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

1.	Find the equation of the perpendicular bisector of the line segment joining the points A(2, 3) and B(6, – 5).
Answer» The mid-point of A(2, 3) and B(6, – 5), = \((\frac{2+6}{2},\frac{3-5}{2})=(4, -1)\) And slope of AB, m =\(\frac{-5-3}{6-2}=-2\) ∴ Slope of the Perpendicular bisector of AB, m’ = \(\frac{-1}{m}=\frac{1}{2}\). Since the bisector passes through (4, – 1), so the equation of the Perpendicular bisector is Y – (– 1) = \(\frac{1}{2}\)( – 4) ⇒ y + 1 = \(\frac{1}{2}\)(x − 4) ⇒ x – 4 = 2y + 2 ⇒ x – 2y – 6 = 0, is the required equation.

Find the equation of the perpendicular bisector of the line segment joining the points A(2, 3) and B(6, – 5).

Answer»

The mid-point of A(2, 3) and B(6, – 5),

= \((\frac{2+6}{2},\frac{3-5}{2})=(4, -1)\)

And slope of AB, m =\(\frac{-5-3}{6-2}=-2\)

∴ Slope of the Perpendicular bisector of

AB, m’ = \(\frac{-1}{m}=\frac{1}{2}\).

Since the bisector passes through (4, – 1), so the equation of the Perpendicular bisector is

Y – (– 1) = \(\frac{1}{2}\)( – 4)

⇒ y + 1 = \(\frac{1}{2}\)(x − 4)

⇒ x – 4 = 2y + 2

⇒ x – 2y – 6 = 0, is the required equation.