Question 1005247
The slopes,  {{{(b^2-1)/(b+1)}}} and {{{-1/2}}}, must be able to form a product of {{{-1}}}.  The solved value for b will make the two lines perpendicular.


{{{(b^2-1)/(b+1)=2}}}


{{{b^2-1=2b+2}}}


{{{b^2-2b-3=0}}}, factorable.


{{{(b+1)(b-3)=0}}}


You WANT the slope of the first equation to be 2.
b=3 will work, but b=-1 will not work.