### Author Topic: Q1: TUT 0201 and TU 0202  (Read 3274 times)

#### Victor Ivrii ##### Q1: TUT 0201 and TU 0202
« 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 $1 - 2i$.

#### Xier Li

• Newbie
• • Posts: 2
• Karma: 2 ##### Re: Q1: TUT 0201 and TU 0202
« Reply #1 on: September 28, 2018, 05:38:47 PM »
|z-(-1+2i)| = |z-(1-2i)|

#### Junya Zhang

• Full Member
•   • Posts: 27
• Karma: 29 ##### Re: Q1: TUT 0201 and TU 0202
« Reply #2 on: September 28, 2018, 06:00:01 PM »
Let $z=x+iy$ denote an arbitrary point on the perpendicular bisector. Then we have $$|z-(-1+2i)| = |z-(1-2i)|$$
Simplify this we get $$|(x+1)+i(y-2)| = |(x-1) + i(y+2)|$$
Square both sides we have $$|(x+1)+i(y-2)|^2 = |(x-1) + i(y+2)|^2$$
Which is $$(x+1)^2+(y-2)^2=(x-1)^2+(y+2)^2$$
Expand the squares we get $$x^2+2x+1+y^2-4y+4=x^2-2x+1+y^2+4y+4$$
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 $z\in\mathbb{C}$

#### Vedant Shah

• Jr. Member
•  • Posts: 13
• Karma: 8 ##### Re: Q1: TUT 0201 and TU 0202
« Reply #3 on: September 28, 2018, 06:01:36 PM »
Let $p = -1+2i$ and $q=1-2i$
The perpendicular bisector is the set of points equidistant to points $p$ and $q$. The distance between some point, $z$ and $p$ is $|z-p|$. Similarly, The distance between some point, $z$ and $q$ is $|z-q|$.
Thus the set of points equidistant to both $p$ and $q$ is given by the equation:
$|z-p| = |z-q|$

Another way to do this is solving for the equation of the line in x-y 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:
$Re[(-\frac{1}{2} + i)z]$

#### Victor Ivrii ##### Re: Q1: TUT 0201 and TU 0202
« Reply #4 on: September 29, 2018, 03:42:45 PM »
Xier, it is not a proper equation by a line.