Find a relation between *x* and *y* such that the point (*x*, *y*) 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