Question: Find the equation of the perpendicular bisector of the straight line segment joining the points (3,2) and (7,6).

Solution:

Let the straight line be AB.
Coordinates of point A are (3,2)
Coordinates of point B are (7,6)
Let (3,2) = ($x_1,y_1$) and (7,6) = ($x_2,y_2$)

Slope of line AB ($m_1$) = $\dfrac{y_2-y_1}{x_2-x_1}$
$= \dfrac{6-2}{7-3}$
$= \dfrac{4}{4}$
$= 1$

Given,
A straight line bisects AB making an angle of 90°.
If the line bisects AB, the point where AB is bisected is the midpoint of AB
$or, (x,y) = \left ( \frac{x_1+x_2}{2} , \frac{y_1+y_2}{2} \right )$
$or, (x,y) = \left ( \frac{3+7}{2} , \frac{2+6}{2} \right )$
$or, (x,y) = (5,4)

Also, line AB and the bisector are perpendicular so the product of their slopes is negative 1.
$or, m_1 × m_2 = -1$

$or, 1 × m_2 = -1$

$\therefore m_2 = -1$

We have,
For the perpendicular bisector,
Slope ($m_2$) = -1
Points ($x_1,y_1$) = (5,4)

Equation of the line is given by,

$or, y -y_1 = m_2(x - x_1)$
$or, y -4 = -1 (x -5)$
$or, y -4 = -x +5$
$or, x + y -4 -5= 0$
$or, x + y -9 = 0$ is the required equation.

Therefore, the required equation of the perpendicular bisector is x+y-9=0.