Q.

If  the  equation  of  the  locus  of a point equi-distant from the points (a1,b1) and (a2,b2) is (a1–a2)x+(b1–b2)y+K= 0 then the value of K is

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a

a12+b12−a22−b22

b

12(a12+b12−a22−b22)

c

a12−a22+b12−b22

d

12(a12+a22+b12+b22)

answer is B.

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Detailed Solution

A(a1, b1),B(a2,b2),PA=PB⇒PA2=PB2   (x−a1)2+(y−b1)2=(x−a2)2+(y−b2)2 ⇒2(a1−a2)x+2(b1−b2)y+a12+b12−a22−b22=0 (a1−a2)x+(b1−b2)y+12(a12+b12−a22−b22)=0 k=12(a12+b12−a22−b22)
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If  the  equation  of  the  locus  of a point equi-distant from the points (a1,b1) and (a2,b2) is (a1–a2)x+(b1–b2)y+K= 0 then the value of K is