Answer:
The required equations are
and
.
Step-by-step explanation:
The given parametric equation of the line,
, is
so, an arbitrary point on the line is 
The vector equation of the line passing through the points
and
is



The given equation can also be written as

The standard equation of the line passes through the point
and having direction
is

Here, The value of the parameter,
, of any point
at a distance
from the point,
, can be determined by

Comparing equations
and 
The line is passing through the point
having direction
.
Note that the point
is the same as
obtained above.
Now, the value of the parameter,
, for point
at distance
from the point
can be determined by equation
, we have





Put the value of
in equation
, the required equations are as follows:
For 


and for
,

