Find the equation of the perpendicular bisector of the line segment whose

Solution:

Perpendicular bisector: A perpendicular bisector is a line segment which is perpendicular to the given line segment and passes through its mid - point (or we can say bisects the line segment).

Now to find the equation of perpendicular bisector first, we will find mid - point of the given line using mid - point formula (call it midpoint as M),

$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right)$

coordinates of $\mathrm{M}=\left(\frac{10+(-4)}{2}, \frac{4+9}{2}\right) \Rightarrow\left(3, \frac{13}{2}\right)$

Now we will calculate the slope of the given line and since lines are perpendicular, so the slope of two is related as m1.m2 = - 1.

Slope of $\mathrm{AB}: \mathrm{m}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}} \Rightarrow \frac{9-4}{-4-10}=-\frac{5}{14}$

Now the slope of perpendicular bisector is

$\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1 \Rightarrow-\frac{5}{14} \cdot \mathrm{m}_{2}=-1$

$m_{2}=\frac{14}{5}$

Now equation of perpendicular bisector using two point form,

$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$

$y-\frac{13}{2}=\frac{14}{5}(x-3) \Rightarrow 5(2 y-13)=28(x-3)$

$10 y-65=28 x-84$

$28 x-10 y-84+65=0$

$28 x-10 y-19=0$

So, required equation of perpendicular bisector 28x - 10y - 19 = 0.