algebra for IITJEE « Mathematics Hothouse

Equation 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)