Question 10991
We can use this formula:
{{{(y-y1)/(y2-y1)=(x-x1)/(x2-x1)}}}
Suppose (-3,2) as (x1,y1) and (-6,3) as (x2,y2). So,
{{{(y-2)/(3-2)=(x-(-3))/((-6)-(-3))}}}
{{{(y-2)/1=(x+3)/(-6+3)}}}
{{{(y-2)/1=(x+3)/-3}}}
{{{-3(y-2)=1(x+3)}}}
{{{-3y+6=x+3}}}
{{{-3y-x+6-3=0}}}
{{{-3y-x+3=0}}}
Thus, the equation of the line containing the points (-3,2) and (-6,3) is {{{-3y-x+3=0}}}

Or,

{{{-3y=x-3}}}
{{{y=(-x/3)+1}}}

OK