Answer:
a)
b)
Step-by-step explanation:
Given two points:
and 
Since in both questions,a and b, we're asked to find lines that are perpendicular to ST. So, we'll do that first!
Perpendicular to ST:
the equation of any line is given by:
where, m is the slope(also known as gradient), and c is the y-intercept.
to find the perpendicular of ST <u>we first need to find the gradient of ST, using the gradient formula.</u>

the coordinates of S and T can be used here. (it doesn't matter if you choose them in any order: S can be either x_1 and y_1 or x_2 and y_2)


to find the perpendicular of this gradient: we'll use:

both
and
denote slopes that are perpendicular to each other. So if
, then we can solve for
for the slop of ther perpendicular!

:: this is the slope of the perpendicular
a) Line through S and Perpendicular to ST
to find any equation of the line all we need is the slope
and the points
. And plug into the equation: 
side note: you can also use the
to find the equation of the line. both of these equations are the same. but I prefer (and also recommend) to use the former equation since the value of 'c' comes out on its own.

we have the slope of the perpendicular to ST i.e 
and the line should pass throught S as well, i.e
. Plugging all these values in the equation we'll get.




this is the equation of the line that is perpendicular to ST and passes through S
a) Line through T and Perpendicular to ST
we'll do the same thing for 




this is the equation of the line that is perpendicular to ST and passes through T