If a point (x, y) in a OXY-plane is equidistant from (-1, 1) and (4, 3) then

Options

A. `10x+4y=23`
B. `6x+4y=23`
C. `6x+4y=23`
D. `-x+y=7`

Answer

To solve the problem of finding the equation of the points (x, y) that are equidistant from the points (-1, 1) and (4, 3), we can follow these steps:

### Step 1: Define the Points
Let point P be (x, y), point Q be (-1, 1), and point R be (4, 3).

### Step 2: Use the Distance Formula
Since point P is equidistant from points Q and R, we can set the distances PQ and PR equal to each other.

The distance formula is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

### Step 3: Calculate Distance PQ
Using the distance formula for PQ:
\[ PQ = \sqrt{(x - (-1))^2 + (y - 1)^2} = \sqrt{(x + 1)^2 + (y - 1)^2} \]

### Step 4: Calculate Distance PR
Using the distance formula for PR:
\[ PR = \sqrt{(x - 4)^2 + (y - 3)^2} \]

### Step 5: Set the Distances Equal
Set the two distances equal to each other:
\[ \sqrt{(x + 1)^2 + (y - 1)^2} = \sqrt{(x - 4)^2 + (y - 3)^2} \]

### Step 6: Square Both Sides
To eliminate the square roots, square both sides:
\[ (x + 1)^2 + (y - 1)^2 = (x - 4)^2 + (y - 3)^2 \]

### Step 7: Expand Both Sides
Expand both sides:
\[ (x^2 + 2x + 1 + y^2 - 2y + 1) = (x^2 - 8x + 16 + y^2 - 6y + 9) \]

### Step 8: Simplify the Equation
Combine like terms:
\[ x^2 + y^2 + 2x - 2y + 2 = x^2 + y^2 - 8x - 6y + 25 \]

### Step 9: Cancel Common Terms
Subtract \(x^2\) and \(y^2\) from both sides:
\[ 2x - 2y + 2 = -8x - 6y + 25 \]

### Step 10: Rearrange the Equation
Rearranging gives:
\[ 2x + 8x + 6y - 2y = 25 - 2 \]
\[ 10x + 4y = 23 \]

### Final Equation
Thus, the equation of the locus of points (x, y) that are equidistant from (-1, 1) and (4, 3) is:
\[ 10x + 4y = 23 \]