10th samacheer kalvi math solution for exercise 5.3 part 4
This page 10th samacheer kalvi math solution for exercise 5.3 part 4 is going to provide you solution for every problems that you find in the exercise no 5.3
10th samacheer kalvi solution for exercise 5.3 part 4
9. If the points (a, 1), (1, 2) and (0, b+1) are collinear, then show that (1/a) + (1/b) = 1
Solution:
Let A (a ,1) B (1,2) and C (0,b+1) be the given points
Since the given points are collinear,then
Slope of AB = Slope of BC
Slope of AB (m) = (y₂-y₁)/(x₂-x₁)
x₁ = a y₁ = 1
x₂ = 1 y₂ = 2
m = (2 - 1)/(1-a)
= 1/(1-a)
Slope of BC (m) = (y₂-y₁)/(x₂-x₁)
x₁ = 1 y₁ = 2
x₂ = 0 y₂ = b + 1
m = (b + 1 - 2)/(0-1)
= (b - 1)/(-1)
1/(1-a) = (b - 1)/(-1)
-1 (1) = (b - 1)(1-a)
- 1 = b - ab - 1 + a
a + b - ab = -1 + 1
a + b = 0 + ab
a + b = ab
dividing the sole equation by ab =>
(a/ab) + (b/ab) = ab/ab
(1/b) + (1/a) = 1
(1/a) + (1/b) = 1
10. The line joining the points A(-2 , 3) and B(a , 5) is parallel to the line joining the points C(0 , 5) and D(-2 , 1). Find the value of a.
Solution:
Slope of the line AB = Slope of the line CD
Slope of AB (m) = (y₂-y₁)/(x₂-x₁)
x₁ = -2 y₁ = 3
x₂ = a y₂ = 5
m = (5 - 3)/(a-(-2))
= 2/(a + 2)
Slope of CD (m) = (y₂-y₁)/(x₂-x₁)
x₁ = 0 y₁ = 5
x₂ = -2 y₂ = 1
m = (1 - 5)/(-2-0)
= -4/(-2)
= 2
2/(a + 2) = 2
2 = 2(a + 2)
2 = 2 a + 4
2 - 4 = 2 a
2 a = -2
a = -1
In the page 10th samacheer kalvi math solution for exercise 5.3 part 4 we are going to see the solution of next problem
11. The line joining the points A(0, 5) and B(4, 2) is perpendicular to the line joining the points C(-1, -2) and D(5, b). Find the value of b.
Solution:
Since the line joining the points AB and CD are perpendicular
slope of AB x Slope of CD = -1
Slope of AB
x₁ = 0 y₁ = 5
x₂ = 4 y₂ = 2
m = (2 - 5)/(4-0)
= -3/4
Slope of CD
x₁ = -1 y₁ = -2
x₂ = 5 y₂ = b
m = (b - (-2))/(5-(-1))
= (b + 2)/(5 + 1)
= (b + 2)/6
slope of AB x Slope of CD = -1
(-3/4) x (b + 2)/6 = -1
(-1/4) x (b + 2)/2 = -1
-1(b + 2)/8= -1
- b - 2 = -8
-b = -8 + 2
-b = -6
b = 6
