Finding Perpendicular Bisector

**adam182** · November 11th 2006, 10:27 AM

Question: Find the perpendicular bisector of the line segment with endpoints (5,2) and (-3,6).

I don't remember the procedure to do this.

**topsquark** · November 11th 2006, 10:50 AM

Originally Posted by adam182

Question: Find the perpendicular bisector of the line segment with endpoints (5,2) and (-3,6).

I don't remember the procedure to do this.

There is more than one bisector, so there is no unique solution to this. The key is that slope of the perpendicular bisector (m2) is the negative inverse of the slope of the line (m1), ie. $m_2 = -\frac{1}{m_1}$ .

For example, the slope of your line is:
$m = \frac{6 - 2}{-3 -5} = \frac{3}{-8} = -\frac{3}{8}$

So the slope of the perpendicular bisector is
$m' = - \frac{1}{m} = - \frac{1}{- \frac{3}{8} } = \frac{8}{3}$

-Dan

**earboth** · November 11th 2006, 12:35 PM

Originally Posted by adam182

Question: Find the perpendicular bisector of the line segment with endpoints (5,2) and (-3,6).
I don't remember the procedure to do this.

Hello, adam,

I'm not quite sure if I found the right explanation:
Perpendicular bisector passes through the midpoint of the line and has the perpendicular direction:

M((5+(-3))/2,(2+6)/2). So M(1,4)

The slope of the perpendicular bisector has been calculated by topsquark.
Now use point-slope-formula of a straight line:

$\frac{y-4}{x-1}=\frac{8}{3}$ . Solve for y:

$y = \frac{8}{3} \cdot x+\frac{4}{3}$

EB

**topsquark** · November 11th 2006, 02:01 PM

Perpendicular bisector. The key word was in my face the whole time! Oops! Thank you, earboth, for finishing that.

-Dan

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