Find the locus of the point such that the sum of the squares of its distances from the points (2, 4) and (-3, -1) is 30.

Answer

To find the locus of the point such that the sum of the squares of its distances from the points (2, 4) and (-3, -1) is 30, we can follow these steps:

### Step 1: Define the point and distances
Let the point be \( P(h, k) \). The distances from \( P \) to the points \( A(2, 4) \) and \( B(-3, -1) \) can be expressed using the distance formula.

### Step 2: Write the distance formulas
The distance from \( P(h, k) \) to \( A(2, 4) \) is:
\[ d_1 = \sqrt{(h - 2)^2 + (k - 4)^2} \]
The distance from \( P(h, k) \) to \( B(-3, -1) \) is:
\[ d_2 = \sqrt{(h + 3)^2 + (k + 1)^2} \]

### Step 3: Write the equation for the sum of squares of distances
According to the problem, the sum of the squares of these distances is equal to 30:
\[ d_1^2 + d_2^2 = 30 \]
Substituting the distance formulas, we have:
\[ (h - 2)^2 + (k - 4)^2 + (h + 3)^2 + (k + 1)^2 = 30 \]

### Step 4: Expand the squares
Now, we expand each term:
\[ (h - 2)^2 = h^2 - 4h + 4 \]
\[ (k - 4)^2 = k^2 - 8k + 16 \]
\[ (h + 3)^2 = h^2 + 6h + 9 \]
\[ (k + 1)^2 = k^2 + 2k + 1 \]

### Step 5: Combine all terms
Adding these together:
\[ (h^2 - 4h + 4) + (k^2 - 8k + 16) + (h^2 + 6h + 9) + (k^2 + 2k + 1) = 30 \]
This simplifies to:
\[ 2h^2 + 2k^2 + ( -4h + 6h) + (-8k + 2k) + (4 + 16 + 9 + 1) = 30 \]
\[ 2h^2 + 2k^2 + 2h - 6k + 30 = 30 \]

### Step 6: Simplify the equation
Subtract 30 from both sides:
\[ 2h^2 + 2k^2 + 2h - 6k = 0 \]
Dividing the entire equation by 2:
\[ h^2 + k^2 + h - 3k = 0 \]

### Step 7: Rearranging the equation
Rearranging gives us the final equation of the locus:
\[ h^2 + k^2 + h - 3k = 0 \]

### Step 8: Substitute back to original variables
Since \( h \) and \( k \) represent \( x \) and \( y \) respectively, we can write:
\[ x^2 + y^2 + x - 3y = 0 \]

### Final Answer
The locus of the point is given by:
\[ x^2 + y^2 + x - 3y = 0 \]