10th samacheer kalvi math solution for exercise 5.3 part  3

This page 10th samacheer kalvi math solution for exercise 5.3 part 3 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 3

7. The side BC of an equilateral Δ ABC is parallel to x-axis. Find the slope of AB and the slope of BC

Solution:

The side BC is parallel to x axis.

            slope of the side BC = 0

Since this is an equilateral triangle each angle is 60.

             m = tan θ

             θ = 60

             m = tan 60

                 = √3


(i) (2 , 3), (3 , -1) and (4 , -5)

Solution:

Let A (2,3) B (3 , -1) and C (4,-5) be the given points

 If the given points are collinear then,

    slope of AB = slope of BC

     slope = (y-y)/(x-x)

x₁ = 2     x₂ = 3

y₁ = 3     y₂ = -1

                = (-1-3)/(3-2)

                = -4/1

                = -4

slope of AB = -4

x₁ = 3     x₂ = 4

y₁ = -1     y₂ = -5

                = (-5-(-1))/(4-3)

                = (-5+1)/1

                = -4

slope of BC = -4

Therefore the given points are collinear.


(ii) (4 , 1), (-2 , -3) and (-5 , -5)

Solution:

Let A (4,1) B (-2 , -3) and C (-5,-5) be the given points

 If the given points are collinear then,

    slope of AB = slope of BC

     slope = (y-y)/(x-x)

x₁ = 4     x₂ = -2

y₁ = 1     y₂ = -3

                = (-3-1)/(-2-4)

                = -4/(-6)

               = 2/3

slope of AB = 2/3

x₁ = -2     x₂ = -5

y₁ = -3     y₂ = -5

                = (-5-(-3))/(-5-(-2))

                = (-5+3)/(-5+2)

                = -2/(-3)

                = 2/3

slope of BC = 2/3

Therefore the given points are collinear.


In the page 10th samacheer kalvi math solution for exercise 5.3 part 3 we are going to see the solution of next problem

(iii) (4 , 4), (-2 , 6) and (1 , 5)

Solution:

Let A (4,4) B (-2 , 6) and C (1,5) be the given points

 If the given points are collinear then,

    slope of AB = slope of BC

     slope = (y-y)/(x-x)

x₁ = 4     x₂ = -2

y₁ = 4     y₂ = 6

                = (6-4)/(-2-4)

                = 2/(-6)

               = -1/3

slope of AB = -1/3

x₁ = -2     x₂ = 1

y₁ = 6     y₂ = 5

                = (5-6)/(1-(-2))

                = (-1)/(1+2)

                = -1/3

slope of BC = -1/3

Therefore the given points are collinear.




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