Question

Find the equation of all lines having slope −1 that are tangents to the curve y=1/x-1,x≠1 .

Answer 1

it is given that y = 1/(x - 1) , x ≠ 1
differentiate y with respect to x,
dy/dx = -1/(x - 1)² , x ≠ 1
so, slope of tangent of the curve at (x,y) = dy/dx
= -1/(x - 1)²

A/C to question,
we equation of all lines having slope -1 that are tangents to the curve, y = 1/(x -1) , x ≠ 1.
so, slope of tangent = - 1
=> -1/(x - 1)² = -1
=> (x - 1)² = 1
square root both sides,
x - 1 = ± 1,
x = 1 ± 1 = 0, 2

put x = 0, in the curve y = 1/(x - 1)
y = 1/(0 - 1) = -1
now equation of line passing through (0,-1) :
(y + 1) = -1(x - 0)
=> y + 1 + x = 0
hence, equation is $\bf{x + y + 1 = 0}$

put x = 2, in the curve y = 1/(x - 1)
y = 1/(2 - 1) = 1
now equation of line passing through (2,1) :
(y - 1) = -1(x - 2)
=> y - 1 + x - 2 = 0
=> x + y - 3 = 0
hence, equation is $\bf{x + y - 3 = 0}$