Given:
End points a line segment are (9,-8) and (-1,-4).
To find:
The equation of perpendicular bisector of given line.
Solution:
Slope of given line is
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
[tex]m_1=\dfrac{-4-(-8)}{-1-9}[/tex]
[tex]m_1=\dfrac{-4+8}{-10}[/tex]
[tex]m_1=\dfrac{4}{-10}[/tex]
[tex]m_1=-\dfrac{2}{5}[/tex]
Product of slopes of two perpendicular line is -1.
[tex]m_1\times m_2=-1[/tex]
[tex](-\dfrac{2}{5})\times m_2=-1[/tex]
[tex]m_2=\dfrac{5}{2}[/tex]
So, slope of perpendicular bisector is [tex]\dfrac{5}{2}[/tex].
Perpendicular bisector passes through the midpoint of given endpoints.
[tex]Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)[/tex]
[tex]Midpoint=\left(\dfrac{9+(-1)}{2},\dfrac{(-8)+(-4)}{2}\right)[/tex]
[tex]Midpoint=\left(\dfrac{8}{2},\dfrac{-12}{2}\right)[/tex]
[tex]Midpoint=\left(4,-6\right)[/tex]
So, perpendicular bisector passes through (4,-6) and having slope [tex]\dfrac{5}{2}[/tex]. The equation of perpendicular bisector is
[tex]y-y_1=m(x-x_1)[/tex]
where, m is slope.
[tex]y-(-6)=\dfrac{5}{2}(x-4)[/tex]
[tex]y+6=\dfrac{5}{2}(x)+\dfrac{5}{2}(-4)[/tex]
[tex]y+6=\dfrac{5}{2}x-10[/tex]
[tex]y=\dfrac{5}{2}x-10-6[/tex]
[tex]y=\dfrac{5}{2}x-16[/tex]
Therefore, the equation of perpendicular bisector is [tex]y=\dfrac{5}{2}x-16[/tex].
