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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!

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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 $\rm\; 6y =\; 4x+23$

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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


  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.

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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....

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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 ( ) 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 $x_ax+y_ay=x_ax_m+y_ay_m$.
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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 $(y-\frac{9}{2})=\frac{2}{3}(x-1)$.

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Notice, the slope of the line joining the given points $(2, 3)$ & $(0, 6)$ $$=\frac{6-3}{0-2}=\frac{-3}{2}$$ Now, the slope of the perpendicular bisector $$=\frac{-1}{\text{Slope of line}\space AB}=\frac{-1}{\frac{-3}{2}}=\frac{2}{3}$$ Now, the perpendicular bisector will pass through the mid-point of the line joining points $(2, 3)$ & $(0, 6)$ i.e. $\left(\frac{2+0}{2}, \frac{3+6}{2}\right)\equiv\left(1, \frac{9}{2}\right)$ Hence, the cartesian equation of the perpendicular bisector having slope $\frac{2}{3}$ & passing through the point $\left(1, \frac{9}{2}\right)$ is given as $$y-\frac{9}{2}=\frac{2}{3}(x-1)$$ $$\implies 6y-27=4x-4$$ $$\implies \color{blue}{4x-6y+23=0}$$

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