The points lie on the same line are called collinear points. Let us consider A(x1, y1), B(x2, y2) and C(x3, y3) be three points lie on the line. Then,
Slope of AB = Slope of BC = Slope of CA
Problem 1 :
For what value of x are the points
A(-3, 12), B(7, 6) and C(x, 9)
collinear ?
a)1 b) -1 c) 2 d) -2
Solution :
Given points A(-3, 12), B(7, 6) and C(x, 9) are collinear.
Slope = (y2 – y1)/(x2 – x1)
A (x1, y1) = (-3, 12) B (x2, y2) = (7, 6) Slope of AB : = (6 - 12)/(7 + 3) = -6/10 = -3/5 --- (1) |
B (x1, y1) = (7, 6) C (x2, y2) = (x, 9) Slope of BC : = (9 - 6)/(x - 7) = 3/(x – 7) --- (2) |
(1) = (2)
-3/5 = 3/(x – 7)
-3(x – 7) = 5(-3)
-3x + 21 = -15
-3x = -15 – 21
-3x = -6
x = 2
So, the value of x is 2. Hence, option c) is correct.
Problem 2 :
For what value of y are the points
A(1, 4), B(3, y) and C(-3, 16)
collinear ?
a)1 b) -1 c) 2 d) -2
Solution :
Given points A(1, 4), B(3, y) and C(-3, 16) are collinear.
Slope = (y2 – y1)/(x2 – x1)
Slope of AB : = (y - 4)/(3 - 1) = (y – 4)/2 --- (1) |
Slope of BC : = (16 - y)/(-3 - 3) = (16 – y)/-6 --- (2) |
(1) = (2)
(y – 4)/2 = (16 – y)/-6
-6(y – 4) = 2(16 - y)
-6y + 24 = 32 – 2y
-6y + 2y = 32 – 24
-4y = 8
Dividing 4 on each sides.
y = -2
So, the value of y is -2. Hence, option d) is correct.
Problem 3 :
Find the value of p for which the points
A(-1, 3), B(2, p) and C(5, -1)
collinear ?
a)1 b) -1 c) 2 d) -2
Solution :
Given points A(-1, 3), B(2, p) and C(5, -1) are collinear.
Slope = (y2 – y1)/(x2 – x1)
Slope of AB : = (p - 3)/(2 + 1) = (p - 3)/3 --- (1) |
Slope of BC : = (-1 - p)/(5 - 2) = (-1 – p)/3 --- (2) |
(1) = (2)
(p - 3)/3 = (-1 – p)/3
3(p – 3) = 3(-1 - p)
3p - 9 = -3 – 3p
3p + 3p = -3 + 9
6p = 6
Dividing p on each sides.
p = 1
So, the value of p is 1. Hence, option a) is correct.
Problem 4 :
If the points
(1, x), (5, 2) and (9, 5)
are collinear then the value of x is
a) 5/2 b) -5/2 c) -1 d) 1
Solution :
Given points A(1, x), B(5, 2) and C(9, 5) are collinear.
Slope of AB : = (2 - x)/(5 - 1) = (2 – x)/4 --- (1) |
Slope of BC : = (5 - 2)/(9 - 5) = 3/4 --- (2) |
(1) = (2)
(2 – x)/4 = 3/4
4(2 – x) = 3(4)
8 – 4x = 12
-4x = 12 – 8
-4x = 4
x = -1
So, the value of x is -1. Hence, option c) is correct.
Problem 5 :
If the points (-1, 1), (2, p) and (8, 11) are collinear, find the value of p using section formula.
Solution :
Given points, (-1, 1), (2, p) and (8, 11) are collinear.
Let k : 1 be the ratio that divides the line segment joining the points (-1, 1) and (8, 11) at (2, p).
(lx2 + mx1)/(l + m), (ly2 + my1)/(l + m) = (2, p)
(x1, y1) = (-1, 1)
(x2, y2) = (8, 11)
(k(8) + 1(-1))/(k + 1), (k(11) + 1(-1))/(k + 1) = (2, p)
(8k - 1)/(k + 1) = 2 --- (1)
(11k - 1)/(k + 1) = p --- (2)
(8k - 1)/(k + 1) = 2
8k - 1 = 2(k + 1)
8k – 1 = 2k + 2
8k – 2k = 2 + 1
6k = 3
k = 3/6
k = 1/2
k = 1/2 substitute in equation (2).
(11(1/2) – 1)/(1/2 + 1) = p
(11/2 – 1)/(3/2) = p
(9/2)/(3/2) = p
9/3 = p
p = 3
So, the value of p is 3.
Problem 6 :
If the points (2, 3), (4, t) and (6, -3) are collinear, find the value of k using section formula.
Solution :
Given points, (2, 3), (4, t) and (6, -3) are collinear.
Let k : 1 be the ratio that divides the line segment joining the points (2, 3) and (4, k) at (6, -3).
(lx2 + mx1)/(l + m), (ly2 + my1)/(l + m) = (4, t)
(x1, y1) = (2, 3)
(x2, y2) = (6, -3)
(k(6) + 1(2))/(k + 1), (k(-3) + 1(3))/(k + 1) = (4, t)
(6k + 2)/(k + 1) = 4 --- (1)
(-3k + 3)/(k + 1) = t --- (2)
(6k + 2)/(k + 1) = 4
6k + 2 = 4(k + 1)
6k + 2 = 4k + 4
6k – 4k = 4 - 2
2k = 2
k = 1
So, the value of k is 1.
Applying the value of k in (2), we get
(-3 + 3)/(3 + 1) = t
t = 0
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM