Question 810421
Slope for line AB is {{{(y-1)/(6-3)=(y-1)/3}}}.
The slope wanted for line BC is {{{-3/(y-1)}}}.


What we should know is that the product of two slopes must be -1.

{{{((y-1)/3)(-3/(y-1))=-1}}} but this does not solve anything.
We still want an expression for slope of BC, which is {{{(-2-y)/(12-6)=-(2+y)/6}}}.


Now use this in the "-1" equation.
{{{((y-1)/3)*(-(2+y)/6)=-1}}}
{{{((y-1)/3)((2+y)/6)=1}}}
{{{(y-1)(y+2)/(18)=1}}}
{{{(y-1)(y+2)=18}}}
{{{y^2+y-2=18}}}
{{{y^2+y-20=0}}} Fortunately, factorable.
(y+5)(y-4)=0


{{{highlight(y=-5)}}} or {{{highlight(y=4)}}}