10th samacheer kalvi math solution for exercise 5.3 part 2

This page 10th samacheer kalvi math solution for exercise 5.3 part 2 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 2

4. Find the angle of inclination of the line passing through the points

(i) (1, 2) and (2 , 3)

Solution:

x₁ = 1     x₂ = 2

y₁ = 2     y₂ = 3

        (y-y)/(y-y) = (x-x)/(x-x)

   (y - 2)/(3 - 2) = (x - 1)/(2 - 1)

   (y - 2)/1 = (x - 1)/1

    y - 2 = x - 1

   x - y - 1 + 2 = 0

   x - y + 1 = 0

(ii) (3 , 3) and (0 , 0)

x₁ = 3     x₂ = 0

y₁ = 3     y₂ = 0

        (y-y)/(y-y) = (x-x)/(x-x)

   (y - 3)/(0 - 3) = (x - 3)/(0 - 3)

   (y - 3)/(-3) = (x - 3)/(-3)

    - 3 (y - 3) = -3 (x - 3)

   - 3 y + 9 = - 3 x + 9

     3 x - 3 y + 9 - 9 = 0

      3 x - 3 y = 0

 divide the whole equation by 3

      x - y = 0

(iii) (a , b) and (-a , -b)

x₁ = a     x₂ = -a

y₁ = b     y₂ = -b

        (y-y)/(y-y) = (x-x)/(x-x)

   (y - b)/(-b - b) = (x - a)/(-a - a)

   (y - b)/(-2b) = (x - a)/(-2a)

    - 2a (y - b) = -2b (x - a)

   -a y + ab = -b x + ab

       b x - a y + ab - ab = 0

       b x - a y = 0


5. Find the slope of the line which passes through the origin and the midpoint of the line segment joining the points (0 ,- 4) and (8 , 0).

Solution:

       midpoint of the line segment joining the points (0,-4) and (8 , 0)

            midpoint = (x₁ + x₂)/2 , (y₁ + y₂)/2

x₁ = 0  x₂ = 8

y₁ = -4  y₂ = 0

                     = (0 + 8)/2 , (-4 + 0)/2

                     = 8/2 , -4/2

                    = (4 , -2)

now we need to find the slope of the line which passes through origin

x₁ = 4  x₂ = 0

y₁ = -2  y₂ = 0

        m = (y₂ - y₁)/(x₂ - x₁)

           = (0 - (-2))/(0 - 4)

           = 2/(-4)

           = -1/2


In the page 10th samacheer kalvi math solution for exercise 5.3 part 2 we are going to see the solution of next problem

6. The side AB of a square ABCD is parallel to x-axis . Find the
(i) slope of AB (ii) slope of BC (iii) slope of the diagonal AC

Solution:

(i) Since the side AB is parallel to x axis,

    slope of the side AB = 0

(ii) The angle formed by the side BC is 90.

            m = tan θ

             θ = 90

             m = tan 90

                 =

(iii) The diagonal AC is the angle bisector of the angle.

             m = tan θ

             θ = 45

             m = tan 45

                 = 1




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