Equation of 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 finding as

C\left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)


The slope m of any perpendicular to the segment AB is

m =  - \frac{1}{{{m_1}}} =  - \frac{{{x_2} - {x_1}}}{{{y_2} - {y_1}}}


Equation of perpendicular bisector of the segmentAB, being the equation of line through C and perpendicular to AB using 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 perpendicular bisector of A\left( {1,2} \right) and B\left( {5, - 1} \right).
Equation of 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}

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