Question

The locus of a point which is always equidistant from two fixed points A and B is:

• A. A straight line which is parallel to the fine AB
• B. A straight line which is perpendicular to the line AB
• C. A straight line which passes through the midpoint of AB
• D. The perpendicular bisector of the line AB

Answer 1

The correct option is D The perpendicular bisector of the line AB

R.E.F image

Let the mid pt.of 'AB' option

Let 'P' be only any pt.has $(x,y)$ coordinate

$AP=\sqrt{(x+a{)}^2+y^2}$ & $BP=\sqrt{(x-a{)}^2+y^2}$

then $AP=BP$ iff $x=0$

$\Rightarrow$ locus of pt.'P' is $\perp$ bisector of AB.