Question 322081
A line with a slope of -1 would have a slope-intercept form of,
{{{y=mx+b}}}
{{{y=-x+b}}}
Since {{{y=1/(x-1)}}}, then solve for {{{b}}} using {{{x=0}}},
{{{-0+b=1/(0-1)}}}
{{{b=1/-1=-1}}}
{{{highlight(y=-x-1)}}}
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{{{graph(300,300,-5,5,-5,5,1/(x-1),-x-1)}}}
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As you can see from graphing the equations, that only solves for one line.
You can use a symmetry argument to find the other line, {{{y=-x+b2}}}, since the perpendicular line, {{{y=x-1}}} is perpendicular to both and intersects at both intersection points.
{{{graph(300,300,-5,5,-5,5,1/(x-1),-x-1,x-1)}}}
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{{{x-1=1/(x-1)}}}
{{{(x-1)^2=1}}}
{{{x-1=0 +- 1}}}
Two solutions:
{{{x=2}}} and {{{x=0}}}
When {{{x=2}}}, then 
{{{-2+b=1/(2-1)}}}
{{{b=3}}}
The other line is then,
{{{highlight_green(y=-x+3)}}}
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{{{graph(300,300,-5,5,-5,5,1/(x-1),-x-1,x-1,-x+3)}}}
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What's k?