Distance formula and Equation of perpendicular bisector - Practice Questions & MCQ

Distance between two points $\mathrm{A}\left(\mathrm{z}_1\right)$ and $\mathrm{B}\left(\mathrm{z}_2\right)$ is

$
A B=\left|z_2-z_1\right|=\mid \text { Affix of } B-\text { Affix of } A \mid
$

Let $\mathrm{z}_1=\mathrm{x}+\mathrm{iy}$ and $\mathrm{z}_2=\mathrm{x}+\mathrm{iy}$ Then,$\left|z_1-z_2\right|=\left|\left(x_1-x_2\right)+i\left(y_1-y_2\right)\right|$$=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2}$$=$ distance between points $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\left(\mathrm{x}_2, \mathrm{y}_2\right)=$ distance between between $\mathrm{z}_1$ and $\mathrm{z}_2$ where $\mathrm{z}_1=\mathrm{x}_1+\mathrm{iy}_1$ and $\mathrm{z}_2=\mathrm{x}_2+\mathrm{iy}_2$

The distance of a point from the origin is $|z-0|=|z|$
Three points $A\left(z_1\right), B\left(z_2\right)$ and $C\left(z_3\right)$ are collinear, then $A B+B C=A C$

i.e. $\left|z_2-z_1\right|+\left|z_3-z_2\right|=\left|z_3-z_1\right|$

Perpendicular bisector

We can use the distance formula to find the equation of perpendicular bisector

Let two fixed points A(z₁) and B(z₂) and a moving point C(z) which lies on the perpendicular bisector of AB

As any point on the perpendicular bisector of AB will be equidistant from A and B, so

$
\begin{aligned}
& A C=B C \\
& \left|z-z_1\right|=\left|z-z_2\right|
\end{aligned}
$
This is the equation of perpendicular to the bisector of $A B$, where $A\left(z_1\right)$ and $B\left(z_2\right)$.

Equation of Circle

The equation of the circle whose centre is at the point $z_0$ and has radius $r$ is given by

$
\left|z-z_0\right|=r
$
If the center is origin then, $z_0=0$, hence equation reduces to $|z|=r$
Interior of the circle is represented by $\left|z-z_0\right|<r$
The exterior is represented by $\left|z-z_0\right|>r$
Here z can be represented as $\mathrm{x}+\mathrm{iy}$ and $z_0$ is represented by $x_0+i y_0$

Equation of Circle in second form

$\frac{\left|z-z_1\right|}{\left|z-z_2\right|}=k \quad(k \neq 1, k>0)$

This equation also represents a circle. This can be verified by putting z = x+iy, z₁ = p+iq, z₂ = a+ib

Equation of Ellipse

$\left|z-z_1\right|+\left|z-z_2\right|=k \quad\left(k>\left|z_1-z_2\right|\right)$

This represents an ellipse as the sum of distances of point z from z1 and z2 is constant, which is the locus of an ellipse.

Equation of Hyperbola

$\left|\left|z-z_1\right|-\left|z-z_2\right|\right|=k \quad\left(k<\left|z_1-z_2\right|\right)$

This represents a hyperbola as the difference of distances of point z from z1 and z2 is constant, which is the locus of a hyperbola.

Section Formula

The complex number z dividing z₁ and z₂ internally in ratio m: n is given by 

$
\mathrm{z}=\frac{m z_2+n z_1}{m+n}
$
And
The complex number $z$ dividing $z_1$ and $z_2$ externally in ratio $m$ : $n$ is given by

$
\mathrm{z}=\frac{m z_2-n z_1}{m-n}
$

Centroid of the triangle with vertices z₁, z₂ and z₃ is given by $\left(z_1+z_2+z_3\right) / 3$