Find the equation of all lines having slope `- 1`that are tangents to the curve `y=1/(x-1), x!=1`.

Explanation

Slope of tangnet =- 1
Equation of curve ` y= 1/(x-1)`
` (dy)/(dx) = 1/((x-1)^(2))` ….(1)
Slope of tangent at point `(x, y) = 1/((x-1)^(2))`
`:. (-1)/((x-1)^(2)) =-1`
` rArr (x-1)^(2)=1`
`rArr x-1 = pm 1`
` rArr x= 2 or x = 0`
put x = 2 in equation (1)
`y=1/(2-1) = 1`
put x= 0 in equation (1)
`y = 1/(0-1) =- 1`
Therefore, a tangent passes through the point (2, 1) and tangent passes through the point (0, 1) with slope -1.
Equation of tangent passing through (2, 1)
`y - 1 = - (x - 2)`
` rArr x+ y = 3`
Equation of tangent passing through (0, -1)
` y+1 =- 1 (x-0)`
` rArr x+ y + 1 = 0`