When a point x Y is equidistant from the point 7 1 and 3/5 Then the relation between X and Y is?

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Solution:

The distance between any two points can be measured using the distance formula which is given by √ [(x₂ - x₁)2 + (y₂ - y₁)2]

Let point P (x, y) be equidistant from points A (3, 6) and B (- 3, 4).

Since they are equidistant, PA = PB

Hence by applying the distance formula for PA = PB, we get

√(x - 3)² + (y - 6)² = √(x - (- 3))² + (y - 4)²

√(x - 3)² + (y - 6)² = √(x + 3)² + (y - 4)²

By squaring, we get

PA2 = PB2

(x - 3)2 + (y - 6)2 = (x + 3)2 + (y - 4)2

x2 + 9 - 6x + y2 + 36 - 12y = x2 + 9 + 6x + y2 + 16 - 8y

6x + 6x + 12y - 8y = 36 - 16 [On further simplifying]

12x + 4y = 20

3x + y = 5

3x + y - 5 = 0

Thus, the relation between x and y is given by 3x + y - 5 = 0

☛ Check: NCERT Solutions for Class 10 Maths Chapter 7

Video Solution:

NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.1 Question 10

Summary:

The relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4) is 3x + y - 5 = 0.

☛ Related Questions:

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:(i) (- 1, - 2), (1, 0), (- 1, 2), (- 3, 0) (ii) (- 3, 5), (3, 1), (0, 3), (- 1, - 4) (iii) (4, 5), (7, 6), (4, 3), (1, 2)
Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9).
Find the values of y for which the distance between the points P (2, - 3) and Q (10, y) is 10 units.
If Q (0, 1) is equidistant from P (5, - 3) and R (x, 6), find the values of x. Also, find the distances QR and PR.

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Suggest Corrections

If P (x , y ) is equidistant from the points A (7,1) and B (3,5) find the relation between x and y

Let the point P(x, y) be equidistant from the points A(7, 1) and B(3, 5) Then

PA = PB

`⇒ PA^2 = PB^2`

`⇒(x-7)^2 +(y-1)^2 = (x-3)^2 +(y-5)^2`

`⇒ x^2 +y^2 -14x-2y +50 = x^2 +y^2 -6x -10y +34`

`⇒ 8x -8y = 16 `

`⇒ x-y =2`