Locus of the points equidistant from the point (-1, -1) and (4, 2) is

Options

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

Answer

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

### Step 1: Define the Points
Let P(x, y) be a point on the locus. The two fixed points are:
- A(-1, -1)
- B(4, 2)

### Step 2: Set Up the Distance Equations
The distance from point P to point A is equal to the distance from point P to point B. Therefore, we can write:
\[ PA = PB \]

Using the distance formula, we have:
\[ \sqrt{(x + 1)^2 + (y + 1)^2} = \sqrt{(x - 4)^2 + (y - 2)^2} \]

### Step 3: Square Both Sides
To eliminate the square roots, we square both sides:
\[ (x + 1)^2 + (y + 1)^2 = (x - 4)^2 + (y - 2)^2 \]

### Step 4: Expand Both Sides
Now, we expand both sides of the equation:
- Left Side:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y + 1)^2 = y^2 + 2y + 1 \]
So,
\[ x^2 + 2x + 1 + y^2 + 2y + 1 = x^2 + y^2 + 2x + 2y + 2 \]

- Right Side:
\[ (x - 4)^2 = x^2 - 8x + 16 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
So,
\[ x^2 - 8x + 16 + y^2 - 4y + 4 = x^2 + y^2 - 8x - 4y + 20 \]

### Step 5: Set the Expanded Equations Equal
Now we set the left side equal to the right side:
\[ x^2 + y^2 + 2x + 2y + 2 = x^2 + y^2 - 8x - 4y + 20 \]

### Step 6: Simplify the Equation
Subtract \(x^2 + y^2\) from both sides:
\[ 2x + 2y + 2 = -8x - 4y + 20 \]

Now, combine like terms:
\[ 2x + 8x + 2y + 4y = 20 - 2 \]
\[ 10x + 6y = 18 \]

### Step 7: Rearrange to Standard Form
Divide the entire equation by 2 to simplify:
\[ 5x + 3y = 9 \]

### Conclusion
The locus of points equidistant from the points (-1, -1) and (4, 2) is given by the equation:
\[ 5x + 3y - 9 = 0 \]