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 find the value of `c`.

Options

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

Answer

B

Explanation

Let P (h,k) be a point equidistant from the points `A(a_(1) , b_(1))` and `B (a_(2) , b_(2))` . Then ,
PA = PB
`implies PA^(2) = PB^(2)`
`implies (h-a_(1))^(2) + (k-b_(1))^(2) = (h- a_(2))^(2) + (k-b_(2))^(2)`
`implies 2h (a_(1) - a_(2)) + 2k(b_(1) - b_(2)) + (a_(2)^(2) - a_(1)^(2)) + (b_(2)^(2) - b_(1)^(2)) = 0`
Hence , the locus of (h,k) is
`2x (a_(1) - a_(2)) + 2y(b_(1) + b_(2)) + (a_(2)^(2) - a_(1)^(2)) + (b_(2)^(2) - b_(1)^(2)) = 0`
or , `x (a_(1) - a_(2)) + y (b_(1) - b_(2)) + (1)/(2) {(a_(2)^(2) - a_(1)^(2)) + (b_(2)^(2) - b_(1)^(2))} = 0`
Hence ,c = `(1)/(2) = (a_(2)^(2) - a_(1)^(2) + b_(2)^(2) - b_(1)^(2))`