Toronto Math Forum
MAT3342018F => MAT334Tests => Quiz1 => Topic started by: Victor Ivrii on September 28, 2018, 04:10:39 PM

$\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}$
Write (in complex number notation) the equation of the perpendicular bisector of the line segment joining $1+2i$ and $1  2i$.

z(1+2i) = z(12i)

Let $z=x+iy$ denote an arbitrary point on the perpendicular bisector. Then we have $$z(1+2i) = z(12i) $$
Simplify this we get $$(x+1)+i(y2) = (x1) + i(y+2)$$
Square both sides we have $$(x+1)+i(y2)^2 = (x1) + i(y+2)^2$$
Which is $$(x+1)^2+(y2)^2=(x1)^2+(y+2)^2$$
Expand the squares we get $$x^2+2x+1+y^24y+4=x^22x+1+y^2+4y+4$$
Cancel same terms from both sides we have $$4x=8y$$
That is $$y=\frac{1}{2}x$$
According to the formula given on page 13 of the textbook, equation of this perpendicular bisector is $$Re\left(\left(\frac{1}{2}+i\right)z\right) = 0$$ where $z\in\mathbb{C}$

Let $p = 1+2i$ and $q=12i$
The perpendicular bisector is the set of points equidistant to points $p$ and $q$. The distance between some point, $z$ and $p$ is $zp$. Similarly, The distance between some point, $z$ and $q$ is $zq$.
Thus the set of points equidistant to both $p$ and $q$ is given by the equation:
$zp = zq$
Another way to do this is solving for the equation of the line in xy coordinates first. The points are $(1, 2) and (1, 2)$, the midpoint is $(0, 0)$ and the slope is $m_1=2$. Thus the perpendicular bisector has the slope $m_2 = \frac{1}{m_1} = \frac{1}{2}$. Therefore, the perpendicular bisector has the equation:
$Re[(\frac{1}{2} + i)z]$

Xier, it is not a proper equation by a line.