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

Let the point P(x,y) be equidistant from the points A(3,6) and B(-3,4)
∴ PA = PB
By using the distance formula = √(x₂ - x₁)² + (y₂ - y₁)²
we have
⇒ √(x - 3)² + (y - 6)² = √(x + 3)² + (y - 4)²
⇒ (x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²
⇒ x² - 6x + 9 + y² - 12y + 36 = x² + 6x + 9 + y² - 8y + 16
⇒ -6x - 12y + 8y + 36 - 16 = 6x
⇒ -12x - 4y + 20 = 0
⇒ 3x + y - 5 = 0
Hence this is a relation between x and y.