Perpendicular bisector question

[mindauggas]

Homework Statement

(b) Show that the equation of the perpendicular bisector of the line segment with endpoints (x$_{1}$, y$_{1}$) and (x$_{2}$,y$_{2}$) can be written as $\frac{x-x^{-}}{y_{2}-y_{1}}$+$\frac{y-y^{-}}{x_{2}-x_{1}}$= 0, where (x$^{-}$,y$^{-}$)are coordinates of the midpoint of the segment

The Attempt at a Solution

Because:

(1) $\frac{x-x^{-}}{y_{2}-y_{1}}$+$\frac{y-y^{-}}{x_{2}-x_{1}}$= 0

We have:

(2)$\frac{x-x^{-}}{y_{2}-y_{1}}$=-$\frac{y-y^{-}}{x_{2}-x_{1}}$

Does this the minus sing in front of the term on the right side of the equation express the relative slope of the perpendicular bisector? (Relative to the line segment).

I also realized that the perpendicular bisector formula does not work, or it seems to me, when line segment is parallel to x axis, hence y=y$^{-}$ which makes the first term undefined. Hence the formula does not work when the line segment is perpendicular to the y-axis for similar reason.

Yeat still I don't know how to show the equation to be true in special cases where y$\neq$y$^{-}$

I reason than symmetry has to be somehow involved, but is it?

 

[ehild]

That perpendicular bisector is a line which goes through a point and perpendicular to a given direction. (which point, what direction?) What is the general formula of a straight line?

ehild

 

[mindauggas]

That perpendicular bisector is a line which goes through a point and perpendicular to a given direction. (which point, what direction?) What is the general formula of a straight line?

ehild

(1) it goes through the mid point of the given line segment ... in my case it's coordinates are (x$^{-}$,y$^{-}$)

(2) it is perpendicular to the direction of the line segment (x$_{1}$,y$_{1}$) to (x$_{2}$,y$_{2}$)

(3)the general formula of a straight line has many forms, but, presumably, you are talking about $y=mx+b$ (the slope-intercept form)

Probably i would get more use out of the "intercept form":
$\frac{x}{a}$+$\frac{y}{b}$=1 ... ?

 

[ehild]

I vote for the first one. Can you find m?

And you were right, the given bisector equation does not work when the segment is parallel with one of the axis. So handle these cases separately.


ehild

 

[mindauggas]

$y=mx+b$

$y-b=mx$

$\frac{y-b}{x}$$=m$

But m is just $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

So we have $\frac{y-b}{x}$=$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Don't know what to do next ...

Seems like this equation $\frac{x-x^{-}}{y_{2}-y_{1}}$+$\frac{y-y^{-}}{x_{2}-x_{1}}$= 0 is a amalgam of two slope expresions. This is confirmed by noticing that they are equal but opposite in sign ... but how to SHOW it ?

Also, the terms have to have the same numerical value ... also don't know how to show it.

 

[ehild]

(y2-y1)/(x2-x1) is the slope of the original line segment. The bisector is perpendicular to it. What do you know about the slopes of perpendicular lines?

ehild

 

[mindauggas]

Two non-vertical lines with slopes m1 and m2 are perpendicular if and only if the product their slopes is -1, that is:

(1) $m1m2=-1$

But then I also know that the slope of P.B. is

(2) $\frac{y-y^{-}}{x-x^{-}}$ (since I only know that (y$^{-}$,x$^{-}$) and (x,y) are points on it)

Combining (1), (2) and the slope of the line segment we get:

$\frac{y-y^{-}}{x-x^{-}}$*$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$=-1

Diving both sides by $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ we get

$\frac{y-y^{-}}{x-x^{-}}$=$-1/\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

But this is the same as:

$\frac{y-y^{-}}{x-x^{-}}$=-$\frac{x_{2}-x_{1}}{y_{2}-y_{1}}$

 

[ehild]

But this is the same as:

$\frac{y-y^{-}}{x-x^{-}}$=-$\frac{x_{2}-x_{1}}{y_{2}-y_{1}}$

and the same as the formula you had to derive. You have done it!

You can also write it in the form (x₂-x₁)(x-x^-)=(y₂-y₁)(y-y^-), and this is valid for any case. If y₂=y₁, the equation of the bisector is = (x-x^-)=0, and it is (y-y^-)=0 in case x₂=x₁.

ehild

 

[mindauggas]

Yes, I see now ... they are the same, because you can get one from the other just by algebraic manipulation.

Thank you very much :)

 

[ehild]

You are welcome.


ehild