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If the equation of the locus of a point ...

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_2+b_2)y+c=0`, then the value of C is

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)`

Text Solution

Verified by Experts

The correct Answer is:
D

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)`
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