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