Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (−3, 4).
\(3x + y - 5 = 0\)
Let the point \((x, y)\) be equidistant from the points \((3, 6)\) and \((-3, 4)\).
Distance from \((x, y)\) to \((3, 6)\) is \( \sqrt{(x - 3)^2 + (y - 6)^2} \).
Distance from \((x, y)\) to \((-3, 4)\) is \( \sqrt{(x + 3)^2 + (y - 4)^2} \).
Since the point is equidistant from both, these distances are equal: \( \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2} \).
Squaring both sides, we get: \( (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2 \).
Expand each term: \( x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16 \).
Combine like terms: \( x^2 + y^2 - 6x - 12y + 45 = x^2 + y^2 + 6x - 8y + 25 \).
Cancel \(x^2\) and \(y^2\) from both sides to get: \( -6x - 12y + 45 = 6x - 8y + 25 \).
Bring all terms to one side: \( -6x - 12y + 45 - 6x + 8y - 25 = 0 \), which simplifies to \( -12x - 4y + 20 = 0 \).
Divide the whole equation by \(-4\): \( 3x + y - 5 = 0 \).
Thus, the required relation between \(x\) and \(y\) is \( 3x + y - 5 = 0 \).