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

The right bisector of a line segment bisects the line segment at 90°.

End-points of the line segment AB are given as A (3, 4) and B (–1, 2).

Let mid-point of AB be (x, y)

x = \( \dfrac{3-1}{2}\)= \( \dfrac{2}{2}\) = 1

y = \( \dfrac{4+2}{2}\) = \( \dfrac{6}{2}\) = 3

(x, y) = (1, 3)

Let the slope of line AB be m₁

m₁ = \( \dfrac{2-4}{-1-3}\)

= \( \dfrac{-2}{-4}\)

= \( \dfrac{1}{2}\)

And let the slope of the line perpendicular to AB be m₂

m₂ =\( \dfrac{-1}{\dfrac{1}{2}}\)

= -2

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

(y – 3) = -2 (x – 1)

y – 3 = – 2x + 2

2x + y = 5

The required equation of the line is 2x + y = 5