Equation of the Right Bisector of a Line

Let $$A\left( {{x_1},{y_1}} \right)$$ and $$B\left( {{x_2},{y_2}} \right)$$ be the ends of a segment, then the slope $${m_1}$$ of the line joining $$A$$ and $$B$$ is
\[{m_1} = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}\]

The midpoint of the segment $$AB$$ can be found as \[C\left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)\]

The slope $$m$$ of any line perpendicular to the segment $$AB$$ is
\[m = – \frac{1}{{{m_1}}} = – \frac{{{x_2} – {x_1}}}{{{y_2} – {y_1}}}\]

The equation of the perpendicular bisector of the segment$$AB$$ being the equation of a line through $$C$$ and perpendicular to $$AB$$ using the slope-point form is
\[y – \frac{{{y_1} + {y_2}}}{2} = – \frac{{{x_2} – {x_1}}}{{{y_2} – {y_1}}}\left( {x – \frac{{{x_1} + {x_2}}}{2}} \right)\]

 

Example: Find the equation of the perpendicular bisector of $$A\left( {1,2} \right)$$ and $$B\left( {5, – 1} \right)$$.

The equation of the perpendicular bisector of $$AB$$ is
\[y – \frac{{{y_1} + {y_2}}}{2} = – \frac{{{x_2} – {x_1}}}{{{y_2} – {y_1}}}\left( {x – \frac{{{x_1} + {x_2}}}{2}} \right)\]

Putting all these values in the equation, we have
\[\begin{gathered} y – \frac{{2 – 1}}{2} = – \frac{{5 – 1}}{{ – 1 – 2}}\left( {x – \frac{{1 + 5}}{2}} \right) \\ \Rightarrow y – \frac{1}{2} = \frac{4}{3}\left( {x – 3} \right) \\ \Rightarrow 8x – 6y – 21 = 0 \\ \end{gathered} \]