Worksheet on Collinearity of Three Points with step-by-step solutions is available on this page. Thus the students who are interested to learn the concept of conditions of collinearity of three points and solving the problems can refer to this article. By practicing the problems from the given Collinearity of Three Points worksheet you can know different types of questions asked in the exams and know how to solve them.
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Collinearity of Three Points Worksheet with Answers
Example 1.
Find the value of p for which the points (p, -1), (4, 1), and (3, 5) are collinear.
Solution:
Let the given points be:
A(p, -1) = (x1, y1)
B(4, 1) = (x2, y2)
C(3, 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,
(1 + 1)/(4 – p) = (5 – 1)/(3 – 2)
2/(4 – p) = 4/1
2/(4 – p) = 4
4 – p = 8
p = 4 – 8
p = -4
Hence, the value of p is -4.
Example 2.
Using the equation method, check the collinearity of the points A(6,-4), B(1,1), and C(-1,6).
Solution:
We know that the equation of a line passing through the points (x1, y1) and (x2, y2) is:
y – y1 = [(y2 – y1) /(x2 – x1)] (x – x1)
Let A(6, -4) = (x1, y1) and B(1,1) = (x2, y2).
So, the equation of a line passing through the points A(6, -4) and B(1, 1) is given by:
y + 2 = [(1 + 4)/(1 – 6)] (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 the two of the given three points.
Therefore, the given points A, B and C are collinear.
Example 3.
Prove that the three points A(2,2), B(6,4), and C(10,6) are collinear.
Solution:
Given three points A(2,2),B(6,4), and C(10,6) are collinear, then the slopes of any two pairs of points will be equal.
That is,
Slope of AB = 6 – 2/4 – 2 = 4/2 = 2
Slope of BC=10 – 6/6 – 4 = 4/2 = 2
Slope of AC=10 -2/6 – 2 = 8/4 = 2
Thus, slope of AB = Slope of BC = Slope of AC
Hence, proved.
Example 4.
The points A(10,1),B(15,7) and C(a,3) are collinear. Find a.
Solution:
The given points A,B and C are collinear
We know that the slope of the line segment is slope = m = y2−y1/x2−x1
The slope of AB= slope of BC
7−1/15−10 = 3−7/a−15
6/5 = -4/a−15
6(a−15) = -4×5
6a−90 = -20
6a = -20+90
6a = 70
a=70/6
=70/6
Hence, the value of a is 70/6.
Example 5.
Using the equation method, check the collinearity of the points A(1,-4), B(1,2) and C(-4,6).
Solution:
We know that the equation of a line passing through the points (x1, y1) and (x2, y2) is:
y – y1 = [(y2 – y1) /(x2 – x1)] (x – x1)
Let A(1, -4) = (x1, y1) and B(1,2) = (x2,y2).
So, the equation of a line passing through the points A(1, -4) and B(1, 2) is given by:
y + 2 = [(2 + 4)/(1 – 1)] (x – 7)
y + 2 = (6/0) (x – 7)
y + 2 = 6x + 42
-6x + y + 2 – 42 = 0
-6x + y – 40 = 0
Now, substituting the point C(-4, 6) in the above equation,
-6(-4) + 6 – 5 = 0
-24+1 = 0
-23 = 0
Thus, the third point does not satisfy the equation of the line passing through two of the given three points.
Therefore, the given points A, B, and C are not collinear.
Example 6.
Prove that the three points A(4,2), B(5,4), and C(7,6) are collinear.
Solution:
Given three points A(4,2),B(5,4), and C(7,6) are collinear, then the slopes of any two pairs of points will be equal.
That is,
Slope formula = y2 – y1/x2 – x1
Slope of AB = 4-2/ 5-4 = 2
Slope of BC= 6-4/ 7-5 7 = 1
Slope of AC= 6-2/ 7-4 = 4/3
Thus, slope of AB ≠ Slope of BC ≠ Slope of AC
Hence, the given points are not collinear.
Example 7.
The points A(2,3), B(4,5) and C(a,6) are collinear. Find a.
Solution:
The given points A,B and C are collinear
We know that the slope of the line segment is slope = m = y2 – y1/x2−x1
The slope of AB= slope of BC
5-3/ 4-2= 6 – 5/a – 4
2/2 = 1/a-4
1 = 1/a – 4
a – 4 = 1
a = 4 – 1
a = -3
Hence, the value of a is -3.
Example 8.
Show that the points (1,-1), (5,2) and (9,5) are collinear
Solution:
Given that the points are
A(1,-1)
B(5,2)
C(9,5)
We know that the distance formula is√(x2−x1)²+(y2 – y1)²
Distance of AB = √(5-1)² + (2+1)² = √16 + 9 = √25 = 5
Distance of BC = √(5-9)² + (2-5)² = √16+9 = √25 = 5
Distance of AC = √(1-9)² + (-1-5)² = √64 + 36 = √100 = 10
From these distance od AB+ distance of BC = distance of AC
So the given points are collinear.
Example 9.
Show that the given points (1,2), (3,2), and (6,2) are collinear or not.
Solution:
Given that the points are
A(1,-1)
B(5,2)
C(9,5)
We know that the distance formula is √(x2−x1)²+(y2 – y1)²
Distance of AB = √(3-1)² + (2-2)² = √4 = 2
Distance of BC = √(6-3)² + (2-2)² = √9 = 3
Distance of AC = √(6-1)² + (2-2)² = √25 = 5
From these distance od AB+ distance of BC = distance of AC
So the given points are collinear.
Example 10.
Find the value of p for which the points (k, -2), (3, 4), and (4, 5) are collinear.
Solution:
Let the given points be:
A(k, -2) = (x1, y1)
B(3, 4) = (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,
(4 + 2)/(3 – p) = (5 – 3)/(4 – 4)
6/(3 – p) = 2
6/(3 – p) = 2
3 – p = 12
-p = 12 – 3
p = -9
Hence, the value of p is -9.