Find the equation of the perpendicular bisector of the line segment joining points $(7, 1)$ and $(3, 5)$.
Given:
Given points are $(7, 1)$ and $(3, 5)$.
To do:
We have to find the equation of the perpendicular bisector of the line segment joining points $(7, 1)$ and $(3, 5)$.
Solution:
Let the given points be $A(7,1)$ and $B(3,5)$ and the perpendicular bisector is $PQ$.
By mid-point formula,
Coordinates of the mid-point of \( \mathrm{AB} \) $O$ is,
\( =(\frac{7+3}{2}, \frac{1+5}{2})=(5,3) \)
Slope of \( \mathrm{AB}\left(m_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-1}{3-7} \)
\( =\frac{4}{-4}=-1 \)
Slope of line perpendicular to \( \mathrm{AB}\left(m_{2}\right) \)
\( =\frac{-1}{m_{1}}=\frac{-1}{-1}=1 \)
Therefore,
The equation of the perpendicular bisector is,
\( y-y_{1}=m\left(x-x_{1}\right) \)
\( \Rightarrow y-3=1(x-5) \)
\( \Rightarrow y-3=x-5 \)
\( \Rightarrow x-y=-3+5 \)
\( \Rightarrow x-y-2=0 \)
The equation of the perpendicular bisector of the line segment joining points $(7, 1)$ and $(3, 5)$ is $x-y-2=0$.
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