How do you write an equation in standard form for a line passing through R(3, 3), S(-6, -6)?

Method 1.
Notice that the values of X-coordinate and Y-coordinate are equal in two different points.
So, an intelligent guess would be to consider an equation
#y=x#
as the representation of this line.

Method 2.
Assume that the equation we are looking for is
#y=a*x+b#
where #a# and #b# are unknown coefficients that we have to determine using the data provided.
Substitute the given coordinates into this equation:
#3=a*3+b#
#-6=a*(-6)+b#

Resolve the first equation for #b# in terms of #a# and substitute into the second equation to have only one equation with one unknown #a#:
#b=3-3a#
#-6=-6a+(3-3a)#

From the latter we derive:
#-9=-6a-3a#
#-9=-9a#
#a=1#

Now find #b# that we expressed in terms of #a#:
#b=3-3a=3-3*1=0#

The above resulted in the equation for our line:
#y=1*x+0# or, simply, #y=x#.

Method 3.
Again, looking for an equation in a format
#y=a*x+b#
As it's known, the coefficient #a# is a slope of a line that can be calculated as a ratio between increment of ordinate (Y-coordinate changes from #-6# to #3#) to increment of abscissa (X-coordinate changes from #-6# to #3#)).
In our case this ratio equal to:
#a=(3-(-6))/(3-(-6))=1#

To determine coefficient #b#, we can substitute the values of abscissa and ordinate of one of the points to get an equation for #b#:
#3 = 1*3+b#
from which follows that #b=0#.
So, our equation is
#y=1*x+0# or, simply, #y=x#.