# Find a Relation Between X and Y Such that the Point (X, Y) is Equidistant from the Points (3, 6) and (−3, 4). - Mathematics

Find a relation between x and y such that the point (xy) is equidistant from the points (3, 6) and (-3, 4).

#### Solution

The distance d between two points (x_1, y_1) and (x_2, y_2) is given by the formula

d = sqrt((x_1-x_2)^2 + (y_1 - y_2)^2)

Let the three given points be P(x, y), A(3,6) and B(3,4).

Now let us find the distance between ‘P’ and ‘A’.

PA = sqrt((x - 3)^2 + (y - 3))

Now, let us find the distance between ‘P’ and ‘B’.

PB = sqrt((x + 3)^2 + (y - 6)^2)

It is given that both these distances are equal. So, let us equate both the above equations,

PA = PB

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

Squaring on both sides of the equation we get

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

x^2 + 9 - 6x + y^2 + 36 - 12y = x62 + 9 + 6x + y^2 + 16 - 8y

12x + 4y = 20

3x + y = 5

Hence the relationship between ‘x’ and ‘y’ based on the given condition is 3x + y = 5

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.2 | Q 43 | Page 17