Equation of the line passing through the point
and
:
Ref: Mathematics for Joint Entrance Examination JEE (Advanced), Second Edition, Algebra, G Tewani.
There are two forms of this equation, as given below:

and 
Proof:
Let
and
. Let A and B be the points representing
and
respectively.
Let
be any point on the line joining A and B. Let
. Then
,
and
. 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,
or

is purely real.

or,
call this as Equation 1.
. Call this as Equation 2.
Hence, from (2), if points
,
,
are collinear, then
.
Equation (2) can also be written as


let us call this Equation 3.
where
and
, which in turn equals
, which is a real number.
Slope of the given line
In Equation (3), replacing z by
, we get
,

Hence, the slope 
Equation of a line parallel to the line
is
(where
is a real number).
Equation of a line perpendicular to the line
is
(where
is a real number).
Equation of a perpendicular bisector
Consider a line segment joining
and
. Let the line L be its perpendicular bisector. If
be any point on L, then we have (see attached fig 2)

or 
or 
or
Here,
and 
Distance of a given point from a given line:
(See attached Fig 3).
Let the given line be
and the given point be
. Then,

Replacing z by
in the given equation, we get

Distance of
from this line is

which in turn equals
which is equal to finally
.
More later,
Nalin Pithwa