# 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
$\begin{gathered} y – \frac{{2 – 1}}{2} = – \frac{{5 – 1}}{{ – 1 – 2}}\left( {x – \frac{{1 + 5}}{2}} \right) \\ \Rightarrow y – \frac{1}{2} = \frac{4}{3}\left( {x – 3} \right) \\ \Rightarrow 8x – 6y – 21 = 0 \\ \end{gathered}$