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 = ay₁ = 1

x₂ = 1 y₂ = 2

m = (2 - 1)/(1-a)

= 1/(1-a)

Slope of BC (m) = (y₂-y₁)/(x₂-x₁)

x₁ = 1y₁ = 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 = -2y₁ = 3

x₂ = a y₂ = 5

m = (5 - 3)/(a-(-2))

= 2/(a + 2)

Slope of CD (m) = (y₂-y₁)/(x₂-x₁)

x₁ = 0y₁ = 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