Find the equation to the locus of points equidistant from the points
(-3,2),(0,4)

Answer

To find the equation of the locus of points equidistant from the points (-3, 2) and (0, 4), we can follow these steps:

### Step 1: Define the points and the point (x, y)
Let the points be A(-3, 2) and B(0, 4). We want to find the locus of points (x, y) that are equidistant from both A and B.

### Step 2: Use the distance formula
The distance from point (x, y) to point A(-3, 2) is given by:
\[ d_A = \sqrt{(x + 3)^2 + (y - 2)^2} \]
The distance from point (x, y) to point B(0, 4) is given by:
\[ d_B = \sqrt{(x - 0)^2 + (y - 4)^2} = \sqrt{x^2 + (y - 4)^2} \]

### Step 3: Set the distances equal
Since the point (x, y) is equidistant from both points, we set the distances equal:
\[ \sqrt{(x + 3)^2 + (y - 2)^2} = \sqrt{x^2 + (y - 4)^2} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ (x + 3)^2 + (y - 2)^2 = x^2 + (y - 4)^2 \]

### Step 5: Expand both sides
Expanding the left side:
\[ (x^2 + 6x + 9) + (y^2 - 4y + 4) = x^2 + y^2 + 6x + 13 - 4y \]
Expanding the right side:
\[ x^2 + (y^2 - 8y + 16) = x^2 + y^2 - 8y + 16 \]

### Step 6: Simplify the equation
Now, we can simplify the equation:
\[ x^2 + 6x + 9 + y^2 - 4y + 4 = x^2 + y^2 - 8y + 16 \]
Canceling \(x^2\) and \(y^2\) from both sides:
\[ 6x + 13 - 4y = -8y + 16 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ 6x + 4y = 3 \]

### Step 8: Final equation of the locus
Thus, the equation of the locus of points equidistant from the points (-3, 2) and (0, 4) is:
\[ 6x + 4y - 3 = 0 \]