# How do you write the equation of the perpendicular bisector of the segment with the given endpoints (2,5) and (4,9)?

Jan 6, 2016

The equation is $y = - \left(\frac{1}{2}\right) x + \frac{17}{2}$

#### Explanation:

To to that you should find the middle point (M), the segment's slope, the perpendicular's slope and finally the equation of the line with slope perpendicular to the segment and passing through M.

Middle Point M
${x}_{M} = \frac{{x}_{A} + {x}_{B}}{2} = \frac{2 + 4}{2} = 3$
${y}_{M} = \frac{{y}_{A} + y + B}{2} = \frac{5 + 6}{2} = 7$
=> M (3,7)

Segment's slope
$k = \frac{\Delta y}{\Delta x} = \frac{{y}_{B} - {y}_{A}}{{x}_{B} - {x}_{A}} = \frac{9 - 5}{4 - 2} = 2$

Perpendicular line's slope
$p = - \frac{1}{k} = - \frac{1}{2}$

Equation of perpendicular line passing through point M
$y - {y}_{M} = p \left(x - {x}_{M}\right)$
$y - 7 = \left(- \frac{1}{2}\right) \left(x - 3\right)$
$y = \left(- \frac{1}{2}\right) x + \frac{3}{2} + 7$
$y = - \left(\frac{1}{2}\right) x + \frac{17}{2}$

Oct 7, 2016

$x + 2 y - 17 = 0$.

#### Explanation:

Let $P \left(x , y\right)$ be any arbitrary point on the $\bot$-bisector of segment

$A B$, where, $A \left(2 , 5\right) \mathmr{and} B \left(4 , 9\right) .$

Clearly, dist.$P A$=dist.$P B$

$\therefore P {A}^{2} = P {B}^{2}$

$\therefore {\left(x - 2\right)}^{2} + {\left(y - 5\right)}^{2} = {\left(x - 4\right)}^{2} + {\left(y - 9\right)}^{2}$

$\therefore - 4 x + 4 - 10 y + 25 = - 8 x + 16 - 18 y + 81$

$\therefore 4 x + 8 y - 68 = 0 , \mathmr{and} , x + 2 y - 17 = 0$

Since, this relation holds for any, and, hence for every pt. on the

$\bot -$bisector of the seg, it is the reqd. en. of the $\bot -$bisector.