want A, B, and C to be collinear, the following conditions have to be met.
(i) Slope of AB = Slope of BC
(ii) There must be a common point between AB and BC.
Then we can say that the three points A, B, and C are collinear.
Concept of the distance between the two points
Let A, B, and C be the three points.
We have to find the three lengths AB, BC, and AC among the given three points A, B, and C.
AB + BC = AC
or
AB + AC = BC
or
AC + BC = AB
Concept of the area of a triangle
Let A(x₁, y₁), B(x₂, y₂), and C (x₃, y₃) be the three points.
If the three points, A, B, and C are collinear, they will lie on the same and they cannot form a triangle.
Hence, the area of triangle ABC = 0
How do you Prove the Collinearity of Three Points?
Consider the points A(x1, y1), B(x2, y2), and C(x3, y3) are collinear, then the conditions of collinearity of three points are as follows,
i. Slope of AB = Slope of BC
ii. AB + BC = AC
or
AB + AC = BC
or
AC + BC = AB
Then prove using the distance formula.
iii. Area of triangle ABC = 0
iv. If the third point fulfills the equation using any two of the given three points, then the three points A, B, and C will be collinear.
Collinearity of Three Points Examples
Example 1.
Find the value of p for which the points (p, -1), (1,2), and (4, 5) are collinear.
Solution:
Let the given points be:
A(p, -1) = (x1, y1)
B(1,2) = (x2, y2)
C(4, 5) = (x3, y3)
Given that A, B, and C are collinear.
Slope of AB = Slope of BC
(y2 – y1)/(x2 – x1) = (y3 – y2)/(x3 – x2)
Substituting the values of coordinates of given points,
(2 + 1)/(1 – p) = (5 – 2)/(4 – 1)
3/(1 – p) = 3/3
3/(1 – p) = 1
3 = 1 – p
3 – 1 = p
p = 2
Hence, the value of p is 2
Example 2.
Using the equation method, check the collinearity of the points A(7, -2), B(2, 3), and C(-1, 6).
Solution:
We know that the equation of a line passing through the points (x1, y1) and (x2, y2) is:
y2– y1 = [(y2 – y1) /(x2 – x1)] (x – x1)
Let A(7, -2) = (x1, y1) and B(2, 3) = (x2, y2).
So, the equation of a line passing through the points A(7, -2) and B(2, 3) is given by:
y + 2 = [(3 + 2)/(2 – 7)] (x – 7)
y + 2 = (5/-5) (x – 7)
y + 2 = -x + 7
x + y + 2 – 7 = 0
x + y – 5 = 0
Now, substituting the point C(-1, 6) in the above equation,
-1 + 6 – 5 = 0
0 = 0
Thus, the third point satisfies the equation of the line passing through two of the given three points.
Therefore, the given points A, B, and C are collinear.
Example 3.
Use the distance formula to show the points (6,1), (3, 4), and (5, 4) are collinear.
Solution:
Let the points be A (6, 1), B (3, 4) and C (5, 4). Then,
AB =√(3-6)²+(4-1)²
= √3²+3²
= √9+9
= √18
BC =√(5−3)²+(4−4)²
=√ (2)²
= √4
and
AC = √(5-6)²+(4-1)²
=√1² + 3²
=√1+9
=√10
BC + AC = √4 + √10 = √18= AB
So, the points A, B and C are not collinear with C lying between A and B.
Example 4.
Prove that points A (2, -9), B (1, -6) and (-3, 6) are collinear.
Solution:
Let A (2, -9), B (1, -6) and C (-3, 6) be the given points. Then,
AB = √(2-1)² + (-9+6)²
AB = √1 + 9
= √10
BC =√(2+3)²+(-9−6)²
=√(5)²+(-15)²
= √25+225
= √45
= 3√5 units.
AC = √(1+3)²+(-6-6)²
= √5²+(−10)²
= √16+144
= √160
Thus, AB + AC = BC
√10 + √160 = √250
1√10 + √16.10 = √25.10
1√10 + 4√10 = 5√10
5√10 = 5√10
Hence, the given points A, B, C are collinear.
Example 5.
Use the distance formula to show the points (1, -1), (6, 4), and (4, 2) are collinear.
Solution:
Let the points be A (1, -1), B (6, 4) and C (4, 2). Then,
AB = √(6−1)²+(4+1)²
= √5²+5²
= √25+25
= √50
= 5√2
BC = √(4−6)²+(2−4)²
= √(−2)²+(−2)²
= √4+4
= √8
= 2√2
and
AC = √(4−1)²+(2+1)²
= √3²+3²
= √9+9
= √18
= √32
BC + AC = 2√2+ 3√2 = 5√2 = AB
So, the points A, B and C are collinear with C lying between A and B.
FAQs on Collinearity of Three Points Conditions
1. What is the condition for collinearity of 3 points?
3 points A, B, and C are collinear if the sum of the lengths of any two line segments among AB, BC, and CA is equal to the length of the remaining line segment, i.e., either AB + BC = AC or AC +CB = AB or BA + AC = BC.
2. Which figure is formed by three noncollinear points?
A triangle is a figure formed by three segments joining three noncollinear points.
3. What is the formula of collinear points?
If the points (x1, y1), (x2, y2) and (x3, y3) are collinear,
x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0