10th CBSE math solution for exercise 7.2 part 5

This page 10th CBSE math solution for exercise 7.2 part 5 is going to provide you solution for every problems that you find in the exercise no 7.2

10th CBSE math solution for exercise 7.2 part 5

(7) Find the coordinates of a point A,where AB is the diameter of a circle whose center is (2,-3) and B(1,4)

Solution:

Midpoint of the diameter is center

 center = (2,-3)

Let the coordinate of point A be  (a,b)

Here x₁ = a, y₁ = b, x₂ = 1, y₂ = 4 

Midpoint = (x₁ + x₂)/2 , (y₁ + y₂)/2

          (a+1)/2,(b+4)/2 = (2,-3)

   equating the coordinates of x and y 

          (a+1)/2 = 2       (b+4)/2 = -3

            a + 1 = 4        b + 4 = -6

           a = 4 - 1         b = -6 -4

             a = 3           b = -10

Therefore the coordinate of A are (3,-10).


(8) If A and B are (-2,-2) and (2,-4),respectively,find the coordinates of P such that AP = (3/7) AB and P lies on the line segment AB.

Solution:

AP = (3/7) AB

AP/AB = 3/7

AB is the line segment P is a point lies on the line segment.

 AB = AP + PB

  7 = 3 + PB

 PB = 4

AP:PB = 3:4

x₁ = -2, y₁ = -2, x₂ = 2 , y₂ = -4   m = 3   n = 4

Coordinates of P = [3(2) + 4(-2)]/(3+4) , [3(-4) + 4(-2)]/(3+4)

                         = (6-8)/7,(-12-8)/7

                         = -2/7,-20/7

In the page 10th CBSE math solution for exercise 7.2 part 5 we are going to see the solution of next problem

(9) Find the coordinates of the points which divide the line segment joining A (-2,2) and B(2,8) into four equal parts.

Solution:

Let "C", "D" and "E" be the points which divides the line segment into four equal parts.

length of AC = 1 unit

length of CD = 1 unit

length of DE = 1 unit

then length of EB is also = 1 unit

 So C divides the line segment in the ratio 1 : 3

x₁ = -2, y₁ = 2, x₂ = 2 , y₂ = 8  m = 1  and n = 3

         = [1 (2) + 3 (-2)]/(1+3) , [1 (8) + 3 (2)]/(1+3)

         = (2 - 6)/4, (8+6)/4

          = -4/4, 14/4

          = (-1,7/2)

D divides the line segment in the ratio 2 : 2

x₁ = -2, y₁ = 2, x₂ = 2 , y₂ = 8  m = 2  and n = 2

         = [2 (2) + 2 (-2)]/(2+2) , [2 (8) + 2 (2)]/(2+2)

         = (4 - 4)/4, (16+4)/4

          = 0/4, 20/4

          = (0,5)

E divides the line segment in the ratio 3 : 1

x₁ = -2, y₁ = 2, x₂ = 2 , y₂ = 8  m = 3  and n = 1

         = [3 (2) + 1 (-2)]/(3+1) , [3 (8) + 1 (2)]/(3+1)

         = (6 - 2)/4, (24+2)/4

          = 4/4, 26/4

          = (1,13/2)


(10) Find the area of a rhombus if its vertices are (3,0) (4,5) (-1,4) and (-2,-1) taken in order.[Hint:Area of rhombus = (1/2) product of its diagonals]

Solution:

 Let A(3,0) B(4,5) C(-1,4) and D(-2,-1) are the vertices of the rhombus.

length of diagonal AC =  √(x₂ - x₁)² + (y₂ - y₁)²

x₁ = 3 y₁ = 0  x₂ =-1  y₂ = 4

         = √(-1 - 3)² + (4 - 0)²

         = √(-4)² + (4)²

         = √16 + 16

         = √32

length of diagonal BD =  √(x₂ - x₁)² + (y₂ - y₁)²

x₁ = 4 y₁ = 5  x₂ =-2  y₂ = -1

         = √(-2-4)² + (-1-5)²

         = √(-6)² + (-6)²

         = √36 + 36

         = √72

Area of rhombus = (1/2) x √32 x √72

                   = (1/2)4√2 x 6√2

                   = 24 square units. 




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