The points which lie on the same line are known as collinear points.
In case we have three point and we need to prove that the given points are collinear, we may follow the steps given below.
Step 1 :
Choose the points A and B and find the slope.
Step 2 :
Choose the points B and C and find the slope.
Step 3 :
If the slopes found in step 1 and step 2 are equal, then the points A, B and C are collinear.
Formula to find the slope of a line joining the two points (x1, y1) and (x2, y2) :
m = (y2 - y1) / (x2 - x1)
Example 1 :
Using the concept of slope, prove that each of the following set of points are collinear.
(i) (2 , 3), (3 , -1) and (4 , -5)
(ii) (4 , 1), (-2 , -3) and (-5 , -5)
(iii) (4 , 4), (-2 , 6) and (1 , 5)
Solution :
(i) Let the given points be A (2 , 3) B (3 , -1) and C (4 , -5)
If the points A, B and C are collinear then,
Slope of AB = Slope of BC
Slope of AB : m = (y2 - y1)/(x2 - x1) A (2 , 3) B (3 , -1) m = (-1 - 3)/(3 - 2) m = -4/1 m = -4 ---(1) |
Slope of BC : m = (y2 - y1)/(x2 - x1) B (3 , -1) C (4 , -5) m = (-5 + 1)/(4 - 3) m = -4/1 m = -4 ---(2) |
(1) = (2)
Hence the given points A, B and C are collinear.
(ii) (4 , 1), (-2 , -3) and (-5 , -5)
Solution :
Let the given points be A(4 , 1) B(-2 , -3) and C(-5 , -5)
If the points A, B and C are collinear then,
Slope of AB = Slope of BC
Slope of AB : m = (y2 - y1)/(x2 - x1) A (4, 1) B (-2, -3) m = (-3 - 1)/(-2 - 4) m = -4/(-6) m = 2/3 ---(1) |
Slope of BC : m = (y2 - y1)/(x2 - x1) B (-2, -3) C (-5, -5) m = (-5 + 3)/(-5 + 2) m = -2/(-3) m = 2/3 ---(2) |
(1) = (2)
Hence the given points are collinear.
(iii) (4 , 4), (-2 , 6) and (1 , 5)
Solution :
Let the given points be A(4 , 4) B(-2 , 6) and C(1 , 5)
If the points A, B and C are collinear then,
Slope of AB = Slope of BC
Slope of AB : m = (y2 - y1)/(x2 - x1) A (4, 4) B (-2, 6) m = (6 - 4)/(-2 - 4) m = 2/(-6) m = -1/3 ---(1) |
Slope of BC : m = (y2 - y1)/(x2 - x1) B (-2, 6) C (1, 5) m = (5 - 6)/(1 + 2) m = -1/3 ---(2) |
(1) = (2)
So, the given points are collinear.
Example 2 :
If the points (a, 1), (1, 2) and (0, b+1) are collinear, then prove that (1/a) + (1/b) = 1
Solution :
Let the given points be A(a, 1) B(1, 2) and C(0, b+1). Since the given points are collinear, slope of AB is equal to slope of BC.
Slope of AB : m = (y2 - y1)/(x2 - x1) A (a, 1) B (1, 2) m = (2 - 1)/(1 - a) m = 1/(1 - a) ---(1) |
Slope of BC : m = (y2 - y1)/(x2 - x1) B (1, 2) C (0, b + 1) m = (b + 1 - 2)/(0 - 1) m = (b - 1)/(-1) ---(2) |
1/(1 - a) = -(b - 1)
1 = (1 - b)(1 - a)
1 = 1 - a - b + ab
a + b = ab + 1 - 1
a + b = ab
Dividing by ab, we get
(a/ab) + (b/ab) = (ab/ab)
1/b + 1/a = 1
(1/a) + (1/b) = 1
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Apr 26, 24 12:39 PM
Apr 26, 24 01:51 AM
Apr 25, 24 08:40 PM