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

Options

A. `3x-y-5=0`
B. `3x+y+5=0`
C. `3x+y-5=0`
D. None

Explanation

To find a relation between \( x \) and \( y \) such that the point \( (x, y) \) is equidistant from the points \( (3, 6) \) and \( (-3, 4) \), we can follow these steps:

### Step 1: Use the Distance Formula
The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
We need to set the distances from \( (x, y) \) to both points equal to each other.

### Step 2: Set Up the Equations
The distance from \( (x, y) \) to \( (3, 6) \) is:
\[ d_1 = \sqrt{(x - 3)^2 + (y - 6)^2} \]
The distance from \( (x, y) \) to \( (-3, 4) \) is:
\[ d_2 = \sqrt{(x + 3)^2 + (y - 4)^2} \]
Since the point \( (x, y) \) is equidistant from both points, we have:
\[ d_1 = d_2 \]

### Step 3: Square Both Sides
To eliminate the square roots, we square both sides:
\[ (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2 \]

### Step 4: Expand Both Sides
Now we expand both sides:
- Left side:
\[ (x - 3)^2 = x^2 - 6x + 9 \]
\[ (y - 6)^2 = y^2 - 12y + 36 \]
So,
\[ x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + y^2 - 6x - 12y + 45 \]

- Right side:
\[ (x + 3)^2 = x^2 + 6x + 9 \]
\[ (y - 4)^2 = y^2 - 8y + 16 \]
So,
\[ x^2 + 6x + 9 + y^2 - 8y + 16 = x^2 + y^2 + 6x - 8y + 25 \]

### Step 5: Set the Expanded Forms Equal
Now we set the expanded forms equal:
\[ x^2 + y^2 - 6x - 12y + 45 = x^2 + y^2 + 6x - 8y + 25 \]

### Step 6: Simplify the Equation
Cancel \( x^2 \) and \( y^2 \) from both sides:
\[ -6x - 12y + 45 = 6x - 8y + 25 \]
Now, rearranging gives:
\[ -6x - 12y + 8y + 45 - 25 = 6x \]
This simplifies to:
\[ -12y + 8y + 45 - 25 = 12x \]
\[ -4y + 20 = 12x \]
Now, rearranging gives:
\[ 12x + 4y - 20 = 0 \]

### Step 7: Final Relation
Dividing through by 4 gives:
\[ 3x + y - 5 = 0 \]

### Conclusion
The relation between \( x \) and \( y \) such that the point \( (x, y) \) is equidistant from the points \( (3, 6) \) and \( (-3, 4) \) is:
\[ 3x + y - 5 = 0 \]