Question

The locus of a point equidistance from two given points $a$ and $b$ is given by
A. $\left[ r-\dfrac{1}{2}(a+b) \right]\cdot (a-b)=0$
B. $\left[ r-\dfrac{1}{2}(a-b) \right]\cdot (a+b)=0$
C. $\left[ r-\dfrac{1}{2}(a+b) \right]\cdot (a+b)=0$
D. $\left[ r-\dfrac{1}{2}(a-b) \right]\cdot (a-b)=0$

Answer 1

Hint: In this question, we are to find the locus of a point equidistance from two given points. In order to find this, the given position vectors are used, and applying the dot product, we get the required equidistance.



Formula Used:If $a$ and $b$ are the position vectors of the points $A$, $B$ respectively, then the position vector of the midpoint of $\overline{AB}$ is $\dfrac{a+b}{2}$.
Here $a$ and $b$are two non-zero vectors, then
$(a,b)=90{}^\circ \Leftrightarrow a\cdot b=0$



Complete step by step solution:Consider given points as $A$ and $B$ and their position vectors are $a$ and $b$ respectively.
To find the locus of the equidistance from the given points, the mid-point of the given points is to be calculated.
I.e., the position vector of the mid-point of $\overline{AB}$ is
$M=\dfrac{1}{2}(a+b)$
Consider a point $P$ which is the locus of a point equidistance from the given points.
So, we can draw a perpendicular line from the mid-point of the line $\overline{AB}$ .
I.e., $\overline{PM}$ whose position vectors is $[r-\dfrac{1}{2}(a+b)]$.
Thus, we can write,
\[\begin{align}
  & \overrightarrow{PM}\bot \overrightarrow{BA} \\
 & \Rightarrow \overrightarrow{PM}\cdot \overrightarrow{BA}=0\text{ }...(1) \\
\end{align}\]
On substituting the position vectors, we get
$[r-\dfrac{1}{2}(a+b)]\cdot (a-b)=0$



Option ‘A’ is correct



Note: Here we need to remember that, the given points are position vectors. So, we need to apply the vector theory and we can able to find the required locus of the point equidistance from the given points.