Find the equation of the right bisector of the line segment joining the points

Question:

Find the equation of the right bisector of the line segment joining the points (3, 4) and (1, 2).

Solution:

The right bisector of a line segment bisects the line segment at $90^{\circ}$.

The end-points of the line segment are given as $A(3,4)$ and $B(-1,2)$.

Accordingly, mid-point of $A B=\left(\frac{3-1}{2}, \frac{4+2}{2}\right)=(1,3)$

Slope of $A B=\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}$

$\therefore$ Slope of the line perpendicular to $\mathrm{AB}=\frac{1}{\left(\frac{1}{2}\right)}=-2$

The equation of the line passing through (1, 3) and having a slope of –2 is

$(y-3)=-2(x-1)$

$y-3=-2 x+2$

$2 x+y=5$

Thus, the required equation of the line is 2x + y = 5.