Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the Cartesian equation for the perpendicular bisector of the line joining A(2,3) and B(0,6)

How do I do this?

Thank you!

share|improve this question

5 Answers 5

up vote 2 down vote accepted

HINT $\;$ The equation is $\rm\;\: 2\ (A-B)\cdot (x,y) \;=\; |A|-|B|\;\;\;$ where $\rm\;\;\; |(a,b)| \ =\ a^2 + b^2$

which, if worked out, yields $\rm\;\: (-4,6)\cdot (x,y) \;=\; \;36 \;- 13\;\;\:$ for $\rm\; A = (0,6),\;\; B = (2,3)$

which, after simplifying, yields the equation $\rm\; 6y =\; 4x+23$

share|improve this answer
    
I'm confused where the 36 and 13 came from, could you help with this please? –  Charlotte Sep 18 '10 at 17:12
    
Thank you so much! :) –  Charlotte Sep 18 '10 at 17:27
    
@Charlotte: those two quantities are the squares of the distances of $A$ and $B$ from the origin. –  J. M. Sep 19 '10 at 4:19

Hints:

  1. Get the slope and the midpoint of the segment joining your two given points.

  2. Recall the relationship of the slopes of two perpendicular lines

  3. Use the point-slope form of the equation of a line.

share|improve this answer

The perpendicular bisector of the segment $AB$ is the locus of points $P$ equidistant from $A$ and $B$, that is $|AP|=|BP|$. It's easier to consider the equation $|AP|^2=|BP|^2$ which, when $A=(a,b)$, $B=(c,d)$ and $P=(x,y)$ becomes $$(x-a)^2+(y-b)^2=(x-c)^2+(y-d)^2$$ and can be simplified further....

share|improve this answer
2  
Nice. But you ought to at least mention that the square terms for both x and y will cancel out, so this will indeed turn out to be a linear equation. –  David Lewis Sep 18 '10 at 16:19
    
This has a nice geometric interpretation: the perpendicular bisector of two points $A$ and $B$ is the radical line (mathworld.wolfram.com/RadicalLine.html ) of two intersecting circles with identical radii and centered at $A$ and $B$. –  J. M. Sep 19 '10 at 4:17

Generally, J.M.'s answer is what I'd suggest at the high school level.

However, this is a somewhat common problem on timed math contests (or part of a problem), and in that settings, I'd take advantage of the fact that in the form $ax+by=c$, $\langle a,b\rangle$ is a vector perpendicular to the line:

  1. Find the vector $\overrightarrow{AB}=\langle x_a,y_a\rangle$, which is perpendicular to the line you want.
  2. The midpoint of $\overline{AB}$ is $A+\frac{1}{2}\overrightarrow{AB}=\langle x_m,y_m\rangle$, which is a point on the line.
  3. An equation for the line is $\langle x_a,y_a\rangle\cdot\langle x,y\rangle=\langle x_a,y_a\rangle\cdot\langle x_m,y_m\rangle$ or $x_ax+y_ay=x_ax_m+y_ay_m$.
share|improve this answer

Here is a hint. The mid point of the line joining $A$ and $B$ is $(1, \frac{9}{2})$. Use this to find the equation of the line passing through this point.

As, J.M says the product of the slopes of the perpendiculars is $-1$.

Slope of the line passing through $A$ and $B$ is $\displaystyle m_{1}= \frac{6-3}{0-2} = -\frac{3}{2}$. Therefore the slope of the line perpendicular to it is $m_{2}= \frac{2}{3}$.

Hence the required equation of the line is $(y-\frac{9}{2})=\frac{2}{3}(x-1)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.