Cartesian Equation for the perpendicular bisector of a line 
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! 
 A: 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....
A: Hints:


*

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

*Recall the relationship of the slopes of two perpendicular lines

*Use the point-slope form of the equation of a line.
A: 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 $\rm\; 6y =\; 4x+23$
A: 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:


*

*Find the vector $\overrightarrow{AB}=\langle x_a,y_a\rangle$, which is perpendicular to the line you want.

*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.

*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 $x_ax+y_ay=x_ax_m+y_ay_m$.

A: 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 $(y-\frac{9}{2})=\frac{2}{3}(x-1)$.
A: 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 $$y-\frac{9}{2}=\frac{2}{3}(x-1)$$ $$\implies 6y-27=4x-4$$ $$\implies \color{blue}{4x-6y+23=0}$$
