Question

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

Hint:

When a point is equidistant from two other points, the equation that relates the coordinates of the point is derived by setting the distances equal to each other and solving for the variables.

Answer 1

We are given that the point \( (x, y) \) is equidistant from the points \( (3, 6) \) and \( (-3, 4) \). To find the relation, we use the distance formula. The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] First, find the distance from \( (x, y) \) to \( (3, 6) \): \[ d_1 = \sqrt{(x - 3)^2 + (y - 6)^2} \] Next, find the distance from \( (x, y) \) to \( (-3, 4) \): \[ d_2 = \sqrt{(x + 3)^2 + (y - 4)^2} \] Since the point \( (x, y) \) is equidistant from both points, we set \( d_1 = d_2 \): \[ \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2} \] Now, square both sides to eliminate the square roots: \[ (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2 \] Expand both sides: \[ (x^2 - 6x + 9) + (y^2 - 12y + 36) = (x^2 + 6x + 9) + (y^2 - 8y + 16) \] Simplify: \[ x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16 \] Cancel the common terms \( x^2 \) and \( y^2 \) from both sides: \[ -6x + 9 + 36 - 12y = 6x + 9 - 8y + 16 \] Simplify further: \[ -6x + 45 - 12y = 6x + 25 - 8y \] Now, move all the terms involving \( x \) and \( y \) to one side: \[ -6x - 6x + 45 - 12y + 8y = 25 \] Simplify: \[ -12x - 4y + 45 = 25 \] Move the constants to the other side: \[ -12x - 4y = -20 \] Finally, divide through by -4: \[ 3x + y = 5 \]
Conclusion: The relation between \( x \) and \( y \) is: \[ \boxed{3x + y = 5} \]