MAT334--2020S > Quiz 1

Q1: TUT 0201

(1/1)

**Ziyi Wang**:

Questions:

Write the equation of the perpendicular bisector of the line segment joining $-1+2i$ and $1-2i$.

Answer:

Let $z = x + iy$.

Since $z$ is an arbitrary point on the perpendicular bisector of the line segment joining $-1+2i$ and $1-2i$, we know that the distance between $z$ and $-1+2i$ is same as the distance between $z$ and $1-2i$.

Then, we can calculate that:

$$|z-(-1+2i)| = |z - (1-2i)|$$

$$|(x + iy)-(-1+2i)| = |(x + iy) - (1-2i)|$$

$$|(x + 1)+(y-2)i| = |(x - 1) + (y + 2)i|$$

$$\sqrt{(x+1)^2 + (y-2)^2} = \sqrt{(x - 1)^2 + (y + 2)^2}$$

$$(x+1)^2 + (y-2)^2 = (x - 1)^2 + (y + 2)^2$$

$$x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 - 2x + 1 + y^2 + 4y + 4$$

$$4x = 8y$$

$$y = \frac{1}{2} x$$

Then, we write the equation in complex number notation:

$$Re((\frac{1}{2} + i)z) = 0$$

Navigation

[0] Message Index

Go to full version