10th samacheer kalvi math solution for exercise 5.3 part 1

This page 10th samacheer kalvi math solution for exercise 5.3 part 1 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 1

1. Find the angle of inclination of the straight line whose slope is
(i) 1     

Solution:

  slope (m) = 1    

         m = tan θ

        tan θ = 1

for which angle of tan we get the value 1

        θ = 45°

(ii) √3           

Solution:

  slope (m) = √3   

         m = tan θ

        tan θ = √3

for which angle of tan we get the value √3

         θ = 60°

(iii) 0

Solution:

  slope (m) = 0   

         m = tan θ

        tan θ = 0

for which angle of tan we get the value  0

         θ = 0°


2. Find the slope of the straight line whose angle of inclination is
(i) 30
°      

Solution:

       m = tan θ

       m = tan 30°

          = 1/√3

(ii) 60°    

Solution:

       m = tan θ

       m = tan 60°

          = √3

(iii) 90°

Solution:

       m = tan θ

       m = tan 90°

             =


In the page 10th samacheer kalvi math solution for exercise 5.3 part 1 we are going to see the solution of next problem

3. Find the slope of the straight line passing through the points

(i) (3 , -2) and (7 , 2)

Solution:

   m = (y₂ - y₁)/(x₂ - x₁)

x₁ = 3     x₂ = 7

y₁ = -2     y₂ = 2

    m = (2 - (-2))/(7- 3)

    m = (2 +2)/(7- 3)

    m = 4/4

        = 1

(ii) (2 , -4) and origin

Solution:

   m = (y₂ - y₁)/(x₂ - x₁)

x₁ = 2     x₂ = 0

y₁ = -4     y₂ = 0

    m = (0 - (-4))/(0- 2)

    m = (0 + 4)/(- 2)

    m = 4/(-2)

        = -2

(iii) (1 + √3 , 2) and (3 + √3 , 4)

Solution:

   m = (y₂ - y₁)/(x₂ - x₁)

x₁ = 1 + √3     x₂ = 3 + √3

y₁ = 2           y₂ = 4

    m = [(3 + √3)- (1 + √3)]/(4 - 2)

         = (3 + √3- 1 - √3)/2

         = 2/2

       = 1




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