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

(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 y****₁ = 0 ****x₂ =-1 y****₂ = 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 y****₁ = 5 ****x₂ =-2 y****₂ = -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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