10th CBSE math solution for exercise 7.1 part 4

This page 10th CBSE math solution for exercise 7.1 part 4 is going to provide you solution for every problems that you find in the exercise no 7.1

10th CBSE math solution for exercise 7.1 part 4

(iii) (4,5) (7,6) (4,3) (1,2)

Solution:

Let the given points as A(4,5)  B(7,6)  C(4,3) and D (1,2)

Distance between two points = √(x₂ - x₁) ² + (y₂ - y₁) ²

Length of the side AB

Here x₁ = 4, y₁ = 5, x₂ = 7  and  y₂ = 6

                 = √(7-4)² + (6-5)²

                 = √3² + 1²

                 = √9 + 1

                 = √10

Length of the side BC

Here x₁ = 7, y₁ = 6, x₂ = 4  and  y₂ = 3

                 = √(4-7)² + (3-6)²

                 = √(-3)² + (-3)²

                 = √9 + 9

                 = √18

Length of the side CD

Here x₁ = 4, y₁ = 3, x₂ = 1  and  y₂ = 2

                 = √(1-4)² + (2-3)²

                 = √(-3)² + (-1)²

                 = √9 + 1

                 = √10

Length of the side DA

Here x₁ = 1, y₁ = 2, x₂ = 4  and  y₂ = 5

                 = √(4-1)² + (5-2)²

                 = √3² + 3²

                 = √9 + 9

                 = √18

AB = CD, BC = DA. length of opposite sides are equal.

Length of AC

Here x₁ = 4, y₁ = 5, x₂ = 4  and  y₂ = 3

                 = √(4-4)² + (3-5)²

                 = √0² + (-2)²

                 = √4

                 = 2

Length of BD

Here x₁ = 7, y₁ = 6, x₂ = 1  and  y₂ = 2

                 = √(1-7)² + (2-6)²

                 = √(-4)² + (-4)²

                 = √16 + 16

                 = 32

It can be observed that opposite sides of this quadrilateral are of the same length.However, the diagonals are of different lengths.Therefore,the given points are the vertices of the parallelog


In the page 10th CBSE math solution for exercise 7.1 part 4 we are going to see the solution of next problem

(7) Find the points on the x-axis which is equidistant from (2,-5) and (-2,9)

Solution:

We have to find the point on x-axis. So its y-coordinate will be 0.

Let the point on x axis be (x,0)

Distance between (x,0) and (2,-5) = 
Distance between (x,0) and (-2,9)

Distance between two points = √(x₂ - x₁) ² + (y₂ - y₁) ²

Here x₁ = x, y₁ = 0, x₂ = 2  and  y₂ = -5

                 = √(2-x)² + (5-0)²

                 = √(2 - x)² + 25

Here x₁ = x, y₁ = 0, x₂ = -2  and  y₂ = 9

                 = √(-2-x)² + (9-0)²

                 = √(2+x)² + (9)²

                 = (2+x)² + 81

√(2 - x)² + 25 = (2+x)² + 81

 4 + x² - 4 x + 25 = 4 + x² + 4 x + 81

 x² - x² - 4 x - 4 x + 4 - 4 = 81 - 25

                       -8 x = 56

                          x = -7

Therefore the required point is (-7,0)




HTML Comment Box is loading comments...