Equation of a line : Geometry and Complex Numbers

complexstlineEquation of the line passing through the point z_{1} and z_{2}:

Ref: Mathematics for Joint Entrance Examination JEE (Advanced), Second Edition, Algebra, G Tewani.

There are two forms of this equation, as given below:

\left | \begin{array}{ccc} z & \overline{z_{1}} & 1 \\ z_{1} & \overline{z_{2}} & 1 \\ z_{2} & \overline{z_{2}} & 1 \end{array} \right |=0

and \frac{z-z_{1}}{\overline{z}-\overline{z_{1}}}=\frac{z_{1}-z_{2}}{\overline{z_{}}-\overline{z_{2}}}

Proof:

Let z_{1}=x_{1}+iy_{1} and z_{2}=x_{2}+iy_{2}. Let A and B be the points representing z_{1} and z_{2} respectively.

Let P(z) be any point on the line joining A and B. Let z=x+iy. Then P \equiv (x,y), A \equiv (x_{1}, y_{1}) and B \equiv (x_{2},y_{2}). Points P, A, and B are collinear.

See attached JPEG figure 1.

The figure shows that the three points A, P  and B are collinear.

Shifting the line AB at the origin as shown in the figure; points O, P, Q are collinear. Hence,

\arg(z-z_{2})=\arg(z_{1}-z_{2}) or

\arg {\frac{z-z_{2}}{z_{1}-z_{2}}}=0

\Longrightarrow \frac{z-z_{2}}{z_{1}-z_{2}} is purely real.

\frac{z-z_{2}}{z_{1}-z_{2}}=\frac{\overline{z-z_{2}}}{z_{1}-z_{2}}

or, \frac{z-z_{2}}{z_{1}-z_{2}}=\frac{\overline{z}-\overline{z_{2}}}{\overline{z_{1}}-\overline{z_{2}}} call this as Equation 1.

\left | \begin{array}{ccc} z & \overline{z} & 1 \\ z_{1} & \overline{z_{1}} & 1\\ z_{2} & \overline{z_{2}} & 1 \end{array} \right |=0. Call this as Equation 2.

Hence, from (2), if points z_{1}, z_{2}, z_{3} are collinear, then

\left | \begin{array}{ccc} z_{1} & \overline{z_{1}} & 1 \\ z_{2} & \overline{z_{2}} & 1 \\ z_{3} & \overline{z_{3}} & 1 \end{array} \right |=0.

Equation (2) can also be written as

(\overline{z_{1}}- \overline{z_{2}}) - (z_{1}-z_{2})\overline{z}+z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}}=0

\Longrightarrow i(\overline{z_{1}}-\overline{z_{2}})z-(z_{1}-z_{2})\overline{z} + z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}}=0

\Longrightarrow \overline{a}z + a\overline{z}+b=0 let us call this Equation 3.

where a=-i(z_{1}-z_{2}) and b=i(z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}})=i 2i \times \Im (z_{1} \overline{z_{2}}), which in turn equals

-2 \times \Im(z_{1}\overline{z_{2}}), which is a real number.

Slope  of the given line

In Equation (3), replacing z by x+iy, we get (x+iy)\overline{a} + (x-iy)a+b=0,

\Longrightarrow (a+\overline{a})x + iy(\overline{a}-a)+b=0

Hence, the slope = \frac{a+\overline{a}}{i(a-\overline{a})}=\frac{2 \Re(a)}{2i \times \Im(a)}=-\frac{\Re(a)}{\Im(a)}

Equation of a line parallel to the line z \overline{a}+\overline{z}a+b=0 is z \overline{a} + \overline{z} a + \lambda=0 (where \lambda is a real number).

Equation of a line perpendicular to the line z\overline{a}+\overline{z}a+b=0 is z\overline{a}+\overline{z} a + i \lambda=0 (where \lambda is a real number).

Equation of a perpendicular bisector

Consider a line segment joining A(z_{1}) and B(z_{2}). Let the line L be its perpendicular bisector. If P(z) be any point on L, then we have (see attached fig 2)

PA=PB \Longrightarrow |z-z_{1}|=|z-z_{2}|

or |z-z_{1}|^{2}=|z-z_{2}|^{2}

or (z-z_{1})(\overline{z}-\overline{z_{1}})=(z-z_{2})(\overline{z}-\overline{z_{2}})

or

Here, a= z_{2}-z_{1} and b=z_{1}\overline{z_{1}}-z_{2} \overline{z_{2}}

Distance of a given point from a given line:

(See attached Fig 3).

Let the given line be z \overline{a} + \overline{z} a + b=0 and the given point be z_{c}. Then,

z_{c}=x_{c}+iy_{c}

Replacing z by x+iy in the given equation, we get

x(a+\overline{a})+iy(\overline{a}-a)+b=0

Distance of (x_{c},y_{c}) from this line is

\frac{|x_{c}(a+\overline{a})+iy_{c}(\overline{a}-a)+b|}{\sqrt{(a+\overline{a})^{2}-(a-\overline{a})^{2}}}

which in turn equals

\frac{z_{c}\overline{a}+\overline{z_{c}}a+b}{\sqrt{4(\Re(a))^{2}+4(\Im(a))^{2}}} which is equal to finally

\frac{|z_{c}\overline{a}+\overline{z_{c}}a+b|}{2|a|}.

More later,

Nalin Pithwa

 

 

 

 

 

 

 

 

 

 

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