This page 10th CBSE math solution for exercise 7.2 part 3 is going to provide you solution for every problems that you find in the exercise no 7.2

(4) Find the ratio in which the line segment joining the points (-3,10) and (6,-8) is divided by (-1,6).

Solution:

Here x₁ = -3, y₁ = 10, x₂ = 6, y₂ = -8

[m(6) + n (-3)]/(m+n) , [m (-8) + n (10)]/(m+n) = (-1,6)

(6m - 3n)/(m+n), (-8m+10n)/(m+n) = (-1,6)

equating the coefficients of x and y

(6m-3n)/(m+n) = -1 (-8m+10n)/(m+n) = 6

6m - 3n = -1 (m+n)

6m - 3n = -1 (m+n)

6m-3n=-m-n

6m+m=-n+3n

7m=2n

m/n= 2/7

m:n = 2:7

So the point (-1,6) is dividing the line segment in the ratio 2:7.

(5) Find the ratio in which the line segment joining A (1,-5) and B(-4,5) is divided by the x axis.Also find the coordinates of the point of division.

Solution:

Let m:n be the ratio which divides the line segment joining the points A(1,-5) and B(-4,5)

Since the point of division is on x axis y-coordinate value will be 0.

x₁ = 1, y₁ = -5, x₂ = -4 , y₂ = 5

m(-4) + n(1)/(m+n) , m(5) + n(-5)/(m+n) = (x,0)

(-4m + n) /(m+n) , (5m-5n)/(m+n) = (x,0)

(-4m + n)/(m+n) = x (5m - 5n)/(m+n) = 0

(5m - 5n)/(m+n) = 0

5 m - 5 n = 0

5 m = 5 n

m/n = 5/5

m:n = 1:1

In the page 10th CBSE math solution for exercise 7.2 part 3 we are going to see the solution of next problem

(6) If (1,2),(4,y) (x,6) and (3,5) are the vertices of a parallelogram taken in order,find x and y

Solution:

Let A(1,2),B(4,y) C(x,6) and D(3,5) are the vertices of a parallelogram.

In a parallelogram midpoint of the diagonals will be equal.

Midpoint of diagonal AC = Midpoint of diagonal BD

midpoint = (x₁ + x₂)/2 , (y₁ + y₂)/2

Midpoint of diagonal AC :

x₁ = 1, y₁ = 2, x₂ = x , y₂ = 6

= (1+x)/2,(2+6)/2

= (1+x)/2,8/2

= (1+x)/2,4

Midpoint of diagonal BD :

x₁ = 4, y₁ = y, x₂ = 3 , y₂ = 5

= (4+3)/2,(y+5)/2

= 7/2,(y+5)/2

(1+x)/2,4 = 7/2,(y+5)/2

(1 + x)/2 = 7/2

1 + x = 7 4 = (y + 5)/2

x = 7 - 1 8 = y + 5

x = 6 y = 8 - 5

y = 3

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