Equation of the perpendicular bisector of the line joining the points whose position vectors are…

1. \((2 \mathrm{r}-\mathrm{a}-\mathrm{b}) \cdot(\mathrm{a}-\mathrm{b})=0\)
2. \((2 \mathrm{r}-\mathrm{a}-\mathrm{b}) \cdot(\mathrm{a}+\mathrm{b})=0\)
3. \((2 \mathrm{r}+\mathrm{a}+\mathrm{b}) \cdot(\mathrm{a}-\mathrm{b})=0\)
4. \((2 \mathrm{r}-\mathrm{a}+\mathrm{b}) \cdot(\mathrm{a}+\mathrm{b})=0\)

Solution

The mid-point of line joining points whose position vectors are \(\mathbf{a}\) and \(\mathbf{b}\) is \(M\left(\frac{\mathbf{a}+\mathbf{b}}{2}\right)\) and the direction ratio vector of line joining of given points is \((\mathbf{a}-\mathbf{b})\).
Let a variable point \(p(\mathbf{r})\) on the perpendicular bisector of \(A B\), so \(\mathbf{M P} \perp \mathbf{B A}\) \(\begin{array}{ll} \Rightarrow & \left(\mathbf{r}-\frac{\mathbf{a}+\mathbf{b}}{2}\right) \cdot(\mathbf{a}-\mathbf{b})=0 \\ \Rightarrow & (2 \mathbf{r}-\mathbf{a}-\mathbf{b}) \cdot(\mathbf{a}-\mathbf{b})=0 \end{array}\)

Asked in: AP EAMCET 2020 (18 Sep Shift 1)

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