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 `a a2-a2 2+b1 2-b2 2` `sqrt(a1 2+b1 2-a2 2-b2 2)` `1/2(a1 2+a2 2+b1 2+b2 2)` `1/2(a2 2+b2 2-a1 2-b1 2)`

Options

A. `a_1^2-a_2^2+b_1^2-b_2^2`
B. `sqrt(a_1^2+b_1^2-a_2^2-b_2^2)`
C. `(1)/(2)(a_1^2+a_2^2+b_1^2+b_2^2)`
D. `(1)/(2)(a_1^2+b_2^2+a_1^2+b_2^2)`

Answer

D

Explanation

Let (h,k) be the point on the locus. Then by the given conditions,
`(h-a_1)^1+(k-b_1)^2=(h-a_2)^2+(k-b_2)^2`
or `2h(a_1-a_2)+2k(b_1-b_2)+a_2^2+b_2^2-a_1^2=0`
`h(a_1-a_2 )+k(b_1-b_2)+(1)/(2)(a_2^2+b_2^2-a_1^2-b_1^2)=0` .....(1)
Also, since (h,k) lies on the given locus, we have
`(a_1-a_2)h+(b_1-b_2)k+c=0`
Comparing (1) and (2) , we get
`c=(1)/(2) (a_2^2+b_2^2-a_2^2-b_1^2)`