Question

Find the locus of a point equidistant from two given points

Answer 1

(i) Let a, b be the position vectors of the given points A, B, with reference to any origin, O.
If r be the position vector of any point P on the locus, we have
$P{A}^2=P{B}^2$
$\Rightarrow (r-a{)}^2=(r-b{)}^2$
$\Rightarrow -2r.a+a^2=-2r.b+b^2$
$\Rightarrow r.(a-b)=\frac{1}{2}(a^2-b^2)=\frac{1}{2}(a+b)(a-b)$
$\Rightarrow [r-\frac{1}{2}(a+b\left)\right]\cdot (a-b)=0$
Thus, the required locus is the plane bisecting the line AB normally.