The equation of the locus of points equidistant from `(-1-1)` and `(4,2)` is

Options

A. `3x-5y-7=0`
B. `5x+3y-9=0`
C. `4x+3y+2=0`
D. `x-3y+5=0`

Answer

To find the equation of the locus of points equidistant from the points \((-1, -1)\) and \((4, 2)\), we can follow these steps:

### Step 1: Set Up the Distance Equation
Let \( P = (-1, -1) \) and \( Q = (4, 2) \). Let \( (h, k) \) be any point on the locus. The distance from point \( P \) to point \( (h, k) \) should be equal to the distance from point \( Q \) to point \( (h, k) \).

Using the distance formula, we can write:
\[ \sqrt{(h + 1)^2 + (k + 1)^2} = \sqrt{(h - 4)^2 + (k - 2)^2} \]

### Step 2: Square Both Sides
To eliminate the square roots, we square both sides:
\[ (h + 1)^2 + (k + 1)^2 = (h - 4)^2 + (k - 2)^2 \]

### Step 3: Expand Both Sides
Now, we expand both sides:
- Left side:
\[ (h + 1)^2 + (k + 1)^2 = h^2 + 2h + 1 + k^2 + 2k + 1 = h^2 + k^2 + 2h + 2k + 2 \]

- Right side:
\[ (h - 4)^2 + (k - 2)^2 = h^2 - 8h + 16 + k^2 - 4k + 4 = h^2 + k^2 - 8h - 4k + 20 \]

### Step 4: Set the Expanded Equations Equal
Now we set the expanded forms equal to each other:
\[ h^2 + k^2 + 2h + 2k + 2 = h^2 + k^2 - 8h - 4k + 20 \]

### Step 5: Simplify the Equation
We can cancel \( h^2 \) and \( k^2 \) from both sides:
\[ 2h + 2k + 2 = -8h - 4k + 20 \]

Rearranging gives:
\[ 2h + 8h + 2k + 4k = 20 - 2 \]
\[ 10h + 6k = 18 \]

### Step 6: Divide by 2
To simplify further, we divide the entire equation by 2:
\[ 5h + 3k = 9 \]

### Step 7: Replace Variables
Finally, we replace \( h \) with \( x \) and \( k \) with \( y \) to get the equation of the locus:
\[ 5x + 3y = 9 \]

### Final Answer
The equation of the locus of points equidistant from \((-1, -1)\) and \((4, 2)\) is:
\[ 5x + 3y = 9 \]