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Given that I know the point coordinates of A and B on segment AB and the expected length of a perpendicular segment CD crossing the middle of AB, how do I calculate the point coordinates of segment CD?

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please inform whether the assumption that the midpoint of $[CD]$ belongs to $[AB]$ is correct. –  Américo Tavares Oct 2 '10 at 19:45
    
Yes, it is indeed. –  Quickredfox Oct 5 '10 at 19:35

3 Answers 3

up vote 2 down vote accepted

Remark: Let $M$ be the midpoint of $[CD]$. I assume that $M$ belongs to $[AB]$, as shown above. If it is not the case, then this is only a particular solution. [The assumption is correct, as commented by OP].


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Let $A\left( x_{A},y_{A}\right) ,B(x_{B},y_{B}),C(x_{C},y_{C}),D(x_{D},y_{D}) $ be the four points. By translation and rotation of the $x,y$ axes as shown in the figure, we can center $[A,B,C,D]$. The translation corresponds to the change of variables

$$x=\overline{x}+X,\qquad y=\overline{y}+Y,$$

where

$$\overline{x}=\dfrac{x_{A}+x_{B}}{2}\qquad\overline{y}=\dfrac{y_{A}+y_{B}}{2},$$

and the rotation corresponds to

$$X=x^{\prime }\cos \theta -y^{\prime }\sin \theta,\qquad Y=x^{\prime }\sin \theta +y^{\prime }\cos \theta ,$$

where

$$\theta =\arctan \dfrac{y_{B}-y_{A}}{x_{B}-x_{A}}.$$

Combining both transformations, we have

$$x=\overline{x}+x^{\prime }\cos \theta -y^{\prime }\sin \theta,\qquad y=\overline{y}+x^{\prime }\sin \theta +y^{\prime }\cos \theta .$$

Denoting the distance between the points $C$ and $D$ by $\overline{CD}=d_{CD} $, we have

$$x_{C}^{\prime }=x_{D}^{\prime }=0\qquad y_{C}^{\prime }=\dfrac{d_{CD}}{2}\qquad y_{D}^{\prime }=-\dfrac{d_{CD}}{2}.$$

Using the above transformations we get

$$C=\left( x_{C},y_{C}\right) =\left( \overline{x}-\dfrac{d_{CD}}{2}\sin \theta ,\overline{y}+\dfrac{d_{CD}}{2}\cos \theta \right) $$

$$D=\left( x_{D},y_{D}\right) =\left( \overline{x}+\dfrac{d_{CD}}{2}\sin \theta ,\overline{y}-\dfrac{d_{CD}}{2}\cos \theta \right) ,$$

where $\overline{x}$ and $\theta $ are given above.

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you changed the original question to imply that AB bisects CD, but the original asker had not exactly specified that... –  Eugene Bulkin Oct 2 '10 at 19:23
    
I added a remark concerning this assumption. If it is not correct, then one has to know the coordinates of $C$ or $D$ or other equivalent information. Also, we can change the position of $[CD]$ to adjust to $x_{C}^{\prime }=x_{D}^{\prime }=0$ and $y_{D}^{\prime }=y_{C}^{\prime }-d_{CD}$ –  Américo Tavares Oct 2 '10 at 20:09
    
The assumption is correct, I accept this answer because it's obviously the most detailed one here and of course, works for me. I'm also attributing a point to Eugene for the extra effort. Thank you all! (Sorry Eugene, apparently I cant vote) –  Quickredfox Oct 5 '10 at 19:41

Knowing the coordinates of A and B, you can find the slope of the line AB and the coordinates of the midpoint M of segment AB. The slopes of perpendicular lines have product -1, so you can use the slope of AB to find the slope of CD. With this slope and the point M, you can write an equation for the line CD.

The problem here is that the points C and D could be anywhere on that line, so long as they are the given length apart, so I suspect there is a bit of information missing. If M is the midpoint of CD as well (so that AB and CD are the perpendicular bisectors of each other), then C and D are each half the given length from M, so you can write an equation for the circle with radius half the length and center M, and find where the line CD intersects the circle (solve the system of the two equations).

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Well, I'm not too good at proving things, but here goes my attempt. This is a general result for when we don't know one of the endpoints of $\overline{CD}$.

We have points $A(a_x,a_y)$ and $B(b_x,b_y)$, and length of the perpendicular bisector $d$. Let the midpoint of $\overline{AB}$ be $\displaystyle M\left(\frac{a_x+b_x}{2},\frac{a_y+b_y}{2}\right)$, and let the point $Q$ be the point $d$ units away from $M$ along the perpendicular bisector $\overline{QM}$.

The slope of $\overline{AB}$ is $\displaystyle\frac{b_y-a_y}{b_x-a_x}$, so the slope of the perpendicular bisector is $\displaystyle -\frac{b_x-a_x}{b_y-a_y}$. The equation of the perpendicular bisector in vector form is the collection of all points $P$ where $P(t)=M+\langle a_y-b_y,b_x-a_x \rangle t$, $t \in \mathbb{R}$ ($M$ being the midpoint of $\overline{AB}$). Now, making the vector into a unit vector is much easier, and since the magnitude of the vector ($\sqrt{(a_y-b_y)^2+(b_x-a_x)^2}$) is simply the length $AB$, we can say that the point equation is $P(t)=M+\frac{1}{AB}\langle a_y-b_y,b_x-a_x \rangle t$, and thus the point $Q$ would be the point returned by the equation when $t=d$.

All possibilities for the points $C$ and $D$ can be found by finding the right number for $t$; namely, the first point $P(t)$ on the range of $t \in [-d,0]$ ($t = 0$ making it the midpoint of $\overline{AB}$), and the second point then being $P(t+d)$.

Correct me in any way necessary! I feel like that $P(t)$ equation will give you all the points possible if you use the correct range.

Note: I think an advantage of this method is that if the question were posed something like finding the endpoints of the perpendicular bisector with a certain length, where $C$ was, say, $\frac{2}{7}$ths of the distance along the line away, all it would take is plugging in $t=\pm\frac{2d}{7},\mp\frac{5d}{7}$.

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It seems to me that your ideas still work when $[C,D]$ and $[A,B]$, however perpendicular to each other, have no common point, don't they?. –  Américo Tavares Oct 2 '10 at 21:54
    
that's the reason for the restriction on $t$. they work for any two points $C$ and $D$ that aren't on the line, but it's also simple to find the points that are on the bisector with that length. –  Eugene Bulkin Oct 3 '10 at 5:57

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