I can't seem to find a formula to find the perpendicular bisector of a line. Anyone know what it is?

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- Apr 24th 2013, 07:50 AMalexpasty2013Perpendicular bisectors
I can't seem to find a formula to find the perpendicular bisector of a line. Anyone know what it is?

- Apr 24th 2013, 08:01 AMPlatoRe: Perpendicular bisectors
- Apr 24th 2013, 08:42 AMalexpasty2013Re: Perpendicular bisectors
So we have 2 points A (-2, 2) and B (3 , 1) the the perpendicular bisector is;

AB is y = (-(-2)-3/-2+1) (x- (-2)+3/2) + (-2+1/2) =

y = -5/3(x-1/2)+ (-1/2)

and our final equation is

y= -5x-1

is that right? - Apr 24th 2013, 06:28 PMProve ItRe: Perpendicular bisectors
The slope of the line through A and B is . So the slope of your perpendicular bisector is 5, since the two slopes will multiply to give -1.

The midpoint of AB is

So the perpendicular bisector is of the form with and a point it passes through is . Substituting these gives

and so the equation of your perpendicular bisector is . - Apr 24th 2013, 09:49 PMalexpasty2013Re: Perpendicular bisectors
So I got the right answer by the wrong means?

- Apr 24th 2013, 10:18 PMalexpasty2013Re: Perpendicular bisectors
this is wrong - the point (1/2, 3, 2) does not lie on the line, my original point (0.5, -0.5) is on the line and apparently in the centre

- Apr 25th 2013, 02:20 AMibduttRe: Perpendicular bisectors
You cannot find the bisector of a line for you cannot measure a line.