If the equation of the locus of a point equidistant from the points `(a_(1),b_(1)`, and `(a_(2),b_(2))` is
`(a_(1)-a_(2))x+(b_(1)-b_(2))y+c=0`, then the value of c is

Options

A. `1/2(a_(2)^(2)+b_(2)^2)-(a_(1)^(2)-b_(1)^(2))`
B. `(a_(1)^(2)-a_(2)^(2)+b_(1)^(2)-b_(2)^(2))`
C. `1/2(a_(1)^(2)+a_(2)^(2)+b_(1)^(2)+b_(2)^(2))`
D. `sqrt((a_(1)^(2)+b_(1)^(2)-a_(2)^(2)-b_(2)^(2)))`

Answer

To find the value of \( c \) in the equation of the locus of a point equidistant from the points \( (a_1, b_1) \) and \( (a_2, b_2) \), we will follow these steps:

### Step 1: Understand the Distance Formula
The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

### Step 2: Set Up the Equidistance Condition
Let \( P(h, k) \) be the point that is equidistant from \( (a_1, b_1) \) and \( (a_2, b_2) \). The distances from \( P \) to each point must be equal:
\[ \sqrt{(h - a_1)^2 + (k - b_1)^2} = \sqrt{(h - a_2)^2 + (k - b_2)^2} \]

### Step 3: Square Both Sides
To eliminate the square roots, we square both sides:
\[ (h - a_1)^2 + (k - b_1)^2 = (h - a_2)^2 + (k - b_2)^2 \]

### Step 4: Expand Both Sides
Expanding both sides gives:
\[ (h^2 - 2ha_1 + a_1^2 + k^2 - 2kb_1 + b_1^2) = (h^2 - 2ha_2 + a_2^2 + k^2 - 2kb_2 + b_2^2) \]

### Step 5: Simplify the Equation
Cancelling \( h^2 \) and \( k^2 \) from both sides, we have:
\[ -2ha_1 + a_1^2 - 2kb_1 + b_1^2 = -2ha_2 + a_2^2 - 2kb_2 + b_2^2 \]

### Step 6: Rearrange the Terms
Rearranging gives:
\[ 2h(a_2 - a_1) + 2k(b_2 - b_1) + (a_1^2 + b_1^2 - a_2^2 - b_2^2) = 0 \]

### Step 7: Divide by 2
Dividing the entire equation by 2 gives:
\[ (a_2 - a_1)h + (b_2 - b_1)k + \frac{1}{2}(a_1^2 + b_1^2 - a_2^2 - b_2^2) = 0 \]

### Step 8: Identify the Value of \( c \)
From the standard form of the equation \( (a_1 - a_2)x + (b_1 - b_2)y + c = 0 \), we can identify that:
\[ c = \frac{1}{2}(a_1^2 + b_1^2 - a_2^2 - b_2^2) \]

### Final Answer
Thus, the value of \( c \) is:
\[ c = \frac{1}{2}(a_1^2 + b_1^2 - a_2^2 - b_2^2) \]

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