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

Distance between two points A(z₁) and B(z₂) is

AB = |z₂ - z₁| = | Affix of B - Affix of A |

• The distance of a point from the origin is |z - 0| = |z|

• Three points A(z₁), B(z₂) and C(z₃) are collinear, then AB + BC = AC

i.e. |z₂ - z₁| + |z₃ - z₂| = |z₃ - z₁|

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 perpendicular bisector of AB

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

AC = BC

|z - z₁| = |z - z₂|

This is the equation of perpendicular of bisector of AB, where A(z₁) and B(z₂).

Equation of Circle

The equation of the circle whose center is at the point   and have radius r is given by

If the center is origin then, , hence equation reduces to |z| = r

Interior of the circle is represented by   

The exterior is represented by

Here z can be represented as x + iy and is represented by  

Equation of Circle in second form

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

Equation of Ellipse

|z - z₁| + |z - z₂| = k    (k > |z₁ - z₂|)

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

Equation of Hyperbola

| |z - z₁| - |z - z₂| | = k    (k < |z₁ - z₂|)

This represents a hyperbola as 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 

z = 

And

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

z = 

Centroid of the triangle with vertices z₁, z₂ and z₃ is given by (z₁ + z₂+ z₃)/3