# How do you find the equation of the perpendicular bisector of the segment joining the points A (6, -3) and B (-2, 5)?

Sep 15, 2016

$\text{The eqn. of p.b. is } x - y - 1 = 0.$

#### Explanation:

From Geometry, we know that the perpendicular bisector (p.b.) of a

line segment (sgmt.) is the line passing through its mid-point (m.p.) &

perpendicular to the sgmt.

Let us denote, by $\overline{A B} ,$ the line sgmt. joining the pts.

$A \left(6 , - 3\right) , \mathmr{and} , B \left(- 2 , 5\right) .$

$M$ is the m.p. of $\overline{A B}$

$\Rightarrow M = M \left(\frac{6 - 2}{2} , \frac{- 3 + 5}{2}\right) = M \left(2 , 1\right) .$

Slope of $\overline{A B} = \frac{5 - \left(- 3\right)}{- 2 - 6} = \frac{8}{-} 8 = - 1.$

$\therefore \text{ The Slope of p.b. of "bar(AB)" must be 1.}$

Using the Slope-Point Form of line for the p.b., we get its eqn.

$y - 1 = 1 \left(x - 2\right) \Rightarrow x - y - 1 = 0.$

Method II:=

We use the following Geometrical Property of the p.b. of a line sgmt.

$\text{Any pt. on the p.b. of a line sgmt. is equidistant from the }$

$\text{end-points of the sgmt.}$

Let $P \left(x , y\right)$ be any pt. on the p.b. of $\overline{A B} .$

$\text{Then, dist. PA= dist. PB} \Rightarrow P {A}^{2} = P {B}^{2}$.

$\Rightarrow {\left(x - 6\right)}^{2} + {\left(y + 3\right)}^{2} = {\left(x + 2\right)}^{2} + {\left(y - 5\right)}^{2}$

$\Rightarrow {x}^{2} - 12 x + 36 + {y}^{2} + 6 y + 9$

$= {x}^{2} + 4 x + 4 + {y}^{2} - 10 y + 25$

$\Rightarrow - 16 x + 16 y + 16 = 0$

$\Rightarrow x - y - 1 = 0.$, as before!

Enjoy Maths.!