Find the equation of the line with slope -1 that is the tangent to the curve y=1/(x-1).

To solve, let's consider the given slope of a line.

`m=-1`

Take note that when a line is tangent to the curve, its derivative at the point of tangency is equal to the slope.

`y' = m`

`y'=1`

To determine the point of tangency, take the derivative of the function of the curve.

`y= 1/(x-1)`

`y=(x-1)^(-1)`

`y'=-1(x-1)^(-1-1) * (x-1)'`

`y'=-(x-1)^(-2)*(1)`

`y'=-1/(x-1)^2`

Then, plug-in the value of the derivative.

`-1=-1/(x-1)^2`

`1=1/(x-1)^2`

`(x-1)^2=1`

`x-1=+-1`

`x=+-1+1`

`x_1=0`

`x_2=2`

To solve for the values of y, plug-in x=0 and x=2.

`x_1=0`

`y_1=1/(0-1)=-1`

`x_2=2`

`y_2=1/(2-1)=1`

So, there two points on the curve in which the tangent have a slope of -1. These are:

(0,-1) and (2,1). 

This means that there are two lines that have a slope of -1. To determine the equation of each line, use the point-slope form.

The first line has a slope of -1 and passes the point (0,-1).  So its equation is:

`y - y_1 = m(x - x_1)`

`y - (-1) = -1(x-0)`

`y+1=-x`

`y=-x - 1`

The second line has a slope of -1 and passes the point (2,1). So its equation is:

`y - y_2=m(x- x_2)`

`y-1 = -1(x-2)`

`y-1=-x+2`

`y=-x+3`

Therefore, the equation of the lines with slope -1 and tangent to the given curve are:

`y=-x-1`

and

`y=-x+3` .