Question

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

Answer 1

Point (x, y) is equidistant from (3, 6) and (−3, 4).

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

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

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

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

⇒ 36 - 16 = 6x + 6x + 12y - 8y

⇒ 20 = 12x + 4y

⇒ 3x + y = 5

⇒ 3x + y - 5 = 0

Answer 2

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 )` 

Squaring on both sides of the equation, we get

(x - 3)² + (y - 6)²

= (x + 3)² + (y - 4)²

= x² + 9 - 6x + y² + 36 - 12y

= x² + 9 + 6x + y² + 16 - 8y

= 12x + 4y = 20

= 3x + y = 5

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