1) If the given points are collinear, then the area of the triangle created by those three points will be equal to 0.
2) The slope of the line joining any of the two points will be equal.
Problem 1 :
Find the values of k if the points A (k + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k) are collinear.
Solution :
(k+1)(2k+3)-2k(3k)+3k(5k)-(2k+3)(5k-1)+(5k-1)2k- 5k(k+1) = 0
2k2 + 5k + 3 - 6k2 + 15k2 - (10k2+13k-3) + 10k2 - 2k - 5k2 - 5k = 0
6k2 -15k + 6 = 0
Dividing by 3, we get
2k2 -5k + 3 = 0
(x - 1)(2x - 3) = 0
x = 0 and x = 3/2
Problem 2 :
Find the value of m if the points (5, 1), (–2, –3) and (8, 2m) are collinear
Solution :
-15-(-2) - 4m - (-24) + 8 - 10m = 0
-15 + 2 - 4m + 24 + 8 - 10m = 0
-14m + 19 = 0
m = 19/4
Problem 3 :
If P (9a–2, –b) divides line segment joining A (3a+1, –3) and B (8a, 5) in the ratio 3 : 1, find the values of a and b.
Solution :
Equating x and y-coordinates, we get
(25a + 1)/4 = 9a - 2 25a + 1 = 4(9a - 2) 25a - 36a = -8 - 1 -9a = -9 a = 1 |
3 = - b b = -3 |
Problem 4 :
If (a, b) is the mid-point of the line segment joining the points
A (10, –6) and B (k, 4) and a – 2b = 18
find the value of k and the distance AB.
Solution :
Comparing the y-coordinates,
b = -1
Comparing x-coordinates,
(10 + k)/2 = a
10 + k = 2a
2a - k = 10 -----(1)
Given a – 2b = 18 ----(2)
Applying the value of b in (2)
a - 2(-1) = 18
a+2 = 18
a = 16
Applying the value of a in (1)
2(16) - k = 10
32 - k = 10
k = 32 - 10
k = 22
Problem 5 :
The line segment joining the points A (3, 2) and B (5,1) is divided at the point P in the ratio 1:2 and it lies on the line 3x – 18y + k = 0. Find the value of k.
Solution :
The line joining the points A and B is divided by the point P, means A, P and B are collinear.
Since the point lies on the line 3x – 18y + k = 0
3(11/3) - 18(5/3) + k = 0
11 - 30 + k = 0
-19 + k = 0
k = 19
Problem 6 :
The center of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, –9) and has diameter 10 √2 units.
Solution :
Center of the circle is (2a, a - 7) and it passes through the point (11, -9). By connecting these two points will get a straight line. So, they are collinear.
Distance between these two points will be the radius.
Radius = 10√2/2
= 5√2 units.
(a - 3)(a - 5) = 0
a = 3 and a = 5.
Problem 7 :
If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then
(A) a = b (B) a = 2b (C) 2a = b (D) a = –b
Solution :
(0 + 0 + 2a) - (0+0+b) = 0
2a - b = 0
2a = b
Problem 8 :
A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2, –5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
(A) (0, – 5) and (2, 0) (B) (0, 10) and (– 4, 0)
(C) (0, 4) and (– 10, 0) (D) (0, – 10) and (4, 0)
Solution :
Let P(a, 0) and Q(0, b) are the points on the x and y-axis respectively.
Midpoint of PQ = (2, -5)
(a + 0)/2 = 2 and (0 + b)/2 = -5
a = 4 and b = -10
So, the required points are (4, 0) and (0, -10).
Problem 9 :
If P (a/3, 4) is the mid-point of the line segment joining the points Q (– 6, 5) and R (– 2, 3), then the value of a is
(A) – 4 (B) – 12 (C) 12 (D) – 6
Solution :
Midpoint of QR = (a/3, 4)
Equating x-coordinates, we get
-4 = a/3
a = -12
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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