The locus of a point equidistant from two given points whose position vectors are a and b is equal to

Options

A. `[r-(1)/(2)(a+b)]. (a + b) =0`
B. `[r-(1)/(2)(a+b)]. (a-b) = 0`
C. `[r-(1)/(2) (a + b) ] .a = 0`
D. `[r-(a+b)].b=0`

Answer

To find the locus of a point that is equidistant from two given points with position vectors **a** and **b**, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Position Vector of the Locus Point**:
Let the position vector of the point we are looking for be **R**.

2. **Set Up the Equidistance Condition**:
The condition for the point to be equidistant from points **A** and **B** is given by:
\[ |R - A| = |R - B| \]
This means the distance from **R** to **A** is equal to the distance from **R** to **B**.

3. **Square Both Sides**:
To eliminate the square root, we square both sides:
\[ |R - A|^2 = |R - B|^2 \]

4. **Expand Both Sides**:
Using the property of dot products, we can expand both sides:
\[ (R - A) \cdot (R - A) = (R - B) \cdot (R - B) \]
This expands to:
\[ R \cdot R - 2R \cdot A + A \cdot A = R \cdot R - 2R \cdot B + B \cdot B \]

5. **Simplify the Equation**:
Cancel \(R \cdot R\) from both sides:
\[ -2R \cdot A + A \cdot A = -2R \cdot B + B \cdot B \]

6. **Rearrange the Terms**:
Rearranging gives:
\[ 2R \cdot B - 2R \cdot A = B \cdot B - A \cdot A \]
Simplifying further:
\[ 2R \cdot (B - A) = B \cdot B - A \cdot A \]

7. **Express R**:
Now, we can express **R**:
\[ R \cdot (B - A) = \frac{1}{2}(B \cdot B - A \cdot A) \]

8. **Final Form**:
This can be rewritten in terms of the average of **A** and **B**:
\[ R = \frac{1}{2}(A + B) + k(B - A) \]
where \(k\) is a scalar parameter.

### Conclusion:
The locus of the point **R** that is equidistant from points **A** and **B** is a line that bisects the segment joining **A** and **B** and is perpendicular to it.