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

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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

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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